leetcode 72 edit distance

Welcome! I'm excited to help you master LeetCode 72: Edit Distance, a classic problem that pops up everywhere from spell checkers to DNA sequencing. Before we map out our game plan, I'd love to know a little bit about your background so we can tailor this session perfectly.

How comfortable are you with Dynamic Programming, and which language (e.g., Python, C++, Java) do you prefer for coding this?

I'm a DP beginner and prefer Python.

That's fantastic! Python is perfect for this because its clean syntax lets us focus purely on the logic. Since you are new to DP, we'll make sure to build a rock-solid foundation step-by-step so you never feel lost.

To start, let's address your preference. Before we jump into the math formulas or the 2D grid, would you like to build your intuition using a visual to see how the decisions branch out?

Yes, let's start with the recursive tree visualization

I would love to walk you through that visualization! To build the tree, we first need to define how we track our position in the two words as we compare them.

Setting Up Our Pointers

To break this down recursively, we will use two pointers (or indices): i for word1 and j for word2. Instead of starting from the beginning, it's actually much easier to start comparing from the end of both strings and work our way backward. This is a common pattern in string-based dynamic programming. Let's define our state using a helper function:

i represents our current position in word1 (looking at the prefix word1[0..i]).
j represents our current position in word2 (looking at the prefix word2[0..j]).
solve(i, j) returns the minimum edit distance to transform the prefix word1[0..i] into word2[0..j].

The Character Comparison

At any given step, we look at the characters at our current pointers: word1[i] and word2[j]. This brings us to a fork in the road with two main scenarios.

Scenario 1: The Characters Match

If the characters are already the same (for example, both are 's'), we don't need to do any work! No edit operation is spent. We can simply shrink both prefixes by moving both pointers backward.

Scenario 2: The Characters Mismatch

If the characters don't match, we have to make a choice. Since we want the minimum operations, we must simulate all three permitted edits: Insert, Delete, and Replace. We then choose whichever path ends up costing the fewest total operations.

Visualizing the Decisions

To make this crystal clear, let's look at a visual representation of how the recursive decisions branch out when we hit a mismatch.

A decision tree diagram showing three branching choices for solving a string mismatch state solve(i, j) using insert, delete, and replace operations.

I've sketched out a visual layout of this decision branch to show you exactly how the pointers shift depending on which action we simulate.

Would you like to walk through a concrete example using short words like "cat" and "cut" to see exactly how these three branches calculate their values?

i see so if yo ureplace you just say solve i,j = solve(i-1, j - 1) + 1?

Spot on! That's exactly right.

When you choose to replace, you match the current characters by changing one to the other. This costs you exactly 1 operation, and then you move both pointers backward. Mathematically, that's indeed representing the subproblem as solve(i - 1, j - 1) + 1.

Tracing a Real Example: cat to cut

To make this concrete, let's trace how this works with word1 = "cat" and word2 = "cut".

We start at the end of both strings:

word1[2] = 't'
word2[2] = 't'

Since the characters match, our cost is 0. We don't perform any edit. We simply move both pointers backward, which means we transition to evaluating the prefixes "ca" and "cu": solve(2, 2) = solve(1, 1)

Now, we compare word1[1] = 'a' and word2[1] = 'u':

They do not match! This mismatch triggers our three choices. Each choice carries a cost of 1 plus the result of the new state:

: We pay 1 to change 'a' to 'u'. The remaining prefixes are "c" and "c". 1 + solve(0, 0)
Delete: We pay 1 to delete 'a' from word1. The remaining prefixes are "c" and "cu". 1 + solve(0, 1)
Insert: We pay 1 to insert 'u' into word1. The remaining prefixes are "ca" and "c". 1 + solve(1, 0)

If we choose the Replace path, we land on solve(0, 0), which compares word1[0] = 'c' and word2[0] = 'c'. Since they match, the cost is 0 and we transition to empty strings, which is our base case and costs 0 edits.

So, the Replace path costs us 1 (mismatch) + 0 (match) + 0 (base case) = 1 edit total, which is the optimal edit distance to turn "cat" into "cut"!

The Downside of Pure Recursion

While this recursive branching is highly intuitive, it has a massive problem when strings get longer: overlapping subproblems.

If we draw out the tree for slightly larger words, we find ourselves calculating the exact same solve(i, j) states over and over again from different decision branches. This redundant work makes simple recursion run in exponential time, which will quickly crash LeetCode's time limits.

A recursion tree for solve(3,2) illustrating overlapping subproblems highlighted in orange.

I've set up a visualization showing how these recursive paths collide and repeat the same work. To bypass this, we can store our results in a 2D Dynamic Programming table so we only calculate each state once.

Shall we take a look at how to structure this table and set up its boundaries?

can you annotate on the visual like the state of word1 and word2 or something at each one

That is an excellent suggestion! Adding the actual string prefixes to each node of our makes it much easier to see how our pointers map to real data as they branch.

A recursion tree for solve("cat", "cut") showing string prefix edit distance subproblems.

I've updated our recursive visualization to overlay the exact state of word1 and word2 at each step.

Looking at this labeled layout, you can see how we start at the root comparing "cat" and "cut".

Notice how quickly the branches generate identical pairs of remaining substrings. For example, if we perform a delete followed by an insert, or an insert followed by a delete, we often end up comparing the exact same sub-prefixes!

In a longer word, these identical subproblem nodes multiply exponentially. Because our basic recursion doesn't remember that it already solved these nodes, it blindly recalculates them from scratch every single time. This is what we call .

To solve this, we can introduce memoization (or store the results in a 2D table), which allows us to solve each state (i, j) exactly once and read it in constant time whenever we encounter it again.

Are you ready to see how we build this memory grid and initialize its boundaries?

Continue to the 2D Dynamic Programming table setup

Let's transition from that recursive tree to a structured, 2D dynamic programming grid. This grid acts as a cheat sheet where the computer stores solutions to subproblems so it never has to calculate them twice.

Defining the Grid and Table Dimensions

To store our results, we'll build a 2D grid called dp. If the length of word1 is m and the length of word2 is n, our table needs to have dimensions of (m + 1) x (n + 1). We add that crucial +1 to both dimensions to represent the state of having an empty string. Let's define exactly what each cell in this grid represents:

The cell dp[i][j] stores the minimum edit distance to convert the prefix word1[0..i-1] (the first i characters of word1) into the prefix word2[0..j-1] (the first j characters of word2).
Because the grid is 1-indexed relative to the string lengths, dp[0][0] represents transforming an empty string into another empty string.

Initializing the Base Cases

Before we can fill in the middle of our table, we must set up the boundaries. These boundaries represent our base cases: converting a string prefix to or from an empty string.

1. The Left Column: Transforming to Empty

The first column, where j = 0, represents converting prefixes of word1 of length i into an empty string. The only way to do this is to delete every single character.

Therefore, dp[i][0] = i for all i from 0 to m.

2. The Top Row: Transforming from Empty

The first row, where i = 0, represents converting an empty string into prefixes of word2 of length j. The only way to achieve this is to insert every character of the target prefix.

Therefore, dp[0][j] = j for all j from 0 to n.

To help you visualize this setup, let's look at how our grid boundaries are prefilled when converting word1 = "cat" to word2 = "cut".

A 4x4 dynamic programming matrix initializing the edit distance calculation between 'cat' and 'cut'.

I have set up a visualization of our initial state grid. You can see how the first row and first column naturally increment because transforming a word of length L to or from an empty string always costs exactly L operations. With this empty-string safety net in place, every other cell in the table can now be calculated step-by-step using values that are already sitting in the cells directly above, to the left, or diagonally adjacent. Are you ready to see the recurrence relation that we use to fill in all those empty squares with question marks?

actually wait its just been a while so the idea is you go till the base case and youre carrying tehse plus 1 in recursion but how do you get the final number?

That is such a crucial question, and it's actually the exact bridge that connects recursion to dynamic programming! Let's clear up how this transition works.

Top-Down Unwinding vs. Bottom-Up Building

In our recursive tree, you are 100% correct. We start at the top (with the full words "cat" and "cut") and dive deeper and deeper. We keep carrying that +1 cost down the branches until we hit a base case (like an empty string, which we know costs 0). Once we hit that floor, the recursion "unwinds." It passes those base numbers back up, adding up the +1s on the way back to the top to give us our final answer.

But in (our 2D table), we turn that process completely on its head. Instead of starting from the top and looking down, we start directly at the very bottom — the base cases — and build our way up.

How the Values Flow to the Final Answer

Think of our DP table as a puzzle where you already have the border pieces solved. Those border pieces are our base cases (the 0, 1, 2, 3 in the first row and column).

To find the value for any empty cell, we don't recurse. Instead, we look at the values of its neighbors that are already filled in:

The cell to the left (dp[i][j-1])
The cell above (dp[i-1][j])
The diagonal cell (dp[i-1][j-1])

We grab the minimum of those three neighbors, add 1 to represent the edit operation, and write that new number right into the cell. This means each cell's value is calculated instantly in constant time because its neighbors are already fully computed!

We fill the table row-by-row, from top-left to bottom-right. The very last cell we fill — at the bottom-right corner, representing dp[m][n] — naturally accumulates all those previous step-by-step decisions. That single cell is our final, optimal answer.

Would you like to fill in the very first blank square of our "cat" to "cut" table together to see exactly how this neighbor-checking works in action?

but you have multiple different sums or is it like you do return min(replace, insert, remove)

Exactly, you've hit on the core magic of dynamic programming! It is indeed a minimum, but because we are building bottom-up, we don't have to wait for a recursive return statement to compute it.

Instead of checking paths dynamically at runtime, we look at the neighbors in our grid that have already been filled. Those neighbors hold the total accumulated minimum costs for their respective paths.

This brings us directly to our official recurrence relation, which is the heartbeat of this entire algorithm.

The Recurrence Relation

At any cell dp[i][j] representing the subproblem of converting word1[0..i-1] to word2[0..j-1], we look at the characters word1[i-1] and word2[j-1]:

Case 1: The Characters Match

If the characters match, no operation is needed! We carry the exact value from the diagonal neighbor because our edit cost doesn't change.

dp[i][j] = dp[i-1][j-1]

Case 2: The Characters Mismatch

If they don't match, we must make an edit. We pay a cost of 1 for the edit, plus we take the minimum of our three immediate neighbor states, each representing a different simulated operation.

dp[i][j] = 1 + \min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])

Step-by-Step: Filling Cell (1,1)

Let's put this formula to work and calculate the very first blank cell, dp[1][1], for our target strings word1 = "cat" and word2 = "cut". This cell represents converting the prefix "c" into "c".

First, we look at the actual characters at word1[0] and word2[0]:

Both characters are 'c'!

Since they match, we fall into Case 1. We don't perform any edit. We look directly at the diagonal neighbor, dp[0][0], which holds the value 0: dp[1][1] = dp[0][0] = 0.

Now, let's step it up. What happens when we fill dp[1][2], which represents converting "c" to "cu"?

The characters we are comparing are 'c' from word1 and 'u' from word2.
This is a mismatch! We must use our mismatch formula: dp[1][2] = 1 + min(delete, insert, replace)

Let's pull the values of its three neighbors from our grid:

Delete (above: dp[0][2]): 2
Insert (left: dp[1][1]): 0
Replace (diagonal: dp[0][1]): 1

We grab the minimum of those three values, which is 0 (the insert path), and add our edit cost of 1: dp[1][2] = 1 + 0 = 1.

We just filled two cells in our grid in a fraction of a millisecond, completely bypassing any deep recursion! This is , the engine behind dynamic programming.

Dynamic programming edit distance matrix cell evaluation for transformation of cat to cut.

I've updated our dynamic programming grid above to show how this math visualizes on our board. You can see how the arrows feed into dp[1][2] to compute its final value of 1.

Now that you've seen how the recurrence relation governs our table, are you ready to write the actual Python code to build and run this bottom-up tabulation solver?

im saying no DP just recurinos case

That makes total sense! Let's hit the brakes on the 2D table and focus entirely on the pure, raw recursive code. Writing out the pure recursion first is actually the absolute best way to make sure the transition formulas feel completely natural.

Thinking Recursively with Indices

When we write our recursive function, we'll use two indices, i and j, to represent our current position in word1 and word2. Since we decided to compare starting from the back of the strings, we'll start i at the last index of word1 and j at the last index of word2.

At any point in the recursion, our helper function solve(i, j) has to answer one simple question:

"What is the minimum number of operations to convert the prefix word1[0..i] into the prefix word2[0..j]?"

The Recursive Base Cases

Every recursive function needs an exit strategy to prevent it from spinning into infinite loops. What happens when we run out of characters in one of our words? If i == -1, it means word1 has become an empty string. To turn an empty string into a prefix of word2 of length j + 1, we have no choice but to perform exactly j + 1 insertions. Similarly, if j == -1, word2 has become empty. To turn our remaining word1 prefix into an empty string, we must perform i + 1 deletions.

The Python Recursive Implementation

Let's take those rules — the matching character shortcut, the mismatch choices, and our empty-string base cases — and translate them directly into a clean Python function.

def minDistance(word1: str, word2: str) -> int:
    def solve(i, j):
        # Base Case 1: word1 is empty, insert all remaining characters of word2
        if i < 0:
            return j + 1
        
        # Base Case 2: word2 is empty, delete all remaining characters of word1
        if j < 0:
            return i + 1
        
        # Case 1: Characters match, no edit cost!
        if word1[i] == word2[j]:
            return solve(i - 1, j - 1)
        
        # Case 2: Mismatch! We simulate all 3 operations and take the minimum
        insert_op = solve(i, j - 1)
        delete_op = solve(i - 1, j)
        replace_op = solve(i - 1, j - 1)
        
        return 1 + min(insert_op, delete_op, replace_op)
    
    # Start the recursion from the last character of both words
    return solve(len(word1) - 1, len(word2) - 1)

If you look closely at the mismatch code, you can see how each choice shifts our pointers:

insert_op keeps i in place but decrements j because we've successfully matched the newly inserted character at the end of word1 with the character at word2[j].
delete_op decrements i because we threw away the character at word1[i], but we still need to match the current target word2[j].
replace_op decrements both pointers because we forced a match.

Does this index shifting make sense, or would you like to map one of these operations directly to a quick example?

oh lets go to DP stuff now

I love that decision. Transitioning directly from that recursive model to the bottom-up is the best way to see the magic of dynamic programming in action.

From Pointers to Grid Coordinates

When we transition from recursion to tabulation, our recursive state solve(i, j) maps directly to the grid coordinate dp[i][j].

Because recursion works backward from the end of the strings and tabulation works forward from the beginning, we adjust our indexing slightly. In our tabulation table:

dp[i][j] represents the minimum edit distance for the first i characters of word1 and the first j characters of word2.
Because our grid includes an extra row and column for empty strings (0 characters), the character in word1 we look at when we are at row i is actually at index i - 1 (since strings are 0-indexed in Python).

Writing the Full Tabulation Code

Let's put this together into our complete Python solution. We will initialize our 2D grid, fill in the base cases for empty strings, and then use nested loops to evaluate the rest of the cells.

def minDistance(word1: str, word2: str) -> int:
    m, n = len(word1), len(word2)
    
    # Create a 2D grid of size (m + 1) x (n + 1) initialized with 0s
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # Base Cases: Fill the first column
    for i in range(m + 1):
        dp[i][0] = i
        
    # Base Cases: Fill the first row
    for j in range(n + 1):
        dp[0][j] = j
        
    # Fill the rest of the grid row by row
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            # Case 1: Characters match (remember the 1-based offset!)
            if word1[i - 1] == word2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                # Case 2: Mismatch! Take 1 + min of (Delete, Insert, Replace)
                dp[i][j] = 1 + min(
                    dp[i - 1][j],    # Delete
                    dp[i][j - 1],    # Insert
                    dp[i - 1][j - 1] # Replace
                )
                
    # The bottom-right cell contains the answer for the full strings
    return dp[m][n]

Look at how cleanly that maps to the recursive logic we wrote earlier!

Instead of a call stack going deep and potentially hitting limit thresholds, we just have two simple loops moving across the grid. Every single cell is computed in constant time because the three neighbors it depends on are guaranteed to be solved already.

Would you like to step through exactly how this code executes using our "cat" and "cut" example to see how the numbers change row by row?

i still need to visualize it

I hear you! Seeing a flat block of code is one thing, but watching the numbers actually step through the grid row-by-row is where the intuition truly locks in.

Row-by-Row Tabulation: cat to cut

Let's watch exactly how our outer loop for i in range(1, m + 1) and inner loop for j in range(1, n + 1) fill our grid step-by-step.

We start with our prefilled boundaries. The first row (representing an empty word1) is [0, 1, 2, 3], and the first column (representing an empty word2) is [0, 1, 2, 3]. All other cells start as question marks.

Now, let's step through the calculations row-by-row.

A dynamic programming matrix computing the edit distance between cat and cut.

I have created an interactive visualization of the grid filling up so you can step through each row's progress yourself.

Row 1: Analyzing prefix "c"

Cell dp[1][1] ("c" to "c"): The characters match ('c' == 'c'). We copy the diagonal value from dp[0][0], which is 0.
Cell dp[1][2] ("c" to "cu"): Mismatch ('c' != 'u'). We check the three neighbors: Left (0), Up (2), Diagonal (1). The minimum is 0, so we add 1 to get 1.
Cell dp[1][3] ("c" to "cut"): Mismatch ('c' != 't'). Neighbors are Left (1), Up (3), Diagonal (2). The minimum is 1, so we add 1 to get 2.

Row 2: Analyzing prefix "ca"

Cell dp[2][1] ("ca" to "c"): Mismatch ('a' != 'c'). Neighbors are Left (2), Up (1), Diagonal (0). The minimum is 0 (Replace), so we add 1 to get 1.
Cell dp[2][2] ("ca" to "cu"): Mismatch ('a' != 'u'). Neighbors are Left (1), Up (1), Diagonal (1). The minimum is 1, so we add 1 to get 2.
Cell dp[2][3] ("ca" to "cut"): Mismatch ('a' != 't'). Neighbors are Left (2), Up (2), Diagonal (1). The minimum is 1, so we add 1 to get 2.

Row 3: Analyzing prefix "cat"

Cell dp[3][1] ("cat" to "c"): Mismatch ('t' != 'c'). Neighbors are Left (3), Up (1), Diagonal (2). The minimum is 1 (Delete), so we add 1 to get 2.
Cell dp[3][2] ("cat" to "cu"): Mismatch ('t' != 'u'). Neighbors are Left (2), Up (2), Diagonal (2). The minimum is 2, so we add 1 to get 3.
Cell dp[3][3] ("cat" to "cut"): Match ('t' == 't')! We copy the diagonal value from dp[2][2], which is 1.

Our final destination cell at dp[3][3] contains 1, which is our final answer. Just 1 replacement ('a' to 'u') gets us from "cat" to "cut"!

Now that you can see how the values flow row-by-row directly into the final bottom-right answer, how do you feel about this transition? Are you ready to talk about how efficient this is compared to our old recursion?

so this is start at [0,0] and then return [3,3]? but recusion was the other way

You have hit on the absolute core conceptual difference between these two strategies!

The Direction of Evaluation

Your observation is spot on. In Recursion (Top-Down), we start at the ultimate goal, which is solve(m - 1, n - 1) (the very end of our strings, representing the full comparison). The program recursively calls itself with smaller and smaller indices, drilling down until it hits the floor—our base cases of empty strings (where indices are -1). Once it hits that bottom floor, it bubbles those return values back up to solve the main problem.

In Tabulation (Bottom-Up), we flip the script. We don't start at the top and look down. Instead, we start directly at the absolute simplest possible subproblems—the base cases represented by our grid boundaries at dp[0][0] (empty string to empty string). We then use simple iterative loops to build the solutions up row-by-row, column-by-column, until we reach the bottom-right corner, dp[m][n].

We don't need to recursively "return" values back up the chain because we literally build the final answer brick by brick directly inside dp[m][n]!

ok hmm so i guess for the dp solve you want to return dp[m,n] and you do that by just being dumb you just say ok at dp[0,0] we check if theyre the same, if so we just move along the diagonal for free, if they are different, then we have 3 choices where dp[1,0] = 1 dp [0,1] = 1 or dp[1,1] = 1

You have hit on the exact spot where a lot of people first get tripped up! Let's unpack your intuition, because you are very close, but there is one crucial pivot to make about how we build the grid.

The Real Meaning of the Coordinates

When you say "at dp[0][0] we check if they are the same," that is actually where we have to be careful.

Remember, dp[0][0] represents comparing an empty string to an empty string. There are no characters to compare there! We already know the answer to that subproblem is 0 edits because they are already identical.

Similarly, dp[1][0] represents converting "c" to "" (empty). We don't make a dynamic choice there; the only way to convert "c" to empty is to delete it, which costs exactly 1 operation. We don't check neighbors here; we initialize these boundary cells with our pre-computed base case numbers.

How We Actually Calculate dp[1][1]

Instead of branching out from dp[0][0] to see where we can go, we sit in our destination cell, say dp[1][1] (comparing "c" and "c"), and we look backward at the neighbors that were already computed.

Because the characters at word1[0] and word2[0] are both 'c', they match! We don't pay anything extra. We just look diagonally back to dp[0][0], grab its value (which is 0), and write 0 into dp[1][1].

Now look at dp[1][2] (comparing "c" and "cu"). The characters 'c' and 'u' do not match. We must make a choice from the three adjacent blocks that have already been filled in:

Delete (above: dp[0][2]): 2
Insert (left: dp[1][1]): 0
Replace (diagonal: dp[0][1]): 1

We grab the absolute minimum of those three pre-calculated paths, which is 0 (the insert path), and add 1 for our mismatch operation. So, dp[1][2] = 1 + 0 = 1.

We repeat this mechanical neighbor-checking for every single cell, moving systematically across the rows. By the time we arrive at dp[3][3], we don't have to trace any branches. The cell simply holds the minimum cost of all optimal pathways combined, and we return it.

So dp[i][j] doesn't choose where to go, it just pulls from where it came?

You've summarized it beautifully! That is the ultimate lightbulb moment for understanding bottom-up dynamic programming.

The Pull-Based Mindset

In a pure recursive solution, the code is proactive: it actively branches forward, searching down multiple paths to find the optimal route.

In tabulation, the cell dp[i][j] is entirely passive. It doesn't look ahead, and it doesn't make active branching choices. Instead, it is a static container that simply inherits the best possible history from its neighboring cells. It pulls the already-calculated results from the left, above, and top-left, filters for the minimum, adds its own transaction fee (the +1 edit cost if there is a mismatch), and stores the result.

Because every single cell only pulls from solved states that are directly adjacent, we completely bypass the overhead of managing a massive, deeply nested execution stack.

so are we buiilding from top down or down up for this solve?

That is the ultimate distinction right there! You've nailed the mental model.

Bottom-Up vs. Top-Down Summarized

For this tabulation code, we are building Bottom-Up.

We start at the absolute bottom floor — the simplest base cases where one or both string prefixes are empty (length 0). We fill in those concrete numbers first. Then, we work our way up to the full length of the strings.

Here is a quick comparison to lock this in:

Feature	Top-Down (Recursion)	Bottom-Up (Tabulation)
Starting Point	The ultimate goal: `solve(m-1, n-1)`	The base cases: `dp[0][0]`
Direction	Drills down to empty strings, then unwinds back up	Builds sequentially from empty strings up to full strings
Mechanism	Active branching decisions that pull from child branches	Passive grid cells that pull from already-solved neighbors
Memory	Call stack space (and a hash map if memoized)	A physical 2D matrix (array)

Now that you have mastered both the recursive design and the 2D tabulation grid, you have fully conquered the core logic of this problem!

Are you ready to explore how we can optimize this further, or would you like to do a quick review of the time and space complexity first?

ok so dp[0,0] then you can brnach out thats why im confused you start at dp[0,0] wishic is to say word1[:0] and word2[:0] but if word1 = "house" now you have to do something, so you are branching to get to the next cell

This is a incredibly common point of confusion when learning dynamic programming, and I am so glad you called this out! Let's clear this up completely, because once this clicks, you will have a superpower for designing DP algorithms.

The Crucial Difference: Branching vs. Populating

When you think of "branching out" from dp[0][0] to find a route to dp[3][3], you are still thinking in terms of active pathfinding (like a maze-solver exploring pathways). But in bottom-up DP, we do not explore routes.

Instead, we are just filling out a grid, one cell at a time, like a simple spreadsheet.

We don't look ahead to see where a path goes. Instead, we use two simple loops that visit every single square in order—row by row, column by column—no matter what.

If we want to turn "house" into "ros", we don't start at dp[0][0] and guess which way to walk. Instead, we run our loops: we calculate row 1, then row 2, then row 3, and so on. Every single cell gets populated. There is no "branching" during the execution; there is only a systematic, deterministic march from the top-left corner of the grid to the bottom-right.

The "Pull" Mechanism in Action

To see why we don't need to branch, let's look at how we populate a cell when we have longer words like word1 = "house" and word2 = "ros".

A 6x4 dynamic programming matrix showing the edit distance computation between house and ros.

I've set up a visualization of our grid for "house" to "ros".

When our loops land on dp[1][1] (which represents comparing "h" from "house" and "r" from "ros"), we don't make a decision about where to go next. We only look backward at the three adjacent cells that already have numbers in them:

Above (dp[0][1]): holds 1
Left (dp[1][0]): holds 1
Diagonal (dp[0][0]): holds 0

Because 'h' and 'r' are different, we apply our mismatch cost: we take the minimum of those three historical values (0), add 1 for the mismatch operation, and write 1 directly into dp[1][1].

Once that cell is written, our loops move on to the next cell, dp[1][2]. We repeat this basic math for every single cell until the entire grid is full of numbers. When the loops finish, we don't search for a path—we simply read the pre-calculated number waiting for us at the bottom-right corner, dp[m][n]!

Does seeing the loops as a flat, mechanical "cell-filling process" (rather than a pathfinding walk) help make sense of why we don't have to branch?

ok do at dp[0,0] its saying what and then next you do what?

I love that you asked this because it exposes the exact moment we stop thinking like a pathfinder and start thinking like a factory worker.

What Happens at dp[0][0]

At dp[0][0], we are comparing an empty string with an empty string. The minimum edits required is 0. We do not branch out from here to make a decision.

Instead, we simply write 0 in that cell, and our nested loops move immediately to the next cell in our strict layout sequence, which is dp[0][1].

The Next Cell in Loop Order

At dp[0][1], we are comparing an empty word1 with the first character of word2 (for example, "h" if the word is "house").

We do not dynamically decide what to do. The nested loop just runs the initialization code we wrote: dp[0][1] = 1

We systematically fill the entire first row and first column this way, cell by cell, with simple incremental values. We are not choosing a path; we are setting up our spreadsheet borders.

Systematic Populating vs. Pathfinding

Once the borders are set, we move to the first inner cell: dp[1][1].

To calculate dp[1][1], we don't look forward to see where to go. We only look back at the solved neighbors (dp[0][1], dp[1][0], and dp[0][0]) to pull our value.

Our code runs a strict, unbending loop sequence that visits every single coordinate in the grid. It evaluates every cell's coordinate one by one, filling in the numbers from left to right, row by row, like reading a book. There is no active choice about which cell to visit next — the loops visit every coordinate regardless of whether it lies on the optimal "path."

so then for like dp[2,2] it is like a min(dp[1,2], dp[2,1], dp[1,1]?

You nailed it! That is 100% correct.

Breaking Down the Formula for dp[2][2]

For dp[2][2], we are looking at the prefix of word1 up to index 1 ("ca") and the prefix of word2 up to index 1 ("cu").

Since the characters at these positions do not match ('a' != 'u'), we look back at our three neighbors and pull the minimum of those values before adding our mismatch penalty of 1:

dp[2][2] = 1 + \min(dp[1][2], dp[2][1], dp[1][1])

Now that you completely understand the mechanics of the 2D grid and how to pull solved history, you have mastered the core logic of this problem!

Are you ready to see how we can optimize this, or would you like to look at the complexity analysis first?

but what cost is there in the example going from 1,1 to 2,2 like if the strings are identical

That is the perfect question to ask because it brings us right back to our Match Case versus our Mismatch Case.

The Cost of Matching Characters

If the characters at our current indices actually match — meaning word1[i-1] == word2[j-1] — the cost is exactly 0 edits.

Because we don't have to spend any operations to force a match, we can just look diagonally back to dp[i-1][j-1] and copy its value directly. No addition, no penalty, no transaction fee! We inherit the exact diagonal value.

dp[i][j] = dp[i-1][j-1]

Contrasting Match vs. Mismatch

Let's put those two side-by-side so you can see how the math naturally handles both pathways in our grid:

Matching Characters: dp[i][j] = dp[i-1][j-1] (cost is 0, we slide down the diagonal for free).
Mismatching Characters: dp[i][j] = 1 + min(Delete, Insert, Replace) (cost is 1, and we check the three surrounding blocks to find the best history).

In our "cat" to "cut" example, if we were transitioning from dp[1][1] (comparing "c" and "c") to another cell, we got to inherit dp[0][0] (which was 0) completely for free because the 'c's matched. That's why dp[1][1] ended up being 0!

Does that distinction between matching for free versus paying 1 for a mismatch feel clear now?

i guess in this way where theres no dynamic anything i dont get why every grid wouldnt be identical

That is a brilliant question, and it shows you are looking deeply at the mechanics. If the process is completely mechanical and unbending, why does every grid look totally different?

The Driver of Diversity: The Comparison Check

Even though our loops visit every single coordinate in the exact same order for every grid, the values we write inside those cells are entirely dictated by the specific characters of our strings.

The unbending loop sequence runs a crucial conditional check at every single coordinate: if word1[i-1] == word2[j-1].

This single comparison acts like a custom valve that alters the flow of values through the grid. Let's look at why identical strings yield an entirely different grid compared to mismatched ones.

Scenario	Comparison Check at Cell `dp[i][j]`	Value Inheritance Rule
Characters Match	`word1[i-1] == word2[j-1]` is True	Inherit diagonal value directly: `dp[i-1][j-1]` (Cost: 0)
Characters Mismatch	`word1[i-1] == word2[j-1]` is False	Take minimum of neighbors + 1: `1 + min(left, up, diagonal)` (Cost: 1)

Comparing 'cat' vs 'cat' to 'cat' vs 'cut'

If we were converting "cat" to "cat" (identical strings), our comparison check would trigger the match rule at almost every step along the main diagonal:

At dp[1][1] ('c' vs 'c'), we match! We pull 0 from dp[0][0] for free.
At dp[2][2] ('a' vs 'a'), we match! We pull 0 from dp[1][1] for free.
At dp[3][3] ('t' vs 't'), we match! We pull 0 from dp[2][2] for free.
The bottom-right final answer is 0 edits.

But because the strings are different for "cat" and "cut", when our loop hits dp[2][2] ('a' vs 'u'), the mismatch check triggers. We pay a penalty of 1 and pull from neighbors, populating dp[2][2] with 2 instead of 0. This altered value then cascades to influence all future coordinates!

So while the coordinate-visiting route is identical, the values flowing into each cell are dynamically shaped by whether the corresponding string characters match. It is the characters of the strings themselves that customize the grid values, not the traversal order.

Does seeing the comparison check as a "custom valve" that changes the cell calculations help explain why the grids turn out so differently?

ok yeah ok so there is a dynamic portion but its just backwards fine basically the loop is like class Solution: def minDistance(self, word1: str, word2: str) -> int: dp = [ [0] * (len(word1)+1)] * (len(word2)+1)

You are so close to having a flawless solution! I love how you've captured the core transition logic, but there are three extremely common Python syntax and indexing pitfalls hiding inside your code snippet that will cause errors or incorrect outputs on LeetCode.

Pitfall 1: The Shallow Copy Matrix Bug

Take a close look at how you initialized your 2D grid: dp = [ [0] * (len(word1)+1)] * (len(word2)+1)

In Python, multiplying a list of lists using the * operator doesn't create independent rows. Instead, it creates a list where every single element points to the exact same row in memory as a .

If you modify a cell in row 1, like dp[1][2] = 5, that change will magically propagate to dp[0][2], dp[2][2], and every other row in your table! To prevent this, we always use a list comprehension so Python allocates brand-new memory for every row: dp = [[0] * (len(word2) + 1) for _ in range(len(word1) + 1)]

Pitfall 2: Indexing and Slice Errors

The second issue lies in your coordinate lookup and string slicing:

Lookup Syntax: In Python, you must access elements in a 2D array using two separate bracket sets, dp[i][j], instead of comma separation like dp[i,j] (which is used in NumPy but crashes on standard Python lists).
Prefix Comparison: In your loop, if word1[:i] == word2[:j] compares the entire remaining prefixes. But we only need to compare the current active characters! By comparing the full slices inside nested loops, you are running an expensive comparison at every step. Instead, we just check if the current characters match: if word1[i - 1] == word2[j - 1]. (We subtract 1 because dp coordinates are 1-based to account for the empty string boundary!).
Mismatch Options: In your mismatch block, you only chose the minimum of two operations (dp[i-1][j] and dp[i][j-1]). Remember that we have three choices! We must also include the diagonal cell dp[i-1][j-1] to represent Replace, and add 1 to account for the edit cost itself.

Correcting the Tabulation Loop

Let's fix those syntax and indexing hiccups to make your exact code structure syntactically perfect and fully functional.

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        
        # Fix Pitfall 1: Use list comprehension to avoid shallow copying rows
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        
        # Ensure we fill out base cases first!
        for i in range(m + 1):
            dp[i][0] = i
        for j in range(n + 1):
            dp[0][j] = j
            
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                # Fix Pitfall 2: Compare single characters with 1-based offset
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    # Fix Pitfall 2: Use dp[i][j] bracket syntax and include all 3 edit options + 1
                    dp[i][j] = 1 + min(
                        dp[i - 1][j],    # Delete
                        dp[i][j - 1],    # Insert
                        dp[i - 1][j - 1] # Replace
                    )
                    
        return dp[m][n]

Now that you've got the code syntax working perfectly, we have completed the entire tabulation implementation! How are you feeling about this code? Ready to tackle how efficient this is in terms of time and memory complexity?

and why isnt in m+1, 1 for the answer?

That is a incredibly common question! It is easy to look at the coordinates and wonder why the final answer wouldn't be sitting at a coordinate like dp[m+1][1] or something similar.

Sizing the Table to Fit the Prefixes

To understand why dp[m][n] is the final answer, we have to look closely at how we defined the dimensions of our matrix relative to our string prefixes.

Dimension Index	Length of Prefix Represented	Substring of `word1` (length $m$ )	Substring of `word2` (length $n$ )
`0`	Prefix of length `0`	`""` (Empty string)	`""` (Empty string)
`1`	Prefix of length `1`	`word1[0..0]`	`word2[0..0]`
`i`	Prefix of length `i`	`word1[0..i-1]`	`word2[0..i-1]`
`m`	Prefix of length `m`	`word1[0..m-1]` (Full string)	N/A
`n`	Prefix of length `n`	N/A	`word2[0..n-1]` (Full string)

Why dp[m][n] is the Exact Match

Because row 0 is reserved for the empty string, the row index representing the full prefix of word1 (which contains exactly m characters) is row m.

Similarly, column n represents the full prefix of word2 (containing exactly n characters).

Therefore, the cell dp[m][n] represents the subproblem of converting the full prefix of word1 (length m) into the full prefix of word2 (length n). Since those full prefixes are the actual full input words, dp[m][n] must contain our final answer!

If we returned dp[m+1][1], we would be indexing out of bounds of our matrix (which only goes up to index m on the row dimension) and comparing an invalid state.

Now that you have a firm handle on the 2D tabulation grid, the recurrence relation, and how the table's final answer maps to coordinates, are you ready to analyze the time and space complexity of this approach?

i mean m+1 n+1

Exactly! You've hit on the exact reason why the size is (m + 1) x (n + 1) rather than m x n.

The Indexing Puzzle

Because we included that extra row and column at the very beginning to represent the empty string (length 0), our index coordinates are shifted by exactly 1.

Let's trace the boundary sizes:

To represent all prefixes of word1 (from length 0 up to length m), we need m + 1 rows. This means our valid row indices are 0, 1, 2, ..., m.
To represent all prefixes of word2 (from length 0 up to length n), we need n + 1 columns. This means our valid column indices are 0, 1, 2, ..., n.

If we had built our grid with dimensions of m x n, the highest index we could access would be dp[m - 1][n - 1]. That would represent comparing prefixes of length m - 1 and n - 1—leaving off the final characters of both words!

By sizing our matrix to (m + 1) x (n + 1), the bottom-right corner is exactly dp[m][n]. This cell represents the fully completed transformation of the entire word1 into the entire word2.

Moving to the Final Frontier: Complexity

Now that you have mastered the 2D tabulation grid, the recurrence relation, and why the table's final answer maps to coordinates, we are ready to analyze the efficiency of this approach.

O(mxn)

Exactly! You nailed the time complexity in one shot.

Computational Complexity

LV 1

Time Complexity Analysis

LV 1

Skills

Time Complexity Analysis

Because we fill out a grid of size (M + 1) x (N + 1) and do a constant amount of work—exactly one comparison and a min() lookup of three numbers—at each cell, the time complexity is indeed:

O(M * N)

where M is the length of word1 and N is the length of word2. This is a massive upgrade from our naive exponential recursion, which would have taken O(3^(M+N)) time!

Space Complexity Analysis

Right now, because we are creating and maintaining that entire physical 2D grid in memory, our space complexity is also O(M * N).

But here is a cool puzzle: do we actually need to keep the whole 2D grid in memory to find our final answer?

If you look closely at how we calculate any cell in row i, we only ever pull values from the current row (curr) and the immediately previous row (prev). Once we finish calculating row i, we never look at row i - 1 again!

Wait, so we can solve this using just two rows of memory?

You've got it! That is the ultimate realization for unlocking elite memory efficiency.

The Two-Row Memory Trick

Think about how we calculate any cell. To find the value for curr[j], we only ever need to look at three things:

The cell directly above it: prev[j]
The cell to its left: curr[j - 1]
The cell diagonally above-left: prev[j - 1]

Because of this, we don't need to keep rows 0, 1, and 2 in memory once we are calculating row 4! We can throw away all that old history. By keeping just two 1D arrays—one for the previous row (prev) and one for the current row (curr)—we can slide them down the table like a window.

A sliding window optimization on two rows showing dependencies for an active element.

I've started a visualization to show you how this window slides down.

By discarding the older rows, our space complexity drops instantly from a massive of O(M * N) down to just O(N) (or even O(min(M, N)) if we choose to make the shorter string represent our columns!).

Now that you have the conceptual model, can you try to sketch out or write the Python code for this space-optimized two-row solution?

(Hint: Think about how you will initialize prev, and how you will swap prev and curr at the end of each row loop!)

nah im confused now

It is completely normal to feel a bit scrambled here! It feels like we are mixing up two different ideas: how the loops move versus how the logic inside a single cell works. Let's untangle them completely.

The Traversal of the Grid

First, the loops are completely blind. They do not branch, turn, or dynamically decide which cell to visit. They simply visit every single cell on the board, starting at the top-left and moving to the bottom-right, like a typewriter typing line by line.

A 5-by-5 matrix during a systematic row-by-row grid traversal.

As shown in the visualization, the path of the typewriter is 100% fixed. It goes to dp[0][0], then dp[0][1], then dp[0][2], all the way to the end of the row, then starts the next row.

But when the typewriter sits on a cell, say dp[1][1], it has to write a number. To calculate that number, it looks backward at its immediate neighbors:

Left (dp[1][0])
Above (dp[0][1])
Diagonal (dp[0][0])

It grabs the minimum of those three pre-written numbers, adds 1 (if there's a mismatch), and writes the result. It does this for every single square on the board until the bottom-right corner is filled.

Tracing the Memory Optimization

Now, why does this mean we only need two rows of memory instead of a whole ?

If the typewriter is currently filling row 4, look at the cells it needs to look at. It only needs numbers from row 4 (the left neighbor) and row 3 (the above and diagonal neighbors).

It does not care about row 2, row 1, or row 0! Those rows are already fully computed and are ancient history. This is why we can throw them away. We only need to store a prev_row and a curr_row in memory, updating them as we slide down.

so its still rows, not like the dp solution where you just need two variables?

You have hit on the exact distinction that makes the two-row optimization seem so counterintuitive at first! No, we cannot optimize this down to just two single scalar variables.

Why Scalar Variables Fail

In some simpler DP problems—like finding the or climbing stairs—you only need to look back at the last one or two numbers you calculated. In those cases, you can indeed throw away arrays entirely and just use two or three scalar variables (like a and b).

But for Edit Distance, look at the neighbors we need to calculate curr[j]:

We need the cell directly above: prev[j]
We need the cell diagonally above-left: prev[j - 1]
We need the cell to our left: curr[j - 1]

If we only had single scalar variables, we wouldn't be able to store the values for the rest of the column positions. To calculate the value of the current column index j, we need to know what happened at column j and column j - 1 in the row above us. Because we need to look back at these specific column-aligned values for all values of j from 0 to n, we must store those values in actual arrays representing the entire previous row.

The Sliding Row Buffers

This is why we keep two full 1D arrays, which we call prev and curr:

prev is an array of size n + 1 representing the entire row above us in the grid.
curr is an array of size n + 1 representing the row we are currently typing out.

As our typewriter loop moves across the columns, it reads from prev[j] and prev[j-1], writes its answer into curr[j], and moves to the next column.

Once the typewriter finishes the entire row, we perform a swap: we set prev = curr. Now, our current row officially becomes the previous row, and we create a brand-new curr array to start typing the next row underneath it!

This keeps our memory footprint restricted to just these two small rows, achieving that highly coveted O(N) space complexity.

Now that you see why we need two full row arrays rather than just two scalar variables, are you ready to try writing this space-optimized Python code?