Edit Distance
A hard Dynamic Programming problem included in Love Babbar 450, Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
- Platform
- LeetCode
The problem
Given two strings, find the minimum number of operations required to convert one string into the other. The allowed operations are inserting a character, deleting a character, or replacing a character.
Example 1
- Input
- word1 = "horse", word2 = "ros"
- Output
- 3
- Why
- horse -> rorse (replace 'h' with 'r') -> rose (remove 'r') -> ros (remove 'e').
Example 2
- Input
- word1 = "intention", word2 = "execution"
- Output
- 5
- Why
- intention -> inention (remove 't') -> enention (replace 'i' with 'e') -> exention (replace 'n' with 'x') -> exection (replace 'n' with 'c') -> execution (insert 'u').
Example 3
- Input
- word1 = "abc", word2 = "abc"
- Output
- 0
- Why
- The strings are identical, so no operations are needed.
Constraints
- 0 <= word1.length, word2.length <= 500
- word1 and word2 consist of lowercase English letters
How to think about it
Updated 2026-09-09Aligning two strings character by character from the ends reduces the edit distance to three local operations: delete (move up in word1), insert (move left in word2), or replace (move diagonally). If characters match, the diagonal move is free; otherwise pay 1 and take the minimum of all three.
Approaches, worst first
Recursive edits
time O(3^(m+n)) · space O(m + n)
Branch into deletion, insertion, and replacement at every mismatch. Re-evaluates identical substring pairs across branches, causing exponential time explosion.
2D table
time O(m * n) · space O(m * n)
Create matrix dp of size (m+1) x (n+1). Base rows and columns represent transforming to and from empty strings. Fill cells by checking character equality or taking 1 + min(up, left, diag).
Two-row rolling bufferWrite this one
time O(m * n) · space O(min(m, n))
Because each row only references cells from its immediate predecessor row, preserve memory by retaining only the previous row and the current row during iteration.
Where people lose marks · 3
- Incorrect base case initialization: transforming an empty prefix to a string of length j requires j insertions, and transforming a prefix of length i to empty requires i deletions.
- Off-by-one indexing when comparing string characters. Character at index i - 1 corresponds to matrix row i.
- Adding 1 when characters match. A matching character costs 0 edits and must carry dp[i-1][j-1] straight forward without incrementing.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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