Move from brute-force thinking to an efficient approach using dynamic programming strategy.
Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.
You have the following three operations permitted on a word:
Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
Constraints:
0 <= word1.length, word2.length <= 500word1 and word2 consist of lowercase English letters.Problem summary: Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2. You have the following three operations permitted on a word: Insert a character Delete a character Replace a character
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
"horse" "ros"
"intention" "execution"
one-edit-distance)delete-operation-for-two-strings)minimum-ascii-delete-sum-for-two-strings)uncrossed-lines)minimum-white-tiles-after-covering-with-carpets)class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();
int[] dp = new int[n + 1];
for (int j = 0; j <= n; j++) dp[j] = j;
for (int i = 1; i <= m; i++) {
int prevDiag = dp[0];
dp[0] = i;
for (int j = 1; j <= n; j++) {
int old = dp[j];
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[j] = prevDiag;
} else {
dp[j] = 1 + Math.min(prevDiag, Math.min(dp[j], dp[j - 1]));
}
prevDiag = old;
}
}
return dp[n];
}
}
func minDistance(word1 string, word2 string) int {
m, n := len(word1), len(word2)
dp := make([]int, n+1)
for j := 0; j <= n; j++ {
dp[j] = j
}
for i := 1; i <= m; i++ {
prevDiag := dp[0]
dp[0] = i
for j := 1; j <= n; j++ {
old := dp[j]
if word1[i-1] == word2[j-1] {
dp[j] = prevDiag
} else {
dp[j] = 1 + min(prevDiag, min(dp[j], dp[j-1]))
}
prevDiag = old
}
}
return dp[n]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
m, n = len(word1), len(word2)
dp = list(range(n + 1))
for i in range(1, m + 1):
prev_diag = dp[0]
dp[0] = i
for j in range(1, n + 1):
old = dp[j]
if word1[i - 1] == word2[j - 1]:
dp[j] = prev_diag
else:
dp[j] = 1 + min(prev_diag, dp[j], dp[j - 1])
prev_diag = old
return dp[n]
impl Solution {
pub fn min_distance(word1: String, word2: String) -> i32 {
let a = word1.as_bytes();
let b = word2.as_bytes();
let n = b.len();
let mut dp: Vec<i32> = (0..=n as i32).collect();
for i in 1..=a.len() {
let mut prev_diag = dp[0];
dp[0] = i as i32;
for j in 1..=n {
let old = dp[j];
if a[i - 1] == b[j - 1] {
dp[j] = prev_diag;
} else {
dp[j] = 1 + prev_diag.min(dp[j]).min(dp[j - 1]);
}
prev_diag = old;
}
}
dp[n]
}
}
function minDistance(word1: string, word2: string): number {
const n = word2.length;
const dp: number[] = Array.from({ length: n + 1 }, (_, j) => j);
for (let i = 1; i <= word1.length; i++) {
let prevDiag = dp[0];
dp[0] = i;
for (let j = 1; j <= n; j++) {
const old = dp[j];
if (word1[i - 1] === word2[j - 1]) {
dp[j] = prevDiag;
} else {
dp[j] = 1 + Math.min(prevDiag, dp[j], dp[j - 1]);
}
prevDiag = old;
}
}
return dp[n];
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.