Problem
Given two strings, return the minimum number of operations (insert, delete, replace) to convert one to the other.
Example
Input: word1 = "horse", word2 = "ros"
Output: 3Solution
2D DP. dp[i][j] = edit distance between first i chars of word1 and first j chars of word2.
def min_distance(word1, word2):
m, n = len(word1), len(word2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
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):
if word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
return dp[m][n]function minDistance(word1, word2) {
const m = word1.length, n = word2.length;
const dp = Array.from({length: m + 1}, () => new Array(n + 1).fill(0));
for (let i = 0; i <= m; i++) dp[i][0] = i;
for (let j = 0; j <= n; j++) dp[0][j] = j;
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
if (word1[i-1] === word2[j-1]) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = 1 + Math.min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]);
}
}
return dp[m][n];
}int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
for (int i = 0; i <= m; i++) dp[i][0] = i;
for (int j = 0; j <= n; j++) dp[0][j] = j;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1[i-1] == word2[j-1]) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = 1 + min({dp[i-1][j], dp[i][j-1], dp[i-1][j-1]});
}
}
return dp[m][n];
}public int minDistance(String w1, String w2) {
int m = w1.length(), n = w2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0; i <= m; i++) dp[i][0] = i;
for (int j = 0; j <= n; j++) dp[0][j] = j;
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (w1.charAt(i-1) == w2.charAt(j-1)) dp[i][j] = dp[i-1][j-1];
else dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1]));
}
}
return dp[m][n];
}Complexity
- Time: O(m * n)
- Space: O(m * n)
Explanation
When chars match, no edit needed. Otherwise pick the cheapest of: delete (dp[i-1][j]), insert (dp[i][j-1]), or replace (dp[i-1][j-1]).
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