LeetCode-in-All

72. Edit Distance

Hard

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

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 <= 500
• word1 and word2 consist of lowercase English letters.

Solution

class Solution {
  int minDistance(String w1, String w2) {
    int n1 = w1.length;
    int n2 = w2.length;

    // If w2 is longer, swap w1 and w2 to ensure w1 is the longer string
    if (n2 > n1) {
      return minDistance(w2, w1);
    }

    List<int> dp = List.filled(n2 + 1, 0);

    // Initialize the first row of the DP table
    for (int j = 0; j <= n2; j++) {
      dp[j] = j;
    }

    for (int i = 1; i <= n1; i++) {
      int pre = dp[0];
      dp[0] = i;
      for (int j = 1; j <= n2; j++) {
        int temp = dp[j];
        dp[j] = w1[i - 1] != w2[j - 1]
            ? 1 + [pre, dp[j], dp[j - 1]].reduce((a, b) => a < b ? a : b)
            : pre;
        pre = temp;
      }
    }

    return dp[n2];
  }
}

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