Edit Distance LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
Share:

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.

Edit Distance Solutions

Time: O(mn)
Space: O(mn)→O(n)

C++

class Solution {
 public:
  int minDistance(string word1, string word2) {
    const int m = word1.length();
    const int n = word2.length();
    // dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
    vector<vector<int>> dp(m + 1, vector<int>(n + 1));

    for (int i = 1; i <= m; ++i)
      dp[i][0] = i;

    for (int j = 1; 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] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;

    return dp[m][n];
  }
};

Java

 class Solution {
  public int minDistance(String word1, String word2) {
    final int m = word1.length();
    final int n = word2.length();
    // dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
    int[][] dp = new int[m + 1][n + 1];

    for (int i = 1; i <= m; ++i)
      dp[i][0] = i;

    for (int j = 1; j <= n; ++j)
      dp[0][j] = j;

    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= n; ++j)
        if (word1.charAt(i - 1) == word2.charAt(j - 1))
          dp[i][j] = dp[i - 1][j - 1];
        else
          dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;

    return dp[m][n];
  }
}

Python

class Solution:
  def minDistance(self, word1: str, word2: str) -> int:
    m = len(word1)
    n = len(word2)
    # dp[i][j] := min # of operations to convert word1[0..i) to word2[0..j)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
      dp[i][0] = i

    for j in range(1, 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] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1

    return dp[m][n]

Watch Tutorial

Checkout more Solutions here

Leave a Comment

Your email address will not be published. Required fields are marked *

x