Problem

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 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')

Solution

/**
 * @param {string} word1
 * @param {string} word2
 * @return {number}
 */
var minDistance = function(word1, word2) {
  var n = word1.length;
  var m = word2.length;
  var dp = Array(n);

  for (var i = 0; i < n; i++) {
    dp[i] = Array(m);
    for (var j = 0; j < m; j++) {
      dp[i][j] = Math.min(
        getDp(i - 1, j, dp) + 1,
        getDp(i, j - 1, dp) + 1,
        getDp(i - 1, j - 1, dp) + (word1[i] === word2[j] ? 0 : 1)
      );
    }
  }

  return getDp(n - 1, m - 1, dp);
};

var getDp = function (i, j, dp) {
  if (i < 0 && j < 0) return 0;
  if (i < 0) return j + 1;
  if (j < 0) return i + 1;
  return dp[i][j];
};

Explain:

dp[i][j] 代表 word1 的 0 ~ i 转为 word2 的 0 ~ j 的最少步骤

dp[i][j] = min(dp[i - 1][j] + 1, dp[i][j - 1] + 1, dp[i - 1][j - 1] + (word1[i] === word2[j] ? 0 : 1));

Complexity:

• Time complexity : O(m*n).
• Space complexity : Om*(n).