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).