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Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

Example 1:


Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Example 2:


Input: word1 = "leetcode", word2 = "etco"
Output: 4

Constraints:

1 <= word1.length, word2.length <= 500

word1 and word2 consist of only lowercase English letters.

풀이

은찬


var minDistance = function(word1, word2) {
    let lcs = 0;
    
    if(word1 === word2){
        return 0;
    }

    // word2.length is short than word1.length
    if(word1.length < word2.length){
        [word1, word2] = [word2, word1];
    }
    
    const word1Len = word1.length;
    const word2Len = word2.length;
    const dp = Array.from({length: word1Len + 1}, () => Array(word2Len + 1).fill(0));

    for(let i = 1; i <= word1Len; i++){
        for(let j = 1; j <= word2Len; j++){
            const char1 = word1[i - 1];
            const char2 = word2[j - 1];
            
            if(char1 === char2){
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else{
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    
    lcs = dp[word1Len][word2Len];
    
    return (word1Len - lcs) + (word2Len - lcs);
};

재영

얼레… 왜 않올렷찌…


const minDistance = (word1, word2) => {
  const word1Length = word1.length;
  const word2Length = word2.length;

  const arr = Array.from({ length: word1Length + 1 }, () =>
    new Array(word2Length + 1).fill(0)
  );
// n x m
 // sea eat
 //   e a t 
 // 0 0 0 0
s// 0 0 0 0
e// 0 1 1 1
a// 0 1 2 2

  for (let i = 1; i <= word1Length; i += 1) {
    for (let j = 1; j <= word2Length; j += 1) {
      const now1 = word1[i - 1];
      const now2 = word2[j - 1];

      if (now1 === now2) {
        arr[i][j] = Math.max(
          arr[i - 1][j - 1] + 1,
          arr[i - 1][j], 
					arr[i][j - 1]
        );
      } else {
        arr[i][j] = Math.max(arr[i - 1][j], arr[i][j - 1]);
      }
      // console.log(now1, now2, arr);
    }
  }
  return word1Length + word2Length - 2 * arr[word1Length][word2Length];
};

효성


var minDistance = function(word1, word2) {
    const lcs = Array.from(Array(word1.length + 1), () => Array(word2.length + 1).fill(0));
    const firstWordLength = word1.length;
    const secondWordLength = word2.length;

    for (let m = 1; m <= firstWordLength; m++) {
        for (let n = 1; n <= secondWordLength; n++) {
            const isSameCharacter = word1[m - 1] === word2[n - 1];
            const maxSubsequenceTillNow = Math.max(lcs[m][n - 1], lcs[m - 1][n]);
            lcs[m][n] = isSameCharacter ? 1 + lcs[m - 1][n - 1] : maxSubsequenceTillNow;
        }
    }
    
    return firstWordLength + secondWordLength - 2 * lcs[firstWordLength][secondWordLength];
};

Delete Operation for Two Strings

풀이