Day56(583, 72)

583. Delete Operation for Two Strings

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

class Solution {  public int minDistance(String word1, String word2) {  int len1 = word1.length();  int len2 = word2.length();  int[][] dp = new int[len1 + 1][len2 + 1];  for (int i = 1; i <= len1; i++) {  for (int j = 1; j <= len2; j++) {  if (word1.charAt(i - 1) == word2.charAt(j - 1)) {  dp[i][j] = dp[i - 1][j - 1] + 1;  } else {  dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);  }  }  }  return len1 + len2 - dp[len1][len2] * 2;  }  
}

72. Edit Distance

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’)

class Solution {  public int minDistance(String word1, String word2) {  int m = word1.length();  int n = word2.length();  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(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;  }  }  }  return dp[m][n];  }  
}