top of page
Search

Delete Operation for Two Strings

Updated: Mar 25, 2021


Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.



Example 1:

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


Note:

  1. The length of given words won't exceed 500.

  2. Characters in given words can only be lower-case letters.

Solution:


Approach 1: Using Longest Common subsequence


class Solution {
    public int minDistance(String word1, String word2) {
      int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        for (int i = 0; i <= word1.length(); i++) {
            for (int j = 0; j <= word2.length(); j++) {
                if (i == 0 || j == 0)
                    continue;
                if (word1.charAt(i - 1) == word2.charAt(j - 1))
                    dp[i][j] = 1 + dp[i - 1][j - 1];
                else
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return word1.length() + word2.length() - 2 * dp[word1.length()][word2.length()];   
    }
}


Approach 2: Without LCS (dynammic programming)


public class Solution {
    public int minDistance(String s1, String s2) {
        int[][] dp = new int[s1.length() + 1][s2.length() + 1];
        for (int i = 0; i <= s1.length(); i++) {
            for (int j = 0; j <= s2.length(); j++) {
                if (i == 0 || j == 0)
                    dp[i][j] = i + j;
                else if (s1.charAt(i - 1) == s2.charAt(j - 1))
                    dp[i][j] = dp[i - 1][j - 1];
                else
                    dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[s1.length()][s2.length()];
    }
}

36 views0 comments

Recent Posts

See All

A string s is called good if there are no two different characters in s that have the same frequency. Given a string s, return the minimum number of characters you need to delete to make s good. The f

The numeric value of a lowercase character is defined as its position (1-indexed) in the alphabet, so the numeric value of a is 1, the numeric value of b is 2, the numeric value of c is 3, and so on.

bottom of page