Here, We see Delete Operation for Two Strings LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Delete Operation for Two Strings LeetCode Solution

Problem Statement

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

Delete Operation for Two Strings Leetcode Solution C++

class Solution {
public:
    int minDistance(string word1, string word2) 
    {
        int m=word1.length(), n=word2.length(); 
        vector<vector<int>> dp(m+1, vector<int> (n+1, 0)); 
        for(int i=0; i<=m; i++)
        {
            for(int j=0; j<=n; j++)
            {
                if(i==0 || j==0) continue;
                else if(word1[i-1]==word2[j-1])
                    dp[i][j] = 1+dp[i-1][j-1];
                else
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
            }
        }
        return m+n-2*dp[m][n];
    }
};Code language: PHP (php)

Delete Operation for Two Strings Leetcode Solution Java

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        if (m < n) {
            String tempStr = word1;
            word1 = word2;
            word2 = tempStr;
            int tempInt = n;
            n = m;
            m = tempInt;
        }
        char[] WA1 = word1.toCharArray(), WA2 = word2.toCharArray();
        int[] dpLast = new int[n+1], dpCurr = new int[n+1];
        for (char c1 : WA1) {
            for (int j = 0; j < n; j++) 
                dpCurr[j+1] = c1 == WA2[j]
                    ? dpLast[j] + 1
                    : Math.max(dpCurr[j], dpLast[j+1]);
            int[] tempArr = dpLast;
            dpLast = dpCurr;
            dpCurr = tempArr;
        }
        return m + n - 2 * dpLast[n];
    }
}Code language: JavaScript (javascript)

Delete Operation for Two Strings Solution JavaScript

var minDistance = function(word1, word2) {
    let m = word1.length, n = word2.length
    if (m < n) [word1, word2, m, n] = [word2, word1, n, m]
    let WA1 = word1.split(""), WA2 = word2.split(""),
        dpLast = new Uint16Array(n + 1),
        dpCurr = new Uint16Array(n + 1)
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) 
            dpCurr[j+1] = WA1[i] === WA2[j]
                ? dpLast[j] + 1
                : Math.max(dpCurr[j], dpLast[j+1]);
        [dpLast, dpCurr] = [dpCurr, dpLast]
    }
    return m + n - 2 * dpLast[n] 
};Code language: JavaScript (javascript)

Delete Operation for Two Strings Solution Python

class Solution(object):
    def minDistance(self, word1, word2):
        m, n = len(word1), len(word2)
        if m < n: word1, word2, m, n = word2, word1, n, m
        dpLast, dpCurr = [0] * (n + 1), [0] * (n + 1)
        for c1 in word1:
            for j in range(n):
                dpCurr[j+1] = dpLast[j] + 1 if c1 == word2[j] else max(dpCurr[j], dpLast[j+1])
            dpLast, dpCurr = dpCurr, dpLast
        return m + n - 2 * dpLast[n]Code language: HTML, XML (xml)