583. Delete Operation for Two Strings

Description

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.

Constraints

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

• word1 and word2 consist of only lowercase English letters.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(M*N), Where M is the size of the word1 and 
 *    N is the size of the word2.
 * Space complexity : O(M*N)
 */

class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null && word2 == null) {
            return 0;
        } else if(word1 == null) {
            return word2.length();
        } else if(word2 == null) {
            return word1.length();
        }
        
        int l1 = word1.length(), l2 = word2.length();
        int[][] dp = new int[l1+1][l2+1];
        
        for(int i = 0; i <= l1; i++) {
            for(int j = 0; j <= l2; j++) {
                if(i == 0 || j == 0) {
                    dp[i][j] = i + j;
                } else if(word1.charAt(i-1) == word2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i][j-1], dp[i-1][j]);
                }
            }
        }
        
        return dp[l1][l2];
    }
}

Solution 2

/**
 * Time complexity : O(m*n), Where m is the size of the word1 and 
 *    n is the size of the word2.
 * Space complexity : O(n). dpdp array of size n is used.
 */

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

Follow up