Delete Characters to Equalize Strings - Google Top Interview Questions


Problem Statement :


Given two lowercase alphabet strings a and b, consider an operation where we delete any character in either string. 
Return the minimum number of operations required such that both strings are equal.

Constraints

0 ≤ n ≤ 1,000 where n is the length of a

0 ≤ m ≤ 1,000 where m is the length of b

Example 1

Input

a = "zyyx"

b = "yfyx"

Output

2

Explanation

We delete the "z" in a and delete "f" in b


Solution :



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                        Solution in C++ :

int solve(string a, string b) {
    int m = a.size();
    int n = b.size();

    int dp[2][n + 1];
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {
            if (i == 0)
                dp[i % 2][j] = j;
            else if (j == 0)
                dp[i % 2][j] = i;
            else if (a[i - 1] == b[j - 1])
                dp[i % 2][j] = dp[1 - i % 2][j - 1];
            else
                dp[i % 2][j] = 1 + min(dp[1 - i % 2][j], dp[i % 2][j - 1]);
        }
    }

    return dp[m % 2][n];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int dp(String a, String b, int i, int j, int[][] memo) {
        if (i == a.length() && j == b.length()) {
            return 0;
        }
        if (i == a.length()) {
            return b.length() - j;
        }
        if (j == b.length()) {
            return a.length() - i;
        }

        if (memo[i][j] != -1) {
            return memo[i][j];
        }
        int ans = Integer.MAX_VALUE;
        if (a.charAt(i) == b.charAt(j)) {
            ans = Math.min(ans, dp(a, b, i + 1, j + 1, memo));
        } else {
            ans = Math.min(ans, 1 + dp(a, b, i + 1, j, memo));
            ans = Math.min(ans, 1 + dp(a, b, i, j + 1, memo));
        }
        return memo[i][j] = ans;
    }
    public int solve(String a, String b) {
        int[][] memo = new int[a.length() + 5][b.length() + 5];
        for (int[] x : memo) {
            Arrays.fill(x, -1);
        }
        return dp(a, b, 0, 0, memo);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a, b):

        m = len(a)
        n = len(b)

        @functools.lru_cache(None)
        def dp(i, j):
            if i == m:
                return n - j
            elif j == n:
                return m - i
            else:
                if a[i] == b[j]:
                    return dp(i + 1, j + 1)
                else:
                    return 1 + min(dp(i + 1, j), dp(i, j + 1))

        return dp(0, 0)


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