Zipped String - Amazon Top Interview Questions


Problem Statement :


Given lowercase alphabet strings a, b, and c return whether there's any way of obtaining c by merging characters in order from a and b.

Constraints

0 ≤ n ≤ 1,000 where n is the length of a
0 ≤ m ≤ 1,000 where m is the length of b
0 ≤ k ≤ 1,000 where k is the length of c

Example 1

Input

a = "abc"
b = "def"
c = "abdefc"

Output

True

Explanation

abdefc is an interleave since abc and def can be interleaved with: abdefc

Example 2

Input

a = "ab"
b = "cd"
c = "abdc"

Output

False

Explanation

Even though ab is in the right order in abdc, cd is backwards, so this is not an interleaving.


Solution :



title-img 




                        Solution in C++ :

bool solve(string a, string b, string c) {
    if (a.size() + b.size() != c.size()) return false;
    vector<vector<bool>> dp(a.size() + 1, vector<bool>(b.size() + 1, false));
    dp[0][0] = true;
    for (int i = 0; i <= a.size(); i++) {
        for (int j = 0; j <= b.size(); j++) {
            if (j > 0) {
                dp[i][j] = dp[i][j - 1] && b[j - 1] == c[i + j - 1];
            }
            if (i > 0) {
                dp[i][j] = dp[i][j] || (dp[i - 1][j] && a[i - 1] == c[i + j - 1]);
            }
        }
    }
    return dp[a.size()][b.size()];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(String a, String b, String c) {
        if (a.length() + b.length() != c.length())
            return false;

        int indexA = 0, indexB = 0, prevA = -1, prevB = -1, prevIndex = -1;
        boolean goBack = false;

        for (int i = 0; i < c.length(); i++) {
            if (indexA < a.length() && a.charAt(indexA) == c.charAt(i)) {
                if (!goBack && indexB < b.length() && b.charAt(indexB) == c.charAt(i)) {
                    prevA = indexA;
                    prevB = indexB;
                    prevIndex = i;
                    goBack = true;
                }
                indexA++;
            } else if (indexB < b.length() && b.charAt(indexB) == c.charAt(i))
                indexB++;
            else if (goBack) {
                indexA = prevA;
                indexB = prevB + 1;
                i = prevIndex;

                goBack = false;
            } else
                return false;
        }

        return true;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, a, b, c):
        if len(a) + len(b) != len(c):
            return False

        def f(a_idx, b_idx, c_idx):
            if c_idx == len(c):
                return True
            possible = False
            if a_idx < len(a) and c[c_idx] == a[a_idx]:
                possible = possible or f(a_idx + 1, b_idx, c_idx + 1)
            if b_idx < len(b) and c[c_idx] == b[b_idx]:
                possible = possible or f(a_idx, b_idx + 1, c_idx + 1)
            return possible

        return f(0, 0, 0)


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