HackerRank Morgan and a String problem solution

In this HackerRank Morgan and a String problem solution, we have given a lexicographically minimal string made of two collections and we need to take a letter from a collection only when it is on the top of the stack and we need to use all of the letters in the collections and form a new collection.

Problem solution in Python.

T = int(input())

def alpha_min(a, b):
    la = len(a)
    lb = len(b)
    a += "z"
    b += "z"
    i = j = 0
    res = ""
    while (i != la and j != lb):
        if a[i:] < b[j:]:
            res += a[i]
            i += 1
        else:
            res += b[j]
            j += 1
        
    res += a[i: -1] + b[j: -1]
    return res
    
for _ in range(T):
    
    a = input().strip()
    b = input().strip()
    print(alpha_min(a, b))

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Problem solution in Java.

import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);

        int t = in.nextInt();
        in.nextLine();

        while(t-- > 0) {
            String A = in.nextLine();
            String B = in.nextLine();

            int i = 0, j = 0;
            StringBuffer sb = new StringBuffer();

            while(i < A.length() && j < B.length()) {
                if (A.charAt(i) < B.charAt(j)) {
                    sb.append(A.charAt(i++));
                } else if (A.charAt(i) > B.charAt(j)) {
                    sb.append(B.charAt(j++));
                } else {
                    int x = i, y = j;
                    char a = A.charAt(i);
                    for(; x < A.length() && y < B.length(); x++, y++) {
                        if (A.charAt(x) != B.charAt(y)) {
                            break;
                        } else if (A.charAt(x) > a) {
                            sb.append(A.substring(i, x)).append(B.substring(j, y));
                            i = x; j = y;
                            a = A.charAt(x);
                        }
                    }

                    if (x == A.length()) {
                        sb.append(B.charAt(j));
                        j++;
                    } else if (y == B.length()) {
                        sb.append(A.charAt(i));
                        i++;
                    } else {
                        if (A.charAt(x) < B.charAt(y)) {
                            sb.append(A.charAt(i));
                            i++;
                        } else {
                            sb.append(B.charAt(j));
                            j++;
                        }
                    }
                }
            }

            sb.append(A.substring(i)).append(B.substring(j));

            System.out.println(sb);
        }
    }

}

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Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int t;
    cin >> t;
    
    string s1, s2, o;
    while(t--) {
        cin >> s1 >> s2;
        int l1 = s1.length();
        int l2 = s2.length();
        int p1 = 0, p2 = 0;
        o = "";
        while(p1 < l1 && p2 < l2) {
            while(p1 < l1 && p2 < l2 && s1[p1] != s2[p2]) {
                if(s1[p1] < s2[p2]) o += s1[p1++];
                else                o += s2[p2++];
            }
            
            int c = 0, e = 0;
            while(p1+c < l1 && p2+c < l2 && s1[p1+c] == s2[p2+c] && s1[p1+c] == s1[p1]) {
                ++e;
                ++c;
            }
            while(p1+c < l1 && p2+c < l2 && s1[p1+c] == s2[p2+c] && s1[p1+c] <= s1[p1]) ++c;
            
            if(p1+c < l1 && p2+c < l2)
                if(s1[p1+c] < s2[p2+c])
                    while(e--) o += s1[p1++];
                else
                    while(e--) o += s2[p2++];
            else
                if(p1+c == l1)
                    while(e--) o += s2[p2++];
                else if(p2+c == l2)
                    while(e--) o += s1[p1++];
        }
        while(p1 < l1) o += s1[p1++];
        while(p2 < l2) o += s2[p2++];
        
        cout << o << endl;
    }
    return 0;
}

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Problem solution in C.

 #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

char str1[100002] = {0};
char str2[100002] = {0};

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%s", str1);
        scanf("%s", str2);

        int len1 = strlen(str1);
        int len2 = strlen(str2);
        str1[len1++] = 'z';
        str2[len2++] = 'z';
        str1[len1] = '\0';
        str2[len2] = '\0';

        int i = 0, j = 0;
        while(str1[i] != 'z' || str2[j] != 'z'){
            if(str1[i] < str2[j]){
                printf("%c",str1[i]);
                i++;
            }else if(str1[i] > str2[j]){
                printf("%c", str2[j]);
                j++;
            }else{
                int res = strcmp(str1 + i +1, str2 + j + 1);
if(res <= 0)
{
printf("%c", str1[i]);
i++;
}else{
printf("%c", str2[j]);
j++;
}
            }
        }
        printf("\n");
    }
    return 0;
}

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