HackerRank Morgan and a String problem solution

HackerRank Morgan and a String problem solution – In this morgan and a string problem, Jack and Daniel are friends. Both of them like letters, especially uppercase ones. They are cutting uppercase letters from newspapers, and each one of them has his collection of letters stored in a stack.

One beautiful day, Morgan visited Jack and Daniel. He saw their collections. He wondered what is the lexicographically minimal string made of those two collections. He can take a letter from a collection only when it is on the top of the stack. Morgan wants to use all of the letters in their collections.

Function Description

Complete the morganAndString function in the editor below.

morganAndString has the following parameter(s):

• string a: Jack’s letters, top at index 0.
• string b: Daniel’s letters, top at index 0.

Returns
– string: the completed string

HackerRank Morgan and a String 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))

Morgan and a String 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);
        }
    }

}

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;
}

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] = '';
        str2[len2] = '';

        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;
}