# 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))

{"mode":"full","isActive":false}

## 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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