Two Strings

Problem Statement :

```Given two strings, determine if they share a common substring. A substring may be as small as one character.

Example

s1 = 'and'
s2  = 'art'

These share the common substring a.

s1 = 'be'
s2 = 'cat'

These do not share a substring.

Function Description

Complete the function twoStrings in the editor below.

twoStrings has the following parameter(s):

string s1: a string
string s2: another string

Returns

string: either YES or NO

Input Format

The first line contains a single integer p, the number of test cases.

The following p pairs of lines are as follows:

The first line contains string s1.
The second line contains string s2.

Constraints

s1 and s2 consist of characters in the range ascii[a-z].
1  <=  p  <= 10
1  <=  | s1 |, | s2 |  <= 10^5

Output Format

For each pair of strings, return YES or NO.```

Solution :

```                            ```Solution in C :

In  C++  :

#include <bits/stdc++.h>
using namespace std;
int main()
{
int a; cin >> a;
for (int g=0;g<a; g++)
{
string b,c; cin >> b >> c; map <char,int> k;
for (int y=0;y<b.length(); y++) k[b[y]]=1; int counter=0;
for (int y=0;y<c.length(); y++)
{
if (k[c[y]]) counter=1;
}
if (counter) cout << "YES" << '\n';
else cout << "NO" << '\n';
}return 0;
}

In   Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-- > 0){
char[] s1 = sc.next().toCharArray();
char[] s2 = sc.next().toCharArray();
int[] chs = new int[1000];
for(int i = 0; i < s1.length; i++){
chs[s1[i]]++;
}
String result = "NO";
for(int i = 0; i < s2.length; i++){
if(chs[s2[i]]>0){
result = "YES";
break;
}
}
System.out.println(result);
}
}
}

In   C  :

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

int main() {
int k;
scanf("%d",&k);
while(k>0)
{
char str[110000],btr[110000];
int cn1[500],cn2[500],i,j,t,a,b;
for(i=0;i<500;i++)
{
cn1[i]=0;
cn2[i]=0;
}
scanf("%s",str);
scanf("%s",btr);
a=strlen(str);
b=strlen(btr);
for(i=0;i<a;i++)
{
cn1[str[i]]++;
}
for(i=0;i<b;i++)
{
cn2[btr[i]]++;
}
t=0;
for(i=0;i<500;i++)
{
if(cn1[i]*cn2[i]>0)
{
t=1;
break;
}
}
if(t)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
k--;
}
return 0;
}

In  Python3  :

a = int(input())
for i in range(a):
i = (input())
j = (input())
setx = set([a for a in i])
sety = set([y for y in j])
if setx.intersection(sety) == set():
print("NO")
else:
print("YES")```
```

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