# Two Strings

### Problem Statement :

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

For example, the words "a", "and", "art" share the common substring a. The words "be" and "cat" do not share a substring.

Function Description

Complete the function twoStrings in the editor below. It should return a string, either YES or NO based on whether the strings share a common substring.

twoStrings has the following parameter(s):

s1, s2: two strings to analyze .

Input Format

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

The following p pairs of lines are as follows:

1. The first line contains string  s1
2. 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 <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
int k;
scanf("%d",&k);
while(k>0)
{
char str,btr;
int cn1,cn2,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;
}```
```

```                        ```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;
}```
```

```                        ```Solution in Java :

In Java :

import java.io.*;
import java.util.*;

public class TwoStrings {

PrintWriter out;
StringTokenizer st;
boolean eof;

void solve() throws IOException {
int t = nextInt();
outer: while (t-- > 0) {
char[] a = nextToken().toCharArray();
char[] b = nextToken().toCharArray();
boolean[] has = new boolean;
for (char c : a) {
has[c - 'a'] = true;
}
for (char c : b) {
if (has[c - 'a']) {
out.println("YES");
continue outer;
}
}
out.println("NO");
}
}

TwoStrings() throws IOException {
out = new PrintWriter(System.out);
solve();
out.close();
}

public static void main(String[] args) throws IOException {
new TwoStrings();
}

String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
} catch (Exception e) {
eof = true;
return null;
}
}
return st.nextToken();
}

String nextString() {
try {
} catch (IOException e) {
eof = true;
return null;
}
}

int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}

long nextLong() throws IOException {
return Long.parseLong(nextToken());
}

double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
}```
```

```                        ```Solution in Python :

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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