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 :



title-img 


                            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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