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.


   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;
        char str[110000],btr[110000];
        int cn1[500],cn2[500],i,j,t,a,b;
    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.util.*;

public class TwoStrings {

	BufferedReader br;
	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[26];
			for (char c : a) {
				has[c - 'a'] = true;
			for (char c : b) {
				if (has[c - 'a']) {
					continue outer;

	TwoStrings() throws IOException {
		br = new BufferedReader(new InputStreamReader(;
		out = new PrintWriter(System.out);

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

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

	String nextString() {
		try {
			return br.readLine();
		} 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():

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