HackerRank Two Strings problem solution

In this HackerRank Two Strings Interview preparation kit problem solution, given two strings, determine if they share a common substring. A substring may be as small as one character.

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

HackerRank Two Strings problem solution in Python.

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the twoStrings function below.
def twoStrings(s1, s2):
    if set(s1) & set(s2):
        return "YES"
    else: 
        return "NO"

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    q = int(input())

    for q_itr in range(q):
        s1 = input()

        s2 = input()

        result = twoStrings(s1, s2)

        fptr.write(result + 'n')

    fptr.close()

Two Strings problem solution in Java.

import java.io.*;
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']) {
					out.println("YES");
					continue outer;
				}
			}
			out.println("NO");
		}
	}

	TwoStrings() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		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 {
				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());
	}
}

Problem solution in C++ programming.

#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; 
}

Problem solution in C programming.

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

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int t;
    scanf("%d", &t);
    int i, j;
    char letter[26];
    char buf[100001];
    for (i=0; i<t; i++) {
        memset(letter, 0, 26);
        scanf("%s", buf);
        for (j=0; buf[j]!=''; j++) {
            letter[buf[j]-'a'] = 1;
        }
        scanf("%s", buf);
        for (j=0; buf[j]!=''; j++) {
            if (letter[buf[j]-'a']) {
                printf("YESn");
                break;
            }
        }
        if (buf[j] == '') {
            printf("NOn");
        }
    }
    return 0;
}

Problem solution in JavaScript programming.

function containsCommonSubstring(a,b) {
    // Since a one character common substring is still a substring, we can just check for
    // a character in common.  A map should be easy way to do that.
    var map = {};
    for (var i = 0; i < a.length; i++) {
        // We could count it, but just having an entry should be sufficient.  Seems like a boolean.
        map[a[i]] = true;
    }
    for (var i = 0; i < b.length; i++) {
        if (map[b[i]]) return true;
    }
    return false;
}

function processData(input) {
    var lines = input.split("n");
    var T = lines[0];
    for (var i = 0; i < T; i++) {
        var a = lines[2*i+1];
        var b = lines[2*i+2];
        if (containsCommonSubstring(a,b)) {
            process.stdout.write("YESn");
        } else {
            process.stdout.write("NOn");
        }        
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});