HackerRank Two Strings Solution

#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;
}

#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; 
}

import java.util.Scanner;

public class Solution {
	public static void main(String[] args) throws Exception {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		testCase:
		for(int i = 0; i < N; i++) {
			String left = in.next();
			String right = in.next();
			boolean[] leftC = new boolean[26];
			for(int j = 0; j < left.length(); j++) {
				leftC[left.charAt(j) - 'a'] = true;
			}
			for(int k = 0; k < right.length(); k++) {
				if(leftC[right.charAt(k) - 'a']) {
					System.out.println("YES");
					continue testCase;
				}
			}
			System.out.println("NO");
		}
	}
}

t=input()
for j in xrange(t):
  a=set(raw_input().strip())
  b=set(raw_input().strip())
  if len(a.intersection(b))>0:
    print "YES"
  else:
    print "NO"

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("YES\n");
        } else {
            process.stdout.write("NO\n");
        }        
    }
} 

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

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

  object Solution {
    def main(args : Array[String]): Unit = {
      val input = scala.io.Source.stdin.getLines drop 1

      input.grouped(2).foreach { twoLines =>
        val List(str1,str2) = twoLines.toList
        val commonChars = str1.toSet.intersect(str2.toSet)
        if (commonChars.isEmpty) println("NO")
        else println("YES")
      }

    }
  }

const fi='';
      fo='';
var a,b:string;
    t:byte;
    f,g:text;
    i:longint;
function kt(a,b:string):boolean;
var j,k:longint;
begin
         kt:=false;
         for j:=1 to length(a) do
          for k:=1 to length(b) do
           if a[j]=b[k] then exit(true);
end;
begin
assign(f,fi);reset(f);
readln(f,t);
assign(g,fo);rewrite(g);
for i:=1 to t do
begin
        readln(f,a);
        readln(f,b);
        if kt(a,b) then writeln(g,'YES') else writeln(g,'NO');

end;
close(g);
close(f);
end.