If you are given two strings s1 and s2 print how many times string s2 occurs in string s1 HackerRank

problem solved and The initial Test cases passed by doing the suggestions given by the commenters:adding break; after "YES" and use substring(i,i+1). However, when I submitted the code after suggested corrections, I got "Time limit exceeded" problem.

public static String twoStrings(String s1, String s2) { String answer = ""; String StringToIterate = ""; String theOtherString = ""; if (s1.length() > s2.length()) { StringToIterate = s2; theOtherString = s1; } else if (s1.length() < s2.length()){ StringToIterate = s1; theOtherString = s2; } else { StringToIterate = s1; theOtherString = s2; } for (int i= 0; i < StringToIterate.length();i++) { String subS = StringToIterate.substring(i,i+1); if (theOtherString.contains(subS)) { answer = "YES"; break; } else { answer = "NO"; } } return answer; } }

Also, I cannot add return inside the for loop/if statement,because the compiler of course wants a return in the method body but outside the for loop. So, any suggestion?

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

Problem solution in Python programming.

#!/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()

Problem solution in Java Programming.

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]!='\0'; j++) { letter[buf[j]-'a'] = 1; } scanf("%s", buf); for (j=0; buf[j]!='\0'; j++) { if (letter[buf[j]-'a']) { printf("YES\n"); break; } } if (buf[j] == '\0') { printf("NO\n"); } } 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("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); });

How To Solve HackerRank’s Plus Minus Code Challenge With JavaScript

Let’s break down the challenge into requirements:

• Link to challenge: HackerRank’s Two Strings Code Challenge
• You will receive two strings, s1 and s2
• s1 and s2 can be a string length of 1 to 100000
• If s1 and s2 have common letters in them return “YES”
• If s1 and s2 do NOT have common letters in them return “NO”

Goal

Write a function or functions that returns “YES” or “NO” if there are common letters between the two strings or not respectively.

Covering Our Bases

Making sure we meet the requirements:

Understanding The Problem

We have two string and we need to traverse each letter to see if at least one letter is the same in each strings. A way that we can loop over things is using arrays, but we first need to convert one of strings to an array.

Converting To Array

Traversing The Array With Filtering

Next, we’ll use filter because instead of map, we want to remove any character that isn’t present in the other string. If there are any left, then we know there are some common letters.

Getting The Result

Lastly we need to return the right message if the length of the array is great than zero.

Solution

The full solution without the comments:

Test Cases

Now let’s validate the code:

Any Feedback?

If you have any tips on how this can be better coded or talk about coding, I would love to talk.

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