How To Solve The Two Strings Code Challenge

How To Solve HackerRank’s Two Strings Code Challenge With JavaScript

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

Problem

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”

// Example 1
// s1 = hello
// s2 = world
// Result, (there is "l" and "o" that are common)
"YES"
// Example 2
// s1 = hi
// s2 = world
// Result, (there are no common letters)
"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:

function twoStrings(s1, s2) {
     let result = "NO";
     const s1len = s1.length;
     const s2len = s2.length;
     if (s1len >= 1
          && s1len <= 100000
          && s2len >= 1
          && s2len <= 100000) {
          // continue with code
     } 
     return result;
} 

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

function twoStrings(s1, s2) {
     let result = "NO";
     const s1len = s1.length;
     const s2len = s2.length;
     if (s1len >= 1
          && s1len <= 100000
          && s2len >= 1
          && s2len <= 100000) {
          
          result = s1.split('');
          // hello > ['h', 'e', 'l', 'l', 'o']
     }
     return result;
}

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.

function twoStrings(s1, s2) {
     let result = "NO";
     const s1len = s1.length;
     const s2len = s2.length;
     if (s1len >= 1
          && s1len <= 100000
          && s2len >= 1
          && s2len <= 100000) {
          
          result = s1.split('')
                   .filter((el, key) => s2.indexOf(el) > -1);
          // s2 = "world"
          // s1 = "hello" > ['h', 'e', 'l', 'l', 'o']
          // result = ["l", "l", "o"]  
     }
     return result;
}

Getting The Result

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

function twoStrings(s1, s2) {
     let result = "NO";
     const s1len = s1.length;
     const s2len = s2.length;
     if (s1len >= 1
          && s1len <= 100000
          && s2len >= 1
          && s2len <= 100000) {
          
          result = (s1.split('')
                   .filter((el, key) => s2.indexOf(el) > -1)
                   .length > 0) ? "YES" : "NO";
          // s2 = "world"
          // s1 = "hello" > ['h', 'e', 'l', 'l', 'o']
          // result = ["l", "l", "o"] 
          // result = "YES"  
     }
     return result;
}

Solution

The full solution without the comments:

function twoStrings(s1, s2) {
     let result = "NO";
     const s1len = s1.length;
     const s2len = s2.length;
     if (s1len >= 1
          && s1len <= 100000
          && s2len >= 1
          && s2len <= 100000) {
          
          result = (s1.split('')
                   .filter((el, key) => s2.indexOf(el) > -1)
                   .length > 0) ? "YES" : "NO";
     }
     return result;
}

Test Cases

Now let’s validate the code:

// "hello", "world", Expected "YES"
// "hi", "world", Expected "NO"
// "wouldyoulikefries", "abcabcabcabcabcabc", Expected "NO" 
// "hackerrankcommunity", "cdecdecdecde", Expected "YES"
twoStrings("hello", "world"); // "YES" ✅
twoStrings("hi", "world"); // "NO" ✅
twoStrings("wouldyoulikefries", "abcabcabcabcabcabc"); // "NO" ✅
twoStrings("hackerrankcommunity", "cdecdecdecde"); // "YES" ✅

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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