HackerRank Sales by Match | JS Solution

Problem

HackerRank detailed problem description can be found here.

Inputs & Outputs

/*
param {number} n
param {array of numbers} ar

returns {number} the total number of pairs
*/

Test Case

// Example used by HackerRank
n = 9;
ar = [10, 20, 20, 10, 10, 30, 50, 10, 20]

// Visual example
n = 9;
ar = ['blue', 'green', 'green', 'blue', 'blue', 'red', 'purple', 'blue', 'green']

Pseudo Code

1. Let's first, validate the ar array
2. Then, loop over ar and save each number (color) in a counterObj object along with its quantity of occurrences, like so { '10': 1, '20': 1 } or  { 'blue': 1, 'green': 1 }
3. If we see that there are 2 occurrences of a certain number (color) in counterObj, we can increment pairs counter and reset that number (color) value inside counterObj to 0 in order to start counting this number (color) again from scratch
4. Finally, we return pairs

JavaScript Solution

function sockMerchant(n, ar) {
  // Array validation
  if (!ar || ar.length !== n) {
    return 0;
  }

  // Counting pairs
  let pairs = 0;
  let counterObj = {};

  for (let sock of ar) {
    if (!counterObj[sock]) {
      counterObj[sock] = 1;
    } else {
      counterObj[sock] = counterObj[sock] + 1;

      if (counterObj[sock] === 2) {
        pairs++;
        counterObj[sock] = 0;
      }
    }
  }
  return pairs;
}

Here is how our counterObj and counterObj[sock] would look like at the beginning of each for/of loop iteration:

{}
undefined
{ blue: 1 }
undefined
{ blue: 1, green: 1 }
1
{ blue: 1, green: 0 }
1
{ blue: 0, green: 0 }
0
{ blue: 1, green: 0 }
undefined
{ blue: 1, green: 0, red: 1 }
undefined
{ blue: 1, green: 0, red: 1, purple: 1 }
1
{ blue: 0, green: 0, red: 1, purple: 1 }
0
3

Resources:

1. Sales by Match algorithm by HackerRank
2. JavaScript For Of Loop by MDN Web Docs
3. Logical OR (||) by MDN Web Docs
4. Working with objects by MDN Web Docs