Codewars: Find the odd int

This problem on codewars asks the following:

Given an array, find the integer that appears an odd number of times. There will always be exactly one integer that appears an odd number of times.

Example: (Input --> Output)

findOdd([20, 1, 1, 2, 2]) = 20 findOdd([5, 4, 3, 2, 1, 5, 4, 3, 2, 10, 10]) = 1 

How to approach it with the PEDAC method:

P: Understand the problem -
We need to identify which number in an array appears an odd number of times. Although each number can appear multiple times, only one will appear an odd number of times.

E: Give Example -
In an array like [8, 8, 7, 7, 7], 7 shows up 3 times (odd), making it our target.
For [1, 1, 2], 2 appears just once (odd), pinpointing it as the solution.

D: What data structure(s) would be needed -
An object seems perfect for mapping each number to its occurrence count within the array.

A: Steps to solve it without language specific information -

- Loop first through the array: This allows me to inspect each number individually.

- If I arrive at a number, add it to a data structure: object

- Check if the number has been encountered before. If not, initialize its count in the object. If the number is already in the object, increment its count

- Loop through the object to inspect counts and return the odd number of counts

C: Code the solution guided by the pseudocode-

- Loop first through the array

for (let i of A) { 

- If I arrive at a number, add it to a data structure: object
- Check if the number has been encountered before. If not, initialize its count in the object. If the number is already in the object, increment its count

 if (!values[i]) { values[i] = 1; } else { values[i] += 1; } } 

- Loop through the object to inspect counts and return the odd number of counts

for (let k in values) { if (values[k] % 2 === 1) { return Number(k); } } 

Putting it all together in the function findOdd:

function findOdd(A) { let values = {}; // Loop first through the array for (let i of A) { // Check if the number has been encountered before if (!values[i]) { values[i] = 1; // If not, initialize its count } else { values[i] += 1; // If it has, increment its count } } // Loop through the object to find and return the number with an odd count for (let k in values) { if (values[k] % 2 === 1) { return Number(k); } } }