347. Top k frequent elements

The trick for this problem is to make an object that has the count of each number as it occurs in the nums array. Then, you convert the object to an array with Object.entries, sorting each entry in descending order by value. Next, you slice off the elements between 0 and k, then lastly use map to return an array of the keys from result array derived from Object.entries.

/*
Given an integer array nums and an integer k, return the k most frequent 
elements. You may return the answer in any order.
*/

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number[]}
 */
var topKFrequent = function (nums, k) {
  // declare cache, an object to hold values and their count.
  const cache = {};
  // iterate over nums, if element is not in cache, add it as key and value at 1, or add 1
  for (let i = 0; i < nums.length; i++) {
    cache[nums[i]] ? (cache[nums[i]] += 1) : (cache[nums[i]] = 1);
  }
  // result is assigned the value of the entries on cache, sorted in descending order by value,
  // then the k elements are sliced, lastly, map is applied to return an array of just the keys
  const result = Object.entries(cache)
    .sort((a, b) => b[1] - a[1])
    .slice(0, k)
    .map((entry) => entry[0]);
  // return result
  return result;
};

console.log(topKFrequent([1, 1, 1, 2, 2, 3], 2)); // [ '1', '2' ]
console.log(topKFrequent([1], 1)); // [ '1' ]
console.log(topKFrequent([1, 2], 2)); // [ '1', '2' ]