Leetcode problem 40 - How Many Numbers Are Smaller Than the Current Number

Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j’s such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3] Output: [4,0,1,1,3] Explanation: For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3). For nums[1]=1 does not exist any smaller number than it. For nums[2]=2 there exist one smaller number than it (1). For nums[3]=2 there exist one smaller number than it (1). For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2). Example 2:

Input: nums = [6,5,4,8] Output: [2,1,0,3] Example 3:

Input: nums = [7,7,7,7] Output: [0,0,0,0]

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var smallerNumbersThanCurrent = function (nums) {
  // Create a variable to store key and value.
  let hash = new Map();
  // Create a variable with value sorted nums array.
  let sorted = Array.from(nums).sort((a, b) => a - b);

  // Loop through all elements in sorted array and store a pair with a value in nums array as a key and an index as a value.
  sorted.forEach((num, i) => (hash.has(num) ? null : hash.set(num, i)));

  // return hash value associated with a key
  return nums.map((num) => hash.get(num));
};

Time complexity: O(n log(n)) Sorting the array ‘nums’ using the sort() has a time complexity O(n log(n)), where ‘n’ is the number of elements in nums array.