leetcode/Easy/Array/1365.how-many-numbers-are-smaller-than-the-current-number.js

/*
 * 1365. How Many Numbers Are Smaller Than the Current Number
 * Difficulty: Easy
 * https://leetcode.com/problems/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]
 *
 *
 *
 * Constraints:
 *
 * 	• 2 <= nums.length <= 500
 *
 * 	• 0 <= nums[i] <= 100
 */

/**
 * @param {number[]} nums
 * @return {number[]}
 */

var smallerNumbersThanCurrent = function (nums) {
  // Step 1: Begin by initializing a [Frequency Map]()
  const freq = {};
  for (let n of nums) freq[n] = (freq[n] || 0) + 1;

  // Step 2: Sort the numbers by ascending order
  const sorted = Object.keys(freq).sort((a, b) => a - b);

  // Step 3: Init a count of numbers smaller than the active number
  let count = 0;

  // Step 4: Init a map to track number of values smaller for each number
  const smaller = {};

  // Step 5: Iterate over the sorted list
  for (let num of sorted) {
    // Set count for active number
    smaller[num] = count;

    // Update the count by frequency
    count += freq[num];
  }

  // Step 6: Use original list and find number of smaller values than it
  return nums.map((n) => smaller[n]);
};