Problem

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

Pre analysis

Will sort the array to automatically get the count of smaller number from length - index. Also will use dictionary to store the count of each number and return the count from dictionary. An offset is required for repeating numbers

Another solution

//Bruteforce method

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var smallerNumbersThanCurrent = function (nums) {
  let finalCount = [];
  for (let i = 0; i < nums.length; i++) {
    let count = 0;
    for (let j = 0; j < nums.length; j++) {
      if (nums[i] > nums[j]) {
        count += 1;
      }
    }
    finalCount.push(count);
  }
  return finalCount;
};