1365. 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.

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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).

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Example 2:

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

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Example 3:

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

Constraints:

2 <= nums.length <= 500
0 <= nums[i] <= 100

Solution

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class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] count = new int[101];
        for (int n: nums) {
            count[n]++;
        }
        for (int i = 1; i < count.length; i++) {
            count[i] = count[i] + count[i - 1];
        }
        int[] order = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) continue;
            order[i] = count[nums[i] - 1];
        }
        return order;
    }
}

Solution#

Solution