1365. How Many Numbers Are Smaller Than the Current Number

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

对数组中每个数统计有多少数小于它，元素取值范围[0, 100]。由于取值范围有限，因此可以统计频数再从头到尾累加遍历一次可得小于等于该数的出现次数，放在数组cnt中。返回时取cnt[nums[i] - 1]就是小于nums[i]所出现的次数和。

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> cnt(101, 0);
        for (const auto num : nums) ++cnt[num];
        for (int i = 1; i <= 100; ++i) cnt[i] += cnt[i - 1];            
        const int N = nums.size();
        vector<int> res(N, 0);
        for (int i = 0; i < N; ++i) res[i] = nums[i] == 0 ? 0 : cnt[nums[i] - 1];
        return res;
    }
};