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.

 

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

这题一开始，直接爆破，时间复杂度O(N^2)，性能不佳。后来想了想，使用一个map提前存排序后在每个数前面的数字即可，勉强优化到O(N)。

class HowManyNumbersAreSmallerThanTheCurrentNumber : public Solution {
public:
    void Exec() {

    }
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> res(nums.size(), 0);
        for (int i = 0; i < nums.size(); i++) {
            int count = 0;
            for (int j = 0; j < nums.size(); j++) {
                if (i == j) {
                    continue;
                }
                if (nums[i] > nums[j]) {
                    count++;
                }
            }
            res[i] = count;
        }
        return res;
    }
    vector<int> smallerNumbersThanCurrent2(vector<int>& nums) {
        vector<int> res = nums;
        std::sort(res.begin(), res.end());

        unordered_map<int, int> less;

        for (int i = 0; i < res.size(); i++) {
            auto it = less.find(res[i]);
            if (it == less.end()) {
                less[res[i]] = i;
            }
        }
        for (int i = 0; i < nums.size(); i++) {
            res[i] = less[nums[i]];
        }
        return res;
    }
};