1365. How Many Numbers Are Smaller Than the Current Number -------E

Q:

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

A:

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> copy = nums;
        sort(copy.begin(), copy.end());
        unordered_map<int, int> myMap;
        for(int i =0;i<copy.size();i++){
            int val = copy[i];
            if( myMap.find(val) == myMap.end()){ // if not found
                myMap[val] = i;
            }
        }
        vector<int> res;
        for(auto k:nums){
            res.push_back(myMap[k]);
        }
        return res;
    }
};