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 <= 5000 <= nums[i] <= 100class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
map<int,int> m;
for(auto e:nums)
m[e]++;
int prev=0;
for (auto e:m){
int ct = e.second;
m[e.first]=prev;
prev+=ct;
}
for(int i=0; i<nums.size(); i++)
nums[i] = m[nums[i]];
return nums;
}
};