# Count Of Smaller Numbers After Self Problem

## Description

LeetCode Problem 315.

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i] is the number of smaller elements to the right of nums[i].

Example 1:

``````1
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Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.
``````

Example 2:

``````1
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Input: nums = [-1]
Output: [0]
``````

Example 3:

``````1
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Input: nums = [-1,-1]
Output: [0,0]
``````

Constraints:

• 1 <= nums.length <= 10^5
• -10^4 <= nums[i] <= 10^4

## Sample C++ Code

``````1
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class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
vector<int> counts;
vector<int> sorted;

int len = nums.size();
int left, right, mid, pos;

if (len == 0)
return counts;

reverse(nums.begin(), nums.end());

sorted.push_back(nums[0]);
counts.push_back(0);

for (int i = 1; i < len; i ++) {
left = 0;
right = sorted.size();
pos = left;
while (left < right) {
mid = (left + right) / 2;
if (sorted[mid] < nums[i]) {
left = mid + 1;
pos = left;
} else {
right = mid;
pos = right;
}
}
sorted.insert(sorted.begin() + pos , nums[i]);
counts.push_back(pos);
}

reverse(counts.begin(), counts.end());

return counts;

}
};
``````