``````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``````

``````class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
int n = nums.size();
vector<int> res;
for(int i = 0;i < n ;i++)
{
int count = 0;
for(int j = 0;j < n ;j++)
{
if(nums[j] < nums[i])
count++;
}
res.push_back(count);
}

return res;
}
};``````

``````Runtime: 44 ms, faster than 46.05% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 10.3 MB, less than 53.70% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.``````

``````class Solution {
public:
vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
int n = nums.size();
vector<int> res(n,0);

int freq[102] = {0};
//we take nums[i] in nums[i] + 1 position,so get freq sum easy
for(int i = 0;i < n;i++)
freq[nums[i] + 1]++;

for(int i = 1;i <102;i++)
freq[i] += freq[i - 1];

//get the res from freq array
for(int i = 0;i < n;i++)
res[i] = freq[nums[i]];

return res;
}
};``````

``````Runtime: 4 ms, faster than 99.70% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.
Memory Usage: 10.5 MB, less than 26.46% of C++ online submissions for How Many Numbers Are Smaller Than the Current Number.``````