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

Hints:
Brute force for each array element.
In order to improve the time complexity, we can sort the array and get the answer for each array element.

We can solve this problem using bruteforce algorithm or the binary search.

Bruteforce Algorithm to Count the Elements smaller than Current

The most intuitive solution is to bruteforce each number and then count the number of elements smaller than the current. This requires O(N^2) square complexity.

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class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> r;
        for (const auto &n: nums) {
            int cnt = 0;
            for (const auto &m: nums) {
                if (m < n) cnt ++;
            }
            r.push_back(cnt);
        }
        return r;
    }
};

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> r;
        for (const auto &n: nums) {
            int cnt = 0;
            for (const auto &m: nums) {
                if (m < n) cnt ++;
            }
            r.push_back(cnt);
        }
        return r;
    }
};

Binary Search Algorithm to Count the Numbers Smaller than Current

To improve the solution, we can sort the numbers, then we can use binary search Algorithm to find the lower bound of the number. This reduces the complexity to O(N LogN).

In C++, we can use the std::lower_bound to return the first position (iterator) of the current number in the sorted array.

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class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> r;
        vector<int> data(begin(nums), end(nums));
        sort(begin(data), end(data));
        for (const auto &n: nums) {
            int cnt = lower_bound(begin(data), end(data), n) - begin(data);
            r.push_back(cnt);
        }
        return r;
    }
};

class Solution {
public:
    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
        vector<int> r;
        vector<int> data(begin(nums), end(nums));
        sort(begin(data), end(data));
        for (const auto &n: nums) {
            int cnt = lower_bound(begin(data), end(data), n) - begin(data);
            r.push_back(cnt);
        }
        return r;
    }
};

–EOF (The Ultimate Computing & Technology Blog) —