Description

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:

Given nums = [5, 2, 6, 1]

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.

Return the array [2, 1, 1, 0].

Solution

Binary search, time O(n log n), space O(n)

这道题可以用Binary segment tree, Binary index tree, Binary search tree去解（which 我还不会写）。但是也可以用一个简简单单的二分查找来做。

从后向前遍历，同时维护一个已访问元素组成的sorted list，每次遍历到一个新元素是，在sorted list中查询其应该插入的位置，这个位置就是所求的smaller elements count。

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        List<Integer> sorted = new ArrayList<>();
        int n = nums.length;
        Integer[] res = new Integer[n];
        
        for (int i = n - 1; i >= 0; --i) {
            int index = findInsertPos(sorted, nums[i]);
            res[i] = index;
            sorted.add(index, nums[i]);
        }
        
        return Arrays.asList(res);
    }
    
    public int findInsertPos(List<Integer> list, int target) {
        int left = 0;
        int right = list.size() - 1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (list.get(mid) < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return left;
    }
}