Count of Smaller Numbers After Self
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] . 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. Solution : class Solution { public List<Integer> countSmaller ( int [] nums) { List<Integer> intList = new ArrayList<Integer>(nums. length ) ; int i , j ; int n = nums. length ; int low[] = new int [n] ; // initialize all the counts in countSmaller array as 0 for (i = 0 ; i < n ; i++) low[i] = 0 ; for (i = 0 ; i < n ; i++) { for (j = i + 1 ; j < n ; j++) { ...