Web# Count the smaller elements after the number. ans, bit= [0] * len (nums), BIT (len (nums) + 1) for i in reversed (xrange (len (nums))): ans [i] = bit.query (places [i]) bit.add (places [i] + 1, 1) return ans # Time: O (nlogn) # Space: O (n) # BST solution. class Solution3 (object): def countSmaller (self, nums): """ :type nums: List [int] WebDec 7, 2024 · But for this coding problem, 2 is smaller than 9, and to the right of 9. We know that 2 is also smaller than, and to the right of, every element in left sub-array after 9. If …
Count of Smaller Numbers After Self - LeetCode
WebMinimum Number of Flips to Convert Binary Matrix to Zero Matrix 1283. Find the Smallest Divisor Given a Threshold 1282. Group the People Given the Group Size They Belong … Web// Count the smaller elements after the number. BIT bit (size (nums)); vector result (size (nums)); for (int i = size (nums) - 1; i >= 0; --i) { result [i] = bit.query (lookup [i] - 1); bit.add (lookup [i], 1); } return result; } private: class BIT { public: BIT (int n) : bit_ (n + 1) { // 0-indexed } void add (int i, int val) { ++i; douglas county ks voting
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WebJul 23, 2024 · The counts array has the property where counts [i] is the number of smaller elements to the right of nums [i]. Example 1: Input: nums = [5,2,6,1] Output: [2,1,1,0] … WebMay 27, 2024 · Count of Smaller Numbers After Self in C Count of Smaller Numbers After Self in C++ C++ Server Side Programming Programming Suppose we have an … WebThe 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). douglas county labor contract oregon