I guess you are trying to find count of inversions in array which is if i < j and a[i] > a[j] This can be done using merge sort along with a simple modification. Link to understand sol → Reply tdpencil 11 months ago, # | 0 Statement's really unclear, can you clarify ...
1. Merge sort array A and create a copy (array B) 2. Take A[1] and find its position in sorted array B via a binary search. The number of inversions for this element will be one less than the index number of its position in B since every lower number that appears after the first...
Let A[0...n - 1] be an array of n distinct positive integers. If i < j and A[i] > A[j] then the pair (i, j) is called an inversion of A. Given n and an array A your task is to find the number of inversions of A. Input The first line contains t, the number of te...
elements in the array. · 50% test cases guarantee that Output Output a single integer which is the number of pairs of significant inversions. Sample Inout 6 13 8 5 3 2 1 Sample Output 6 题意与例题1相同,只不过增加一个限定条件:
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the array is. the more inversions, the more steps we need to take to sort the array by swapping adjacent elements. 2.1. example let’s look at a simple example. suppose we have an array: int[] array = {3, 1, 2}; in this array, there are two inversions: (3, 1) because 3 ...
It appears that the statement is actually asking for the number of permutations of 1... n with k inversions. If In(t) is the number of permutations of size n with t inversions, then . This recurrence can be calculated in O(nt) using prefix sums on In. → Reply Codeforces...
When k=2k=2, the problem simplifies to counting inversions, solvable directly using a Fenwick Tree. For k=3k=3, one can consider a Divide and Conquer (D&C) approach, achieving a complexity of O(nlog2n)O(nlog2n), which is excellent. Alternatively, a K-D Tree approach achieves O...
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