Fenwick | Analysis , Implementation and Applications of Fenwick Trees | Dataset library

Assuming you know the array offline, you can sort to find the rank of each element, e.g., [8, 4, 7, 0] --> [(8, 3), (4, 1), (7, 2), (0, 0)]. Then initialize a Fenwick tree with all zeros, and to process an element, sum the suffix of the Fenwick tree corresponding to higher ranks and then insert the element at its rank. The evolution of the Fenwick tree is

Your problem is coming from incorrect for loop. When you do following:

If the bit pattern of i is like ...0111, the bit pattern of i+1 will be ...1000. Hence, i & (i+1) means i - (2^{number of trailing ones} - 1) and transforming all trailing 1s to zero. For example if i is even, i & (i+1) will be equal to i. On the other hand, -1 means, transforming all trailing zeros to 1 (including all trailing ones to zeros of the previous step) and transforming the successive 1s to zeros.

By doing the -1 for the next step, i & (i+1) will decrease the i by the 2^{number of trailing 1's} of the previous value of the i.

What is the reason? It helps to show that the time complexity of finding the cumulative sum will be O(log n) in the worst case.

To find the reason behind this computation, you need to focus on each node i will responsible to compute index i to (i - (1 << r)) + 1. And "r represents the last 1 bit set in the index i". To find the total sum corresponding to index i, we need to sum all related values from 0 to i. Beware that we do not need to sum all indices because of the specified property. Hence, we need to sum indices i, i - (1 << r), ... and so on so forth.

arr[i] can be at most 40. We can use this to our advantage. What we need is a segment tree. Each node will hold 41 values (A long long int which represents inversions for this range and a array of size 40 for count of each numbers. A struct will do). How do we merge two children of a node. We know inversions for left child and right child. Also know frequency of each numbers in both of them. Inversion of parent node will be summation of inversions of both children plus number of inversions between left and right child. We can easily find inversions between two children from frequency of numbers. Query can be done in similar way. Complexity O(40*qlog(n))

I read this on Quora. Hope you find it useful.

1. There are things that a segment tree can do but a BIT cannot : A BIT essentially works with cumulative quantities. When the cumulative quantity for interval [i..j] is required, it is found as the difference between cumulative quantities for [1...j] and [1...i-1]. This works only because addition has an inverse operation. You cannot do this if the operation is non-invertible (such as max). On the other hand, every interval on a segment tree can be found as union of disjoint intervals and no inverse operation is required
2. A BIT requires only half as much memory as a segment tree : In cases where you have masochistic memory constraints, you are almost stuck with using a BIT
3. Though BIT and segment tree operations are both O(log(n)), the segment tree operations have a larger constant factor : This should not matter for most cases. But once again, if you have masochistic time constraints, you might want to switch from a segment tree to a BIT. The constant factor might become more of a problem if the BIT/Segment tree is multidimensional.

Here is my implementation of a lower_bound-like function for a Fenwick tree with 0-based indexing:

The key is that you don't do a step of 1 in the loop, but a step of the size x&-x. This is equivalent to going upwards in the tree, to the next relevant node that needs to include the current one and thus gives you a worst case of O(log n).

my solution here:

Idea I got: