[Tutorial] Searching Binary Indexed Tree in O(log(N)) using Binary Lifting

Revision en5, by sdnr1, 2018-08-21 18:19:38

NOTE : Knowledge of Binary Indexed Trees is a prerequisite.

Problem Statement

Assume we need to solve the following problem. We have an array, A of length N with only non-negative values. We want to perform the following operations on this array:

  1. Update value at a given position

  2. Compute prefix sum of A upto i, i ≤ N

  3. Search for a prefix sum (something like a lower_bound in the prefix sums array of A)

Basic Solution

Seeing such a problem we might think of using a Binary Indexed Tree (BIT) and implementing a binary search for type 3 operation. Its easy to see that binary search is possible here because prefix sums array is monotonic (only non-negative values in A).

The only issue with this is that binary search in a BIT has time complexity of O(log2(N)) (other operations can be done in O(log(N))). Even though this is naive,


Most of the times this would be fast enough (because of small constant of above technique). But if the time limit is very tight, we will need something faster. Also we must note that there are other techniques like segment trees, policy based data structures, treaps, etc. which can perform operation 3 in O(log(N)). But they are harder to implement and have a high constant factor associated with their time complexity due to which they might be even slower than O(log2(N)) of BIT.

Hence we need an efficient searching method in BIT itself.

Efficient Solution

We will make use of binary lifting to achieve O(log(N)) (well I actually do not know if this technique has a name but I am calling it binary lifting because the algorithm is similar to binary lifting in sparse tables).

What is binary lifting?

In binary lifting, a value is increased (or lifted) by powers of 2, starting with the highest possible power of 2, 2ceil(log(N)), down to the lowest power, 20.

So, we initialize the target position, pos = 0 and also maintain the corresponding prefix sum. We increase (or lift) pos when the v

Implementation :

// This is equivalent to calculating lower_bound on prefix sums array
// LOGN = log(N)

int bit[N]; // BIT array

int bit_search(int v)
	int sum = 0;
	int pos = 0;
	for(int i=LOGN; i>=0; i--)
		if(pos + (1 << i) < N and sum + bit[pos + (1 << i)] < v)
			sum += bit[pos + (1 << i)];
			pos += (1 << i);

	return pos + 1;

Taking this forward

You must have noted that proof of correctness of this approach relies on the property of the prefix sums array that it monotonic. This means that this approach can be used for with any operation that maintains the monotonicity of the prefix array, like multiplication of positive numbers, etc.

Tags #binary-lifting, #binary search, binary indexed tree, bit


  Rev. Lang. By When Δ Comment
en19 English sdnr1 2018-08-22 15:35:12 132
en18 English sdnr1 2018-08-22 13:14:18 53
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en16 English sdnr1 2018-08-22 12:52:42 174
en15 English sdnr1 2018-08-22 12:50:46 390 (published)
en14 English sdnr1 2018-08-22 11:41:30 953 (saved to drafts)
en13 English MikeMirzayanov 2018-08-22 11:10:35 7
en12 English sdnr1 2018-08-22 11:05:32 20
en11 English sdnr1 2018-08-22 10:48:05 25 (published)
en10 English sdnr1 2018-08-22 10:46:44 36 (saved to drafts)
en9 English sdnr1 2018-08-21 22:37:35 39
en8 English sdnr1 2018-08-21 22:28:02 1056 Added some more insight for better understanding of the algorithm
en7 English sdnr1 2018-08-21 21:57:33 96
en6 English sdnr1 2018-08-21 21:25:18 988 (published)
en5 English sdnr1 2018-08-21 18:19:38 867
en4 English sdnr1 2018-08-21 10:48:16 321
en3 English sdnr1 2018-08-21 10:12:15 228
en2 English sdnr1 2018-08-21 10:07:53 214
en1 English sdnr1 2018-08-21 09:55:51 1869 Initial revision (saved to drafts)