What is the time complexity of Number of Integers With Popcount-Depth Equal to K II?

The optimal solution for Number of Integers With Popcount-Depth Equal to K II runs in O(n + q log n) time and uses O(n) space. Precomputation takes O(n), each query takes O(log n) for Fenwick tree operations.

What is the brute force solution for Number of Integers With Popcount-Depth Equal to K II?

The brute force approach for Number of Integers With Popcount-Depth Equal to K II is Brute Force, with O(n²) time complexity and O(1) space complexity. Calculate the popcount-depth for each number in the array for every query. This involves repeatedly calculating the popcount until reaching 1. The optimal Optimal Solution approach improves this to O(n + q log n).

What algorithm or data structure is used to solve Number of Integers With Popcount-Depth Equal to K II?

Number of Integers With Popcount-Depth Equal to K II uses the following concepts: Array, Divide and Conquer, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Integers With Popcount-Depth Equal to K II?

Explain your thought process clearly. Discuss trade-offs between brute force and optimal solutions. Practice coding Fenwick trees and similar data structures.

What are common mistakes when solving Number of Integers With Popcount-Depth Equal to K II?

Not updating the Fenwick tree correctly after an update. Forgetting to handle edge cases in range queries.

#3624

Number of Integers With Popcount-Depth Equal to K II

Hard

ArrayDivide and ConquerBinary Indexed TreeSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q log n)
Space	O(1)	O(n)

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Intuition

Time O(n + q log n)Space O(n)

Precompute the popcount-depth for all numbers and use Fenwick trees to efficiently answer range queries.

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Algorithm

3 steps

1Step 1: Precompute the popcount-depth for each element in nums and store counts in Fenwick trees for depths 0 to 5.
2Step 2: For each query of type [1, l, r, k], use the Fenwick tree to get the count of indices with depth k in the range [l, r].
3Step 3: For updates, adjust the Fenwick tree based on the new depth of the updated number.

solution.py8 lines

1def popcount(n): return bin(n).count('1')
2def popcount_depth(x): d = 0; while x > 1: x = popcount(x); d += 1; return d
3class FenwickTree: def __init__(self, size): self.size = size; self.tree = [0] * (size + 1)
4 def update(self, idx, delta): while idx <= self.size: self.tree[idx] += delta; idx += idx & -idx
5 def query(self, idx): total = 0; while idx > 0: total += self.tree[idx]; idx -= idx & -idx; return total
6 def range_query(self, left, right): return self.query(right) - self.query(left - 1)
7def process_queries(nums, queries): n = len(nums); depths = [popcount_depth(num) for num in nums]; fenwick = [FenwickTree(n) for _ in range(6)]; for i in range(n): fenwick[depths[i]].update(i + 1, 1); results = []
8 for query in queries: if query[0] == 1: results.append(fenwick[query[3]].range_query(query[1] + 1, query[2] + 1)); else: old_depth = depths[query[1]]; new_depth = popcount_depth(query[2]); depths[query[1]] = new_depth; fenwick[old_depth].update(query[1] + 1, -1); fenwick[new_depth].update(query[1] + 1, 1); return results

ℹ

Complexity note: Precomputation takes O(n), each query takes O(log n) for Fenwick tree operations.

1Precomputing values can save time during queries.
2Using data structures like Fenwick trees allows efficient range queries.

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