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Leetcode #2438: Range Product Queries of Powers

In this guide, we solve Leetcode #2438 Range Product Queries of Powers in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

Given a positive integer n, there exists a 0-indexed array called powers, composed of the minimum number of powers of 2 that sum to n. The array is sorted in non-decreasing order, and there is only one way to form the array.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Bit Manipulation, Array, Prefix Sum

Intuition

Range queries become simple once we precompute cumulative sums.

We can transform subarray conditions into prefix comparisons.

Approach

Compute prefix sums and use a map to find matching prefixes.

This avoids nested loops while keeping the logic clear.

Steps:

  • Compute prefix sums.
  • Use a map to find valid ranges.
  • Update the answer.

Example

Input: n = 15, queries = [[0,1],[2,2],[0,3]] Output: [2,4,64] Explanation: For n = 15, powers = [1,2,4,8]. It can be shown that powers cannot be a smaller size. Answer to 1st query: powers[0] * powers[1] = 1 * 2 = 2. Answer to 2nd query: powers[2] = 4. Answer to 3rd query: powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64. Each answer modulo 109 + 7 yields the same answer, so [2,4,64] is returned.

Python Solution

class Solution: def productQueries(self, n: int, queries: List[List[int]]) -> List[int]: powers = [] while n: x = n & -n powers.append(x) n -= x mod = 10**9 + 7 ans = [] for l, r in queries: x = 1 for i in range(l, r + 1): x = x * powers[i] % mod ans.append(x) return ans

Complexity

The time complexity is O(m×log⁡n)O(m \times \log n)O(m×logn), where mmm is the length of the array queries\textit{queries}queries. The space complexity is O(log⁡n)O(\log n)O(logn).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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