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2438. Range Product Queries of Powers
Description
Given a positive integer n
, there exists a 0indexed array called powers
, composed of the minimum number of powers of 2
that sum to n
. The array is sorted in nondecreasing order, and there is only one way to form the array.
You are also given a 0indexed 2D integer array queries
, where queries[i] = [left_{i}, right_{i}]
. Each queries[i]
represents a query where you have to find the product of all powers[j]
with left_{i} <= j <= right_{i}
.
Return an array answers
, equal in length to queries
, where answers[i]
is the answer to the i^{th}
query. Since the answer to the i^{th}
query may be too large, each answers[i]
should be returned modulo 10^{9} + 7
.
Example 1:
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 10^{9} + 7 yields the same answer, so [2,4,64] is returned.
Example 2:
Input: n = 2, queries = [[0,0]] Output: [2] Explanation: For n = 2, powers = [2]. The answer to the only query is powers[0] = 2. The answer modulo 10^{9} + 7 is the same, so [2] is returned.
Constraints:
1 <= n <= 10^{9}
1 <= queries.length <= 10^{5}
0 <= start_{i} <= end_{i} < powers.length
Solutions
Solution 1: Bit Manipulation + Simulation
First, we use bit manipulation (lowbit) to get the powers array, and then simulate to get the answer for each query.
The time complexity is $O(n \times \log n)$, ignoring the space consumption of the answer, the space complexity is $O(\log n)$. Here, $n$ is the length of $queries$.

class Solution { private static final int MOD = (int) 1e9 + 7; public int[] productQueries(int n, int[][] queries) { int[] powers = new int[Integer.bitCount(n)]; for (int i = 0; n > 0; ++i) { int x = n & n; powers[i] = x; n = x; } int[] ans = new int[queries.length]; for (int i = 0; i < ans.length; ++i) { long x = 1; int l = queries[i][0], r = queries[i][1]; for (int j = l; j <= r; ++j) { x = (x * powers[j]) % MOD; } ans[i] = (int) x; } return ans; } }

class Solution { public: const int mod = 1e9 + 7; vector<int> productQueries(int n, vector<vector<int>>& queries) { vector<int> powers; while (n) { int x = n & n; powers.emplace_back(x); n = x; } vector<int> ans; for (auto& q : queries) { int l = q[0], r = q[1]; long long x = 1l; for (int j = l; j <= r; ++j) { x = (x * powers[j]) % mod; } ans.emplace_back(x); } return ans; } };

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 y in powers[l : r + 1]: x = (x * y) % mod ans.append(x) return ans

func productQueries(n int, queries [][]int) []int { var mod int = 1e9 + 7 powers := []int{} for n > 0 { x := n & n powers = append(powers, x) n = x } ans := make([]int, len(queries)) for i, q := range queries { l, r := q[0], q[1] x := 1 for _, y := range powers[l : r+1] { x = (x * y) % mod } ans[i] = x } return ans }