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2438. Range Product Queries of Powers

Description

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.

You are also given a 0-indexed 2D integer array queries, where queries[i] = [lefti, righti]. Each queries[i] represents a query where you have to find the product of all powers[j] with lefti <= j <= righti.

Return an array answers, equal in length to queries, where answers[i] is the answer to the ith query. Since the answer to the ith query may be too large, each answers[i] should be returned modulo 109 + 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 109 + 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 109 + 7 is the same, so [2] is returned.

 

Constraints:

  • 1 <= n <= 109
  • 1 <= queries.length <= 105
  • 0 <= starti <= endi < 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 $queries$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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