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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 $queries$.

  • 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
    }
    
  • function productQueries(n: number, queries: number[][]): number[] {
        const powers: number[] = [];
        while (n > 0) {
            const x = n & -n;
            powers.push(x);
            n -= x;
        }
        const mod = 1_000_000_007;
        const ans: number[] = [];
        for (const [l, r] of queries) {
            let x = 1;
            for (let j = l; j <= r; j++) {
                x = (x * powers[j]) % mod;
            }
            ans.push(x);
        }
        return ans;
    }
    
    
  • impl Solution {
        pub fn product_queries(mut n: i32, queries: Vec<Vec<i32>>) -> Vec<i32> {
            let mut powers = Vec::new();
            while n > 0 {
                let x = n & -n;
                powers.push(x);
                n -= x;
            }
            let modulo = 1_000_000_007;
            let mut ans = Vec::with_capacity(queries.len());
            for q in queries {
                let l = q[0] as usize;
                let r = q[1] as usize;
                let mut x: i64 = 1;
                for j in l..=r {
                    x = x * powers[j] as i64 % modulo;
                }
                ans.push(x as i32);
            }
            ans
        }
    }
    
    

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