README.md

2438. Range Product Queries of Powers

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 (Rust)

1. Solution

impl Solution {
    pub fn product_queries(n: i32, queries: Vec<Vec<i32>>) -> Vec<i32> {
        let mut powers = vec![];
        let mut answers = vec![1_i64; queries.len()];

        for i in 0..30 {
            if n & (1 << i) != 0 {
                powers.push(1 << i);
            }
        }

        for i in 0..queries.len() {
            for j in queries[i][0] as usize..=queries[i][1] as usize {
                answers[i] = (answers[i] * powers[j]) % 1_000_000_007;
            }
        }

        answers.into_iter().map(|x| x as i32).collect()
    }
}