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: Bit Manipulation + Prefix Sum

• Time complexity: O(logn+queries.length)
• Space complexity: O(logn+queries.length)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} queries
 * @return {number[]}
 */
const productQueries = function (n, queries) {
  const MODULO = BigInt(10 ** 9 + 7);
  const powers = [];
  const bits = Math.ceil(Math.log2(n));

  for (let index = 0; index <= bits; index++) {
    if ((n >> index) & 1) {
      const power = BigInt(1 << index);

      powers.push(power);
    }
  }

  const len = powers.length;
  const prefixProd = Array.from({ length: len + 1 }, () => 1n);

  for (let index = 1; index <= len; index++) {
    prefixProd[index] = prefixProd[index - 1] * powers[index - 1];
  }

  return queries.map(([left, right]) => {
    const product = (prefixProd[right + 1] / prefixProd[left]) % MODULO;

    return Number(product);
  });
};