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2438. Range Product Queries of Powers

Difficulty Topics

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

Solution

range-product-queries-of-powers.py
class Solution:
    def productQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        M = 10 ** 9 + 7
        res = []
        P = []

        curr = 0
        while 2 ** curr <= n:
            P.append(2 ** curr)
            curr += 1

        A = []
        curr = n
        for p in P[::-1]:
            if curr >= p:
                A.append(p)
                curr -= p

        A.append(1)
        A.reverse()

        for i in range(1, len(A)):
            A[i] *= A[i - 1]

        for left, right in queries:
            x = A[right + 1] // A[left]
            res.append(x % M)

        return res