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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 $queries$.

• Java
• C++
• Python
• Go

• 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
  }
  

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