Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 11

Description

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

Solution

const int MOD = 1e9 + 7;
constexpr int qmul(long long a, long long b, int m) {
  return a * b % m;
}
constexpr int qpow(int b, int e, int m) {
  int result = 1;
  while(e) {
    if(e & 1) {
      result = qmul(result, b, m);
    }
    b = qmul(b, b, m);
    e >>= 1;
  }
  return result;
}
constexpr int inv(int a) {
  return qpow(a, MOD - 2, MOD);
}

class Solution {
  const int MOD = 1e9 + 7;
public:
  vector<int> productQueries(int n, vector<vector<int>>& queries) {
    vector<int> powers(1);
    for(int i = 0; i < 30; ++i) {
      if((1 << i) & n) powers.push_back(i);
    }
    int len = powers.size();
    for(int i = 1; i < len; ++i) {
      powers[i] += powers[i - 1];
    }
    int qLen = queries.size();
    vector<int> answer(qLen);
    for(int i = 0; i < qLen; ++i) {
      answer[i] = qpow(2, powers[queries[i][1] + 1] - powers[queries[i][0]], MOD);
    }
    return answer;
  }
};

// Accepted
// 70/70 cases passed (8 ms)
// Your runtime beats 90.43 % of cpp submissions
// Your memory usage beats 89.89 % of cpp submissions (144.9 MB)