Powers of two

Simple problem that requires breaking down a number into powers of two. There is a simple algorithm for that, running in LogN. Overall complexity for this problem then becomes NLogN. Code is down below, cheers, ACC.

Range Product Queries of Powers - LeetCode

2438. Range Product Queries of Powers

Medium

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

Accepted

6,117

Submissions

19,018

public int[] ProductQueries(int n, int[][] queries)
{
    List powersList = new List();

    while (n > 0)
    {
        int power = 1;
        while (power * 2 <= n)
            power *= 2;
        powersList.Add(power);
        n -= power;
    }

    long[] powerArr = new long[powersList.Count];
    int index = powerArr.Length - 1;
    foreach (int p in powersList) powerArr[index--] = (long)p;

    int[] retVal = new int[queries.Length];
    for (int i = 0; i < queries.Length; i++)
    {
        long product = 1;
        for (int j = queries[i][0]; j <= queries[i][1]; j++)
            product = (product * powerArr[j]) % (1000000000 + 7);
        retVal[i] = (int)product;
    }

    return retVal;
}