2529. Range Product Queries Of Powers¶

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2529. 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

Solution¶

class Solution {
    private ArrayList<Integer> arr;
    private int mod = (int)(1e9 + 7);
    public int[] productQueries(int n, int[][] queries) {
        arr = new ArrayList<>();
        while (n > 0) 
            n -= change(n);
        Collections.sort(arr);
        long pre[] = new long[arr.size()];
        long prod = 1;
        for (int i = 0; i < arr.size(); i++) {
            prod = (prod * arr.get(i)) % mod;
            pre[i] = prod;
        }
        int res[] = new int[queries.length];
        int k = 0;
        for (int[] temp : queries) {
            int l = temp[0], r = temp[1];
            if (l == 0) {
                res[k++] = (int) pre[r];
            } else {
                long numerator = pre[r];
                long denominator = pre[l - 1];
                long inv = modPow(denominator, mod - 2, mod);
                res[k++] = (int) ((numerator * inv) % mod);
            }
        }
        return res;
    }
    private int change(int n) {
        int current = 1;
        while (current * 2 <= n) {
            current *= 2;
        }
        arr.add(current);
        return current;
    }
    private long modPow(long base, long exp, int m) {
        long result = 1;
        base %= m;
        while (exp > 0) {
            if ((exp & 1) == 1) result = (result * base) % m;
            base = (base * base) % m;
            exp >>= 1;
        }
        return result;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here