LeetCode 2438: Range Product Queries of Powers Solution

Master LeetCode problem 2438 (Range Product Queries of Powers), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2438. Range Product Queries of Powers

Medium

Array Bit Manipulation Prefix Sum

Problem Explanation

Explanation

To solve this problem, we first need to generate the array powers that represents the minimum number of powers of 2 to sum up to n. We then iterate through each query and calculate the product of elements in the range specified by the query in the powers array. Finally, we return the answers modulo 10^9 + 7.

Generate the powers array representing the minimum powers of 2 to sum up to n.
Iterate through each query and calculate the product of elements in the specified range.
Return the answers modulo 10^9 + 7.

Time Complexity:
Generating the powers array takes O(log n) time. Handling each query takes O(1) time. Thus, the overall time complexity is O(log n + q), where q is the number of queries.

Space Complexity:
The space complexity is O(log n) to store the powers array.

Solution Code

class Solution {
    public int[] rangeProductQueriesOfPowers(int n, int[][] queries) {
        int[] powers = new int[(int)(Math.log(n) / Math.log(2)) + 1];
        powers[0] = 1;
        for (int i = 1; i < powers.length; i++) {
            powers[i] = 2 * powers[i - 1];
        }

        int[] answers = new int[queries.length];
        int mod = 1000000007;

        for (int i = 0; i < queries.length; i++) {
            int left = queries[i][0];
            int right = queries[i][1];
            long product = 1;
            for (int j = left; j <= right; j++) {
                product = (product * powers[j]) % mod;
            }
            answers[i] = (int)product;
        }

        return answers;
    }
}

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