# Solving Leetcode Interviews in Seconds with AI: Range Product Queries of Powers


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2438" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	Here's a breakdown of the solution:

*   **Generate Powers of 2:** Convert the given integer `n` into its binary representation. Each '1' in the binary representation indicates a power of 2 that contributes to the sum `n`. Extract these powers of 2 and store them in a list called `powers`.
*   **Process Queries:** Iterate through the `queries` array. For each query `[left, right]`, calculate the product of the elements in the `powers` array within the specified range (inclusive). Apply the modulo operator (10^9 + 7) to prevent integer overflow.
*   **Return Results:** Store the results of each query in an `answers` array and return it.

*   **Time Complexity:** O(log n + q * log n) where n is the input integer and q is the number of queries. The 'log n' factor arises from binary representation conversion, while the 'q * log n' comes from calculating the product for each query over potentially log n powers.
*   **Space Complexity:** O(log n) due to storing the `powers` array, which holds at most log n elements.

	
	# Code
	```python
	def product_of_powers(n: int, queries: list[list[int]]) -> list[int]:
    """
    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:
    n = 15, queries = [[0,1],[2,2],[0,3]]
    powers = [1,2,4,8]
    1st query: powers[0] * powers[1] = 1 * 2 = 2
    2nd query: powers[2] = 4
    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.
    """
    powers = []
    binary_representation = bin(n)[2:]  # Convert to binary and remove "0b" prefix

    for i, bit in enumerate(reversed(binary_representation)):
        if bit == '1':
            powers.append(2**i)

    answers = []
    modulo = 10**9 + 7
    for left, right in queries:
        product = 1
        for j in range(left, right + 1):
            product = (product * powers[j]) % modulo
        answers.append(product)

    return answers
	```