# Solving Leetcode Interviews in Seconds with AI: Sorted GCD Pair Queries


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3312" 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
	> You are given an integer array nums of length n and an integer array queries. Let gcdPairs denote an array obtained by calculating the GCD of all possible pairs (nums[i], nums[j]), where 0 <= i < j < n, and then sorting these values in ascending order. For each query queries[i], you need to find the element at index queries[i] in gcdPairs. Return an integer array answer, where answer[i] is the value at gcdPairs[queries[i]] for each query. The term gcd(a, b) denotes the greatest common divisor of a and b.   Example 1:  Input: nums = [2,3,4], queries = [0,2,2] Output: [1,2,2] Explanation: gcdPairs = [gcd(nums[0], nums[1]), gcd(nums[0], nums[2]), gcd(nums[1], nums[2])] = [1, 2, 1]. After sorting in ascending order, gcdPairs = [1, 1, 2]. So, the answer is [gcdPairs[queries[0]], gcdPairs[queries[1]], gcdPairs[queries[2]]] = [1, 2, 2].  Example 2:  Input: nums = [4,4,2,1], queries = [5,3,1,0] Output: [4,2,1,1] Explanation: gcdPairs sorted in ascending order is [1, 1, 1, 2, 2, 4].  Example 3:  Input: nums = [2,2], queries = [0,0] Output: [2,2] Explanation: gcdPairs = [2].    Constraints:  2 <= n == nums.length <= 105 1 <= nums[i] <= 5 * 104 1 <= queries.length <= 105 0 <= queries[i] < n * (n - 1) / 2  

	# Explanation
	Here's a solution approach, complexity analysis, and Python code for the problem:

*   **Calculate GCDs and Count Frequencies:** Compute the GCD for all unique pairs of numbers in the input array `nums`. Instead of explicitly storing all GCDs in a list and sorting, which can be memory-intensive, we count the frequency of each GCD value using a dictionary or a similar data structure. This avoids storing duplicate GCDs.

*   **Prefix Sum for Efficient Lookup:**  Create a prefix sum array (or cumulative frequency array) from the GCD frequencies. This allows us to efficiently determine the GCD value at a given index (query) by iterating through the sorted unique GCDs and checking the cumulative counts.

*   **Handle Queries:** For each query, use the prefix sum array to find the corresponding GCD value by linearly searching through the unique sorted GCDs.

*   **Complexity:**
    *   Runtime Complexity: O(N^2 + M\*K) where N is the length of `nums`, M is the length of `queries`, and K is the number of unique GCDs, which is bounded by the maximum value in `nums` (5 * 10^4 in the constraints).
    *   Storage Complexity: O(K), where K is the number of unique GCDs.

	
	# Code
	```python
	import math

def solve():
    def gcd(a, b):
        if b == 0:
            return a
        return gcd(b, a % b)

    def solve_queries(nums, queries):
        n = len(nums)
        gcd_counts = {}
        for i in range(n):
            for j in range(i + 1, n):
                val = gcd(nums[i], nums[j])
                gcd_counts[val] = gcd_counts.get(val, 0) + 1

        unique_gcds = sorted(gcd_counts.keys())
        prefix_sums = []
        current_sum = 0
        for val in unique_gcds:
            current_sum += gcd_counts[val]
            prefix_sums.append(current_sum)

        results = []
        for query in queries:
            for i in range(len(unique_gcds)):
                if query < prefix_sums[i]:
                    results.append(unique_gcds[i])
                    break
        return results

    # Read input (if needed - example usage here)
    # nums = list(map(int, input().split()))
    # queries = list(map(int, input().split()))
    # result = solve_queries(nums, queries)
    # print(*result)
    return solve_queries


def main():
    nums1 = [2, 3, 4]
    queries1 = [0, 2, 2]
    print(solve_queries(nums1, queries1))  # Output: [1, 2, 2]

    nums2 = [4, 4, 2, 1]
    queries2 = [5, 3, 1, 0]
    print(solve_queries(nums2, queries2))  # Output: [4, 2, 1, 1]

    nums3 = [2, 2]
    queries3 = [0, 0]
    print(solve_queries(nums3, queries3))  # Output: [2, 2]
    
if __name__ == "__main__":
    main()
	```