Solution Explanation:

This problem asks to calculate the sum of elements in an array nums based on specific queries. Each query [x, y] specifies a starting index x and a step size y. The sum should include nums[j] where j starts at x, and (j - x) is divisible by y.

A naive approach would iterate through the array for each query, resulting in a time complexity of O(n*m), where n is the length of nums and m is the number of queries. This is inefficient for large inputs.

The provided solution employs a technique called block decomposition or sqrt decomposition to optimize the process. The core idea is to precompute certain sums to speed up query processing.

Algorithm:

1. Preprocessing (Block Creation):

   • Determine the block size m as the square root of n ( m = √n). This is a heuristic – other block sizes could be explored.
   • Create a 2D array suf of size (m+1) x (n+1). suf[i][j] will store the sum of elements in nums starting from index j with a step size of i, up to the end of the array. This is a suffix sum, calculated efficiently during preprocessing.

2. Query Processing:

   • For each query [x, y]:
     • Case 1: y ≤ m (Small Step Size): If the step size is smaller than or equal to the block size, we can directly retrieve the precomputed sum from suf[y][x].
     • Case 2: y > m (Large Step Size): If the step size is larger than the block size, a direct iteration is still more efficient than pre-calculating all such sums because there are far fewer elements to sum. We iterate through the array from index x with step y, summing elements until the end.

3. Modulo Operation: The final sum for each query is taken modulo 10^9 + 7 to handle potential integer overflow.

Time Complexity Analysis:

• Preprocessing: The nested loops in the suf array initialization take O(n√n) time.
• Query Processing: For each query:
  • Case 1 (small step): O(1) to access suf[y][x].
  • Case 2 (large step): O(n/y) which is at most O(√n) because y > √n.
• With m queries, the overall time complexity becomes O(n√n + m√n) which simplifies to O((n+m)√n).

Space Complexity Analysis:

The suf array uses O(n√n) space.

Code Explanation (Python):

The Python code directly implements the algorithm described above. The solve function takes the nums array and queries as input. It initializes the suf array using nested loops, calculates the suffix sums, and then processes each query based on the step size, efficiently utilizing the precomputed sums where possible.

Other languages (Java, C++, Go, TypeScript) follow the same logic but with syntax variations specific to each language. The core algorithm remains the same across all implementations.