Introduction

In this blog post, we will explore how to solve the LeetCode problem "1237" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return all positive integer pairs x and y where f(x,y) == z. You may return the pairs in any order. While the exact formula is hidden, the function is monotonically increasing, i.e.: f(x, y) < f(x + 1, y) f(x, y) < f(x, y + 1) The function interface is defined like this: interface CustomFunction { public: // Returns some positive integer f(x, y) for two positive integers x and y based on a formula. int f(int x, int y); }; We will judge your solution as follows: The judge has a list of 9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z. The judge will receive two inputs: a function_id (to determine which implementation to test your code with), and the target z. The judge will call your findSolution and compare your results with the answer key. If your results match the answer key, your solution will be Accepted. Example 1: Input: function_id = 1, z = 5 Output: [[1,4],[2,3],[3,2],[4,1]] Explanation: The hidden formula for function_id = 1 is f(x, y) = x + y. The following positive integer values of x and y make f(x, y) equal to 5: x=1, y=4 -> f(1, 4) = 1 + 4 = 5. x=2, y=3 -> f(2, 3) = 2 + 3 = 5. x=3, y=2 -> f(3, 2) = 3 + 2 = 5. x=4, y=1 -> f(4, 1) = 4 + 1 = 5. Example 2: Input: function_id = 2, z = 5 Output: [[1,5],[5,1]] Explanation: The hidden formula for function_id = 2 is f(x, y) = x y. The following positive integer values of x and y make f(x, y) equal to 5: x=1, y=5 -> f(1, 5) = 1 5 = 5. x=5, y=1 -> f(5, 1) = 5 * 1 = 5. Constraints: 1 <= function_id <= 9 1 <= z <= 100 It is guaranteed that the solutions of f(x, y) == z will be in the range 1 <= x, y <= 1000. It is also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000.

Explanation

• Two-Pointer Approach: Exploit the monotonically increasing nature of f(x, y). Start with x = 1 and y = 1000. If f(x, y) > z, decrease y. If f(x, y) < z, increase x. If f(x, y) == z, add the pair to the result and then increment x and decrement y in the next iteration to find other possible pairs.
  • Optimization: Limit the search space based on the problem constraints (1 <= x, y <= 1000). This avoids unnecessary calls to the custom function.

• Runtime Complexity: O(x+y) which is O(2000) in the worst case hence O(1), Storage Complexity: O(N), where N is the number of pairs found (at most ~O(z)).

Code

    """
   This is the interface for the CustomFunction you are given.
   You do not need to implement it, or speculate about its implementation
   class CustomFunction:
       # Returns f(x, y) for any given positive integers x and y.
       # Note that f(x, y) is increasing w.r.t. both x and y.
       # i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
       def f(self, x, y):

"""

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> list[list[int]]:
        result = []
        x = 1
        y = 1000

        while x <= 1000 and y >= 1:
            val = customfunction.f(x, y)

            if val == z:
                result.append([x, y])
                x += 1
                y -= 1
            elif val < z:
                x += 1
            else:
                y -= 1

        return result