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Find Positive Integer Solution for a Given Equation

Difficulty: Medium

Problem Description

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.

The function is monotonically increasing, meaning:

  • f(x, y) < f(x + 1, y)
  • f(x, y) < f(x, y + 1)

Key Insights

  • The function f(x, y) is guaranteed to produce positive integers.
  • The search space for x and y is limited to the range [1, 1000].
  • Since the function is monotonically increasing, we can use a two-pointer approach to efficiently find all pairs (x, y) that satisfy f(x, y) == z.

Space and Time Complexity

Time Complexity: O(1000)
Space Complexity: O(1)

Solution

To solve the problem, we will use a two-pointer technique. We will initialize one pointer at x = 1 and the other pointer as y = z - x. We will iterate through possible values of x from 1 to z, checking the result of f(x, y) for each y calculated as y = z - x. If f(x, y) is less than z, we will increment x; if it is greater than z, we will decrement y. This continues until we exhaust the possible values for x and y within the defined bounds.

Code Solutions

class CustomFunction:
    # This is just a placeholder for the actual hidden function.
    def f(self, x, y):
        # Hidden formula, replace with actual logic
        pass

def findSolution(customfunction, z):
    result = []
    x = 1
    while x <= z:
        y = z - x
        if y < 1:
            break
        if customfunction.f(x, y) == z:
            result.append([x, y])
        x += 1
    return result
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