Find Positive Integer Solution for a Given Equation

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.

SOLUTION:

"""
   This is the custom function interface.
   You should not 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 with respect to 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]]:
        op = []
        for x in range(1, 1001):
            beg = 1
            end = 1000
            while beg <= end:
                y = (beg + end) // 2
                val =  customfunction.f(x, y)
                if val == z:
                    op.append([x, y])
                    break
                elif beg == end:
                    break
                elif val > z:
                    end = y
                else:
                    beg = y + 1
        return op