'''
1237. Find Positive Integer Solution for a Given Equation
https://leetcode.com/contest/weekly-contest-160/problems/find-positive-integer-solution-for-a-given-equation/
Given a function  f(x, y) and a value z, 
return all positive integer pairs x and y where f(x,y) == z.
The function is constantly 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 positive integer f(x, y) for any given positive integer x and y.
  int f(int x, int y);
};
For custom testing purposes you're given 
an integer function_id and a target z as input, 
where function_id represent one function from an secret internal list, 
on the examples you'll know only two functions from the list.  
You may return the solutions in any order.

Example 1:
Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: function_id = 1 means that f(x, y) = x + y

Example 2:
Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: function_id = 2 means that f(x, y) = x * y

Constraints:
1 <= function_id <= 9
1 <= z <= 100
It's guaranteed that the solutions of f(x, y) == z will be on the range 1 <= x, y <= 1000
It's also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000
'''
from typing import *

class Solution:
    def findSolution(self, customfunction, z: int) -> List[List[int]]:
        ans = []
        for x in range(1, 1001):
            lefty = 1
            righty = 1001
            while lefty < righty:
                y = lefty + (righty-lefty)//2
                value = customfunction.f(x, y)
                if value == z:
                    ans.append([x, y])
                    break
                elif value < z:
                    lefty = y + 1
                elif value > z:
                    righty = y
            else:
                if righty == 1:
                    break
        return ans

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):
        return x * y

print(Solution().findSolution(CustomFunction(), 5))
# [[1,5],[5,1]]