# 1237. 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

# Solution

Approach 1: two pointers in linear time, since the function is monotonic.

Approach 2: binary search.

# Code (Python)

Approach 1:

    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        x, y = 1, 1000
        solutions = []
        while x <= 1000 and y >= 1:
            val = customfunction.f(x, y)
            if val > z:
                y -= 1
            elif val < z:
                x += 1
            else:
                solutions.append([x, y])
                x += 1
                y -= 1
        return solutions

Approach 2:

    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        solutions = []
        for x in range(1, 1001):
            lo, hi = 1, 1000
            while lo <= hi: # use <= to cover case in the end
                mid = (lo + hi) // 2
                val = customfunction.f(x, mid)
                if val < z:
                    lo = mid + 1
                elif val > z:
                    hi = mid - 1
                else:
                    solutions.append([x, mid])
                    break
        return solutions

# Code (C++)

Approach 1:

Approach 2: