# Posts published in “Brute Force”

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

## Solution1 : Brute Force

Time complexity: O(1000*1000)
Space complexity: O(1)