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:
9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z.function_id (to determine which implementation to test your code with), and the target z.findSolution and compare your results with the answer key.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 <= 91 <= z <= 100f(x, y) == z will be in the range 1 <= x, y <= 1000.f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000.