Medium
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./*
* // 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)
* fun f(x:Int, y:Int):Int {}
* };
*/
class Solution {
// This is the custom function interface.
// You should not implement it, or speculate about its implementation
fun interface 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)
fun f(x: Int, y: Int): Int
}
fun findSolution(customfunction: CustomFunction, z: Int): List<List<Int>> {
val result: MutableList<List<Int>> = ArrayList()
var x = 1
var y = 1000
while (x < 1001 && y > 0) {
val functionResult = customfunction.f(x, y)
if (functionResult < z) {
x++
} else if (functionResult > z) {
y--
} else {
result.add(listOf(x++, y--))
}
}
return result
}
}