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 represents one function from a 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
/*
* // This is the custom function interface.
* // You should not implement it, or speculate about its implementation
* public 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)
* public int f(int x, int y);
* };
*/
public class Solution {
public IList<IList<int>> FindSolution(CustomFunction customfunction, int z) {
var res = new List<IList<int>>();
int x = 1;
int y = 1000;
while (x <= 1000 && y > 0) {
int v = customfunction.f(x, y);
if (v > z){
--y;
}
else if (v < z){
++x;
}
else{
var arr = new List<int>();
arr.Add(x++);
arr.Add(y--);
res.Add(arr);
}
}
return res;
}
}
Time Complexity: O(n)
Space Complexity: O(n)