Algorithm
Problem Name: 1237. Find Positive Integer Solution for a Given Equation
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:
- The judge has a list of
9hidden implementations ofCustomFunction, along with a way to generate an answer key of all valid pairs for a specificz. - The judge will receive two inputs: a
function_id(to determine which implementation to test your code with), and the targetz. - The judge will call your
findSolutionand compare your results with the answer key. - If your results match the answer key, your solution will be
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 <= 100- It is guaranteed that the solutions of
f(x, y) == zwill be in the range1 <= x, y <= 1000. - It is also guaranteed that
f(x, y)will fit in 32 bit signed integer if1 <= x, y <= 1000.
Code Examples
#1 Code Example with Python Programming
Code -
Python Programming
from itertools import product as pr
class Solution(object):
def findSolution(self, customfunction, z):
return [
[i, j]
for i, j in pr(range(1, z + 1), repeat=2)
if customfunction.f(i, j) == z
]
Input
function_id = 1, z = 5
Output
[[1,4],[2,3],[3,2],[4,1]]
#2 Code Example with C# Programming
Code -
C# Programming
using System;
using System.Collections.Generic;
namespace LeetCode
{
public class _1237_FindPositiveIntegerSolutionForAGivenEquation
{
public IList < IList<int>> FindSolution(CustomFunction customfunction, int z)
{
var solutions = new List < IList<int>>();
int minX = MinX(customfunction, z);
int maxX = MaxX(customfunction, z);
for (int x = minX; x < = maxX; x++)
{
var solution = BinarySearchY(customfunction, x, z);
if (solution != null)
solutions.Add(solution);
}
return solutions;
}
private int MinX(CustomFunction customfunction, int z)
{
int min = 1;
int max = 1000;
while (min < = max)
{
int mid = (min + max) / 2;
int fMid = customfunction.f(mid, 1000);
if (fMid == z)
return mid;
if (fMid < z)
min = mid + 1;
else
max = mid - 1;
}
return min;
}
private int MaxX(CustomFunction customfunction, int z)
{
int min = 1;
int max = 1000;
while (min < = max)
{
int mid = (min + max) / 2;
int fMid = customfunction.f(mid, 1);
if (fMid == z)
return mid;
if (fMid < z)
min = mid + 1;
else
max = mid - 1;
}
return max;
}
private IList < int> BinarySearchY(CustomFunction customfunction, int x, int z)
{
int min = 1;
int max = 1000;
while (min < = max)
{
int mid = (min + max) / 2;
int fMid = customfunction.f(x, mid);
if (fMid == z)
return new int[] { x, mid };
if (fMid < z)
min = mid + 1;
else
max = mid - 1;
}
return null;
}
public class CustomFunction
{
private Func < int, int, int> func;
public CustomFunction(Func<int, int, int> func)
{
this.func = func;
}
// 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)
{
return func(x, y);
}
};
}
}
Input
function_id = 1, z = 5
Output
[[1,4],[2,3],[3,2],[4,1]]