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 9 hidden implementations of CustomFunction, along with a way to generate an answer key of all valid pairs for a specific z.
  • The judge will receive two inputs: a function_id (to determine which implementation to test your code with), and the target z.
  • The judge will call your findSolution and 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 <= 9
  • 1 <= z <= 100
  • It is guaranteed that the solutions of f(x, y) == z will be in the range 1 <= x, y <= 1000.
  • It is also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= 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
        ]

Copy The Code & Try With Live Editor

Input

x
+
cmd
function_id = 1, z = 5

Output

x
+
cmd
[[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);
            }
        };
    }
}
Copy The Code & Try With Live Editor

Input

x
+
cmd
function_id = 1, z = 5

Output

x
+
cmd
[[1,4],[2,3],[3,2],[4,1]]
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