1237. Find Positive Integer Solution for a Given Equation

Problem

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.

Solution (Java)

/*
 * // 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)
 *     public int f(int x, int y);
 * };
 */

class Solution {

    public List<List<Integer>> findSolution(CustomFunction customfunction, int z) {
        List<List<Integer>> result = new ArrayList<>();
        int x = 1;
        int y = 1000;
        while (x < 1001 && y > 0) {
            int functionResult = customfunction.f(x, y);
            if (functionResult < z) {
                x++;
            } else if (functionResult > z) {
                y--;
            } else {
                result.add(Arrays.asList(x++, y--));
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).