Find Positive Integer Solution for a Given Equation
Tags: leetcode, dsa, theabbie
Published: 2022-06-16 21:00:46

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:**

    """
       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)
           def f(self, x, y):
      
    """
    
    class Solution:
        def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
            op = []
            for x in range(1, 1001):
                beg = 1
                end = 1000
                while beg <= end:
                    y = (beg + end) // 2
                    val =  customfunction.f(x, y)
                    if val == z:
                        op.append([x, y])
                        break
                    elif beg == end:
                        break
                    elif val > z:
                        end = y
                    else:
                        beg = y + 1
            return op