LeetCode //C - 1237. Find Positive Integer Solution for a Given Equation

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:

c 复制代码

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.

From: LeetCode

Link: 1237. Find Positive Integer Solution for a Given Equation

Solution:

Ideas:

use monotonic property with two pointers.

Start at (x=1, y=1000):

If f(x,y) == z, save pair, then x++, y--
If f(x,y) < z, increase x
If f(x,y) > z, decrease y

Code:

c 复制代码

#include <stdlib.h>

int** findSolution(int (*customFunction)(int, int), int z,
                   int* returnSize, int** returnColumnSizes) {
    int** ans = (int**)malloc(sizeof(int*) * 1000);
    *returnColumnSizes = (int*)malloc(sizeof(int) * 1000);
    *returnSize = 0;

    int x = 1, y = 1000;

    while (x <= 1000 && y >= 1) {
        int val = customFunction(x, y);

        if (val == z) {
            ans[*returnSize] = (int*)malloc(sizeof(int) * 2);
            ans[*returnSize][0] = x;
            ans[*returnSize][1] = y;
            (*returnColumnSizes)[*returnSize] = 2;
            (*returnSize)++;

            x++;
            y--;
        } else if (val < z) {
            x++;
        } else {
            y--;
        }
    }

    return ans;
}