1237. Find Positive Integer Solution for a Given Equation

Description

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.

Solutions

Solution 1: Enumeration + Binary Search

According to the problem, we know that the function $f(x, y)$ is a monotonically increasing function. Therefore, we can enumerate $x$, and then binary search $y$ in $[1,...z]$ to make $f(x, y) = z$. If found, add $(x, y)$ to the answer.

The time complexity is $O(n \log n)$, where $n$ is the value of $z$, and the space complexity is $O(1)$.

Python3JavaC++GoTypeScript

"""
   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]]:
        ans = []
        for x in range(1, z + 1):
            y = 1 + bisect_left(
                range(1, z + 1), z, key=lambda y: customfunction.f(x, y)
            )
            if customfunction.f(x, y) == z:
                ans.append([x, y])
        return ans

/*
 * // 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>> ans = new ArrayList<>();
        for (int x = 1; x <= 1000; ++x) {
            int l = 1, r = 1000;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (customfunction.f(x, mid) >= z) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            if (customfunction.f(x, l) == z) {
                ans.add(Arrays.asList(x, l));
            }
        }
        return ans;
    }
}

/*
 * // This is the custom function interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction {
 * public:
 *     // 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)
 *     int f(int x, int y);
 * };
 */

class Solution {
public:
    vector<vector<int>> findSolution(CustomFunction& customfunction, int z) {
        vector<vector<int>> ans;
        for (int x = 1; x <= 1000; ++x) {
            int l = 1, r = 1000;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (customfunction.f(x, mid) >= z) {
                    r = mid;
                } else {
                    l = mid + 1;
                }
            }
            if (customfunction.f(x, l) == z) {
                ans.push_back({x, l});
            }
        }
        return ans;
    }
};

/**
 * This is the declaration of customFunction API.
 * @param  x    int
 * @param  x    int
 * @return      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)
 */

func findSolution(customFunction func(int, int) int, z int) (ans [][]int) {
    for x := 1; x <= 1000; x++ {
        y := 1 + sort.Search(999, func(y int) bool { return customFunction(x, y+1) >= z })
        if customFunction(x, y) == z {
            ans = append(ans, []int{x, y})
        }
    }
    return
}

/**
 * // This is the CustomFunction's API interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction {
 *      f(x: number, y: number): number {}
 * }
 */

function findSolution(customfunction: CustomFunction, z: number): number[][] {
    const ans: number[][] = [];
    for (let x = 1; x <= 1000; ++x) {
        let l = 1;
        let r = 1000;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (customfunction.f(x, mid) >= z) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if (customfunction.f(x, l) == z) {
            ans.push([x, l]);
        }
    }
    return ans;
}

Solution 2: Two Pointers

We can define two pointers $x$ and $y$, initially $x = 1$, $y = z$.

If $f(x, y) = z$, we add $(x, y)$ to the answer, then $x \leftarrow x + 1$, $y \leftarrow y - 1$;
If $f(x, y) \lt z$, at this time for any $y' \lt y$, we have $f(x, y') \lt f(x, y) \lt z$, so we cannot decrease $y$, we can only increase $x$, so $x \leftarrow x + 1$;
If $f(x, y) \gt z$, at this time for any $x' \gt x$, we have $f(x', y) \gt f(x, y) \gt z$, so we cannot increase $x$, we can only decrease $y$, so $y \leftarrow y - 1$.

After the loop ends, return the answer.

The time complexity is $O(n)$, where $n$ is the value of $z$, and the space complexity is $O(1)$.