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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

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 $O(1)$.

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"""
   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

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/*
 * // 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;
    }
}

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/*
 * // 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;