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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:

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.

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

This problem involves finding all pairs of positive integers (x, y) that satisfy a given equation f(x, y) == z, where f is a hidden monotonically increasing function. The challenge lies in the unknown nature of f. Two efficient solutions are presented below.

Solution 1: Enumeration + Binary Search

This approach iterates through possible values of x and uses binary search to find the corresponding y for each x.

Algorithm:

  1. Iterate through x: We loop through all possible values of x from 1 to z (since f is monotonically increasing, if x exceeds z, it's impossible to find a y that satisfies the equation).
  2. Binary Search for y: For each x, we perform a binary search on the interval [1, z] to find a y such that f(x, y) == z. The monotonicity of f guarantees that binary search will work efficiently.
  3. Store Solutions: If a solution (x, y) is found, it's added to the result list.

Time Complexity: O(z log z). The outer loop iterates z times, and the binary search takes O(log z) time for each iteration.

Space Complexity: O(1). We only use a constant amount of extra space.

Solution 2: Two Pointers

This solution leverages the monotonicity of f to efficiently explore the solution space using two pointers.

Algorithm:

  1. Initialize Pointers: Start with x = 1 and y = 1000 (the maximum possible value of y).
  2. Compare f(x, y) with z:
    • If f(x, y) < z, increment x (since increasing x will increase the result of f).
    • If f(x, y) > z, decrement y (since decreasing y will decrease the result of f).
    • If f(x, y) == z, add (x, y) to the result list, then increment x and decrement y to continue searching.
  3. Termination: The loop terminates when x exceeds 1000 or y becomes 0.

Time Complexity: O(z). In the worst case, we might iterate through all possible values of x and y, but this is linear in the range of possible values.

Space Complexity: O(1). We only use a constant amount of extra space.

Code Implementation (Python)

Solution 1 (Enumeration + Binary Search):

import bisect
 
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

Solution 2 (Two Pointers):

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        ans = []
        x, y = 1, 1000
        while x <= 1000 and y > 0:
            t = customfunction.f(x, y)
            if t < z:
                x += 1
            elif t > z:
                y -= 1
            else:
                ans.append([x, y])
                x += 1
                y -= 1
        return ans

The Two Pointers solution is generally preferred due to its better average-case time complexity, although in the worst case they can perform similarly. Both solutions demonstrate how to effectively utilize the monotonicity property of the unknown function. Remember that you'll need a CustomFunction class definition (provided by LeetCode) to run these code snippets.