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.

Code

Binary Search

Time Complexity: $O(x\log y)$, Space Complexity: $O(1)$

/*
 * // 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>> res;
        for(int x = 1; x <= 1000; x++) {
            int l = 1, r = 1000;
            while(l < r) {
                int m = (l + r) / 2;
                if(customfunction.f(x, m) < z)
                    l = m + 1;
                else 
                    r = m; 
            }
            if(customfunction.f(x, l) == z)
                res.push_back({x, l});
        }
 
        return res;
    }
};

2D Matrix

Time Complexity: $O(x+y)$, Space Complexity: $O(1)$

因為在 x, y 上 f(x, y) 都具有單調性，因此當找到一個解後，若x 增加，則 y 必然需要減少，才會再度得到一個解滿足：f(x, y) = z，因此可以寫成以下的 for loop 以及 while loop 的組合。

/*
 * // 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>> res;
        int y = 1000;
        for(int x = 1; x <= 1000; x++) {
            while(y > 1 && customfunction.f(x, y) > z) y--;
            if(customfunction.f(x, y) == z)
                res.push_back({x, y});
        }
 
        return res;
    }
};

Source

• Find Positive Integer Solution for a Given Equation - LeetCode