Problem

Given a callable function f(x, y) with a hidden formula and a value z, reverse engineer the formula and return all positive integer pairsx andy wheref(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.

Examples

Example 1

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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

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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.

Solution

Method 1 – Two Pointers (or Binary Search)

Intuition

Since f(x, y) is monotonically increasing in both x and y, we can efficiently search for all solutions by fixing one variable and searching for the other using two pointers or binary search.

Approach

  1. For each x from 1 to 1000 (or a reasonable upper bound):
    • Use binary search (or increment) for y from 1 to 1000 to find if there exists a y such that f(x, y) == z.
    • If found, add [x, y] to the answer.
  2. Stop if f(x, 1) exceeds z (since f(x, y) increases with x).
  3. Return the list of all valid pairs.

Code

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class Solution {
public:
    vector<vector<int>> findSolution(CustomFunction& f, int z) {
        vector<vector<int>> ans;
        for (int x = 1; x <= 1000; ++x) {
            int l = 1, r = 1000;
            while (l <= r) {
                int y = l + (r - l) / 2;
                int val = f.f(x, y);
                if (val == z) {
                    ans.push_back({x, y});
                    break;
                } else if (val < z) {
                    l = y + 1;
                } else {
                    r = y - 1;
                }
            }
        }
        return ans;
    }
};
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func findSolution(f CustomFunction, z int) [][]int {
    ans := [][]int{}
    for x := 1; x <= 1000; x++ {
        l, r := 1, 1000
        for l <= r {
            y := l + (r-l)/2
            val := f.F(x, y)
            if val == z {
                ans = append(ans, []int{x, y})
                break
            } else if val < z {
                l = y + 1
            } else {
                r = y - 1
            }
        }
    }
    return ans
}
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class Solution {
    public List<List<Integer>> findSolution(CustomFunction f, int z) {
        List<List<Integer>> ans = new ArrayList<>();
        for (int x = 1; x <= 1000; x++) {
            int l = 1, r = 1000;
            while (l <= r) {
                int y = l + (r - l) / 2;
                int val = f.f(x, y);
                if (val == z) {
                    ans.add(List.of(x, y));
                    break;
                } else if (val < z) {
                    l = y + 1;
                } else {
                    r = y - 1;
                }
            }
        }
        return ans;
    }
}
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class Solution {
    fun findSolution(f: CustomFunction, z: Int): List<List<Int>> {
        val ans = mutableListOf<List<Int>>() 
        for (x in 1..1000) {
            var l = 1
            var r = 1000
            while (l <= r) {
                val y = l + (r - l) / 2
                val valf = f.f(x, y)
                if (valf == z) {
                    ans.add(listOf(x, y))
                    break
                } else if (valf < z) {
                    l = y + 1
                } else {
                    r = y - 1
                }
            }
        }
        return ans
    }
}
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class Solution:
    def findSolution(self, f: 'CustomFunction', z: int) -> list[list[int]]:
        ans = []
        for x in range(1, 1001):
            l, r = 1, 1000
            while l <= r:
                y = (l + r) // 2
                val = f.f(x, y)
                if val == z:
                    ans.append([x, y])
                    break
                elif val < z:
                    l = y + 1
                else:
                    r = y - 1
        return ans
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impl Solution {
    pub fn find_solution(f: &CustomFunction, z: i32) -> Vec<Vec<i32>> {
        let mut ans = vec![];
        for x in 1..=1000 {
            let (mut l, mut r) = (1, 1000);
            while l <= r {
                let y = l + (r - l) / 2;
                let val = f.f(x, y);
                if val == z {
                    ans.push(vec![x, y]);
                    break;
                } else if val < z {
                    l = y + 1;
                } else {
                    r = y - 1;
                }
            }
        }
        ans
    }
}
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class Solution {
    findSolution(f: CustomFunction, z: number): number[][] {
        const ans: number[][] = [];
        for (let x = 1; x <= 1000; x++) {
            let l = 1, r = 1000;
            while (l <= r) {
                const y = Math.floor((l + r) / 2);
                const val = f.f(x, y);
                if (val === z) {
                    ans.push([x, y]);
                    break;
                } else if (val < z) {
                    l = y + 1;
                } else {
                    r = y - 1;
                }
            }
        }
        return ans;
    }
}

Complexity

  • ⏰ Time complexity: O(1000 * log 1000) = O(1000 * 10) = O(10^4), since for each x we binary search y.
  • 🧺 Space complexity: O(k), where k is the number of valid pairs found.