# 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 $O(1)$.

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 $O(1)$.

• /*
* // 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) {
}
}
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 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 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;
}