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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 ofCustomFunction
, along with a way to generate an answer key of all valid pairs for a specificz
.  The judge will receive two inputs: a
function_id
(to determine which implementation to test your code with), and the targetz
.  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 range1 <= x, y <= 1000
.  It is also guaranteed that
f(x, y)
will fit in 32 bit signed integer if1 <= 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) { ans.add(Arrays.asList(x, l)); } } 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; }