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
9hidden 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
findSolutionand 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 <= 91 <= z <= 100- It is guaranteed that the solutions of
f(x, y) == zwill 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.
Binary search.
"""
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]]:
res = []
for x in range(1, 1001):
left, right = 1, 1000
while left < right:
mid = (left + right) >> 1
if customfunction.f(x, mid) >= z:
right = mid
else:
left = mid + 1
if customfunction.f(x, left) == z:
res.append([x, left])
return res/*
* // 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>> res = new ArrayList<>();
for (int i = 1; i <= 1000; ++i) {
int left = 1, right = 1000;
while (left < right) {
int mid = (left + right) >> 1;
if (customfunction.f(i, mid) >= z) {
right = mid;
} else {
left = mid + 1;
}
}
if (customfunction.f(i, left) == z) {
res.add(Arrays.asList(i, left));
}
}
return res;
}
}/**
* // 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[][] {
// 二分
let ans = [];
for (let i = 1; i <= 1000; i++) {
let left = 1,
right = 1000;
while (left < right) {
let mid = (left + right) >> 1;
if (customfunction.f(i, mid) >= z) {
right = mid;
} else {
left = mid + 1;
}
}
if (customfunction.f(i, left) == z) {
ans.push([i, left]);
}
}
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>> res;
for (int i = 1; i <= 1000; ++i) {
int left = 1, right = 1000;
while (left < right) {
int mid = left + right >> 1;
if (customfunction.f(i, mid) >= z) {
right = mid;
} else {
left = mid + 1;
}
}
if (customfunction.f(i, left) == z) {
res.push_back({i, left});
}
}
return res;
}
};/**
* 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) [][]int {
res := [][]int{}
for i := 1; i <= 1000; i++ {
left, right := 1, 1000
for left < right {
mid := (left + right) >> 1
if customFunction(i, mid) >= z {
right = mid
} else {
left = mid + 1
}
}
if customFunction(i, left) == z {
res = append(res, []int{i, left})
}
}
return res
}