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
.
SOLUTION:
"""
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]]:
op = []
for x in range(1, 1001):
beg = 1
end = 1000
while beg <= end:
y = (beg + end) // 2
val = customfunction.f(x, y)
if val == z:
op.append([x, y])
break
elif beg == end:
break
elif val > z:
end = y
else:
beg = y + 1
return op
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