# 1237. Find Positive Integer Solution for a Given Equation
Given a function f(x, y) and a value z, return all positive integer pairs x and y where f(x,y) == z.
The function is constantly 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 positive integer f(x, y) for any given positive integer x and y. int f(int x, int y); }; For custom testing purposes you're given an integer function_id and a target z as input, where function_id represent one function from an secret internal list, on the examples you'll know only two functions from the list.
You may return the solutions in any order.
Example 1:
Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: function_id = 1 means that f(x, y) = x + y
Example 2:
Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: function_id = 2 means that f(x, y) = x * y
Constraints:
1 <= function_id <= 9
1 <= z <= 100
It's guaranteed that the solutions of f(x, y) == z will be on the range 1 <= x, y <= 1000
It's also guaranteed that f(x, y) will fit in 32 bit signed integer if 1 <= x, y <= 1000
# Solution
Approach 1: two pointers in linear time, since the function is monotonic.
Approach 2: binary search.
# Code (Python)
Approach 1:
def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
x, y = 1, 1000
solutions = []
while x <= 1000 and y >= 1:
val = customfunction.f(x, y)
if val > z:
y -= 1
elif val < z:
x += 1
else:
solutions.append([x, y])
x += 1
y -= 1
return solutions
Approach 2:
def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
solutions = []
for x in range(1, 1001):
lo, hi = 1, 1000
while lo <= hi: # use <= to cover case in the end
mid = (lo + hi) // 2
val = customfunction.f(x, mid)
if val < z:
lo = mid + 1
elif val > z:
hi = mid - 1
else:
solutions.append([x, mid])
break
return solutions
# Code (C++)
Approach 1:
Approach 2: