1237. Find Positive Integer Solution for a Given Equation
Easy
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 range1 <= x, y <= 1000
- It's also guaranteed that
f(x, y)
will fit in 32 bit signed integer if1 <= x, y <= 1000
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | /* * // 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<>(); int x = 1, y = 1000; while (x <= 1000 && y > 0) { int v = customfunction.f(x, y); if (v > z) --y; else if (v < z) ++x; else res.add(Arrays.asList(x++, y--)); } return res; } } |
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