Formatted question description: https://leetcode.ca/all/1237.html

# 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 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

Related Topics:
Math, Binary Search

## Solution 1. Brute Force

// OJ: https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/

// Time: O(N^2)
// Space: O(1)
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& func, int z) {
vector<vector<int>> ans;
for (int x = 1; x <= 1000; ++x) {
for (int y = 1; y <= 1000; ++y) {
if (func.f(x, y) == z) ans.push_back({ x, y });
}
}
return ans;
}
};

// OJ: https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& func, int z) {
vector<vector<int>> ans;
for (int x = 1; x <= 1000; ++x) {
int L = 1, R = 1000;
while (L <= R) {
int y = (L + R) / 2, val = func.f(x, y);
if (val == z) {
ans.push_back({ x, y });
break;
} else if (val < z) L = y + 1;
else R = y - 1;
}
}
return ans;
}
};


## Solution 3. Two Pointers

// OJ: https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/

// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<vector<int>> findSolution(CustomFunction& func, int z) {
vector<vector<int>> ans;
for (int x = 1, y = 1000; x <= 1000 && y >= 1; ++x) {
while (y >= 1 && func.f(x, y) > z) --y;
if (y >= 1 && func.f(x, y) == z) ans.push_back({ x, y });
}
return ans;
}
};


Java

/*
* // 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>> solutions = new ArrayList<List<Integer>>();
for (int x = 1; x <= 1000; x++) {
for (int y = 1; y <= 1000; y++) {
int function = customfunction.f(x, y);
if (function >= z) {
if (function == z) {
List<Integer> solution = new ArrayList<Integer>();