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>();
                        solution.add(x);
                        solution.add(y);
                        solutions.add(solution);
                    }
                    break;
                }
            }
        }
        return solutions;
    }
}

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