Find Positive Integer Solution for a Given Equation in Java

Problem Description

Ah, the classic “Find Positive Integer Solution for a Given Equation” problem! It’s like trying to find a parking spot in a crowded mall during the holiday season—frustrating, yet oddly satisfying when you finally spot that elusive space.

In this problem, you are given a mysterious function f(x, y) that returns a positive integer. Your mission, should you choose to accept it, is to find all pairs of positive integers (x, y) such that f(x, y) = z. Think of it as a treasure hunt where z is the treasure, and f(x, y) is the map that leads you there.

Code Solution


class Solution {
  public List> findSolution(CustomFunction customfunction, int z) {
    List> ans = new LinkedList<>();
    int x = 1;
    int y = 1000;

    while (x <= 1000 && y >= 1) {
      int f = customfunction.f(x, y);
      if (f < z)
        ++x;
      else if (f > z)
        --y;
      else
        ans.add(Arrays.asList(x++, y--));
    }

    return ans;
  }
}

Approach

The approach here is as straightforward as a Sunday morning stroll. We start with x at 1 and y at 1000, and we keep adjusting these values based on the output of f(x, y). If f(x, y) is less than z, we increase x (because we need a bigger number). If it’s more than z, we decrease y (because we need a smaller number). When we find a match, we add the pair to our answer list and move on. It’s like a game of tug-of-war, but with numbers!

Time and Space Complexity

Time Complexity

O(1000) in the worst case, since we are iterating through possible values of x and y within the range of 1 to 1000.

Space Complexity

O(1) for the variables used, but O(n) for the output list where n is the number of valid pairs found.

Real-World Example

Imagine you’re at a restaurant trying to order a meal that costs exactly $z. You have a menu with various dishes (the function f(x, y)), and you can only order two items (the positive integers x and y). You start with the cheapest item and the most expensive item on the menu. If your total is less than z, you need to order a pricier dish. If it’s more, you need to go for a cheaper one. Eventually, you find the perfect combination that matches your budget—just like finding the right pairs in our problem!