Find Positive Integer Solution for a Given Equation in C++

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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 succeed. The problem essentially asks you to find all pairs of positive integers (x, y) such that a given function f(x, y) equals a target value z.

Imagine you’re at a party, and you need to find the perfect dance partner (let’s call them y) who can match your moves (that’s x). But there’s a catch: you can only dance if the total number of moves equals z. So, you’re on a quest to find all the possible pairs of dance partners that can make this happen.

Code Solution


class Solution {
 public:
  vector> findSolution(CustomFunction& customfunction, int z) {
    vector> ans;
    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.push_back({x++, y--});
    }

    return ans;
  }
};

Approach

The approach here is as straightforward as a game of hide and seek. We start with x = 1 and y = 1000 and use a while loop to explore possible pairs. If the function f(x, y) is less than z, we increase x (because we need to “dance” more). If it’s greater, we decrease y (because we need to “dance” less). When we find a match, we store the pair and move on.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(1000)
Space Complexity	O(k)

Real-World Example

Let’s say you’re organizing a bake sale, and you need to find the right combination of cookies and brownies to reach a total of z sales. Each cookie and brownie has a different sales value (like our function f(x, y)). You start with a certain number of cookies and brownies and adjust based on how close you are to your sales goal. Just like in our problem, you’ll find the perfect combination of treats to maximize your sales!