Maximum Score From Removing Stones

Problem Description

So, you think you can just sit back and relax while the stones do all the work? Think again! The “Maximum Score From Removing Stones” problem is like trying to decide which of your friends to invite to a party based on how many snacks they can bring. You have three piles of stones, and you need to figure out how to maximize your score by removing stones in pairs. But wait, there’s a catch! You can only remove stones from two different piles at a time. It’s like trying to balance your love for pizza and tacos at a buffet—how do you choose?

Code Solution

class Solution {
 public:
  int maximumScore(int a, int b, int c) {
    // the maximum <= the minimum + the middle
    const int x = (a + b + c) / 2;
    // the maximum > the minimum + the middle
    const int y = a + b + c - max({a, b, c});
    return min(x, y);
  }
};
    

Approach

The approach here is as simple as pie (or should we say, as simple as a stone?). The code calculates two potential maximum scores based on the total number of stones and the maximum pile. It then returns the minimum of these two values. This ensures that you’re not overreaching and trying to remove more stones than you can handle. It’s like knowing your limits at an all-you-can-eat buffet—don’t bite off more than you can chew!

Time and Space Complexity

• Time Complexity: O(1) – The solution runs in constant time since it only involves a few arithmetic operations and comparisons.
• Space Complexity: O(1) – The space used does not depend on the input size; it remains constant.

Real-World Example

Imagine you’re at a game night with your friends, and you have three different types of snacks: chips, cookies, and candy. You can only take two types of snacks at a time. The goal? To maximize your snack intake without overloading your plate. Just like in the problem, you need to strategize which snacks to take to ensure you get the most delicious combinations without running out of options.