Maximum Number of Operations With the Same Score I

Ah, the classic problem of finding the maximum number of operations with the same score! It’s like trying to find the perfect pair of socks in a laundry basket—easy in theory, but in practice, you’re left with mismatched pairs and a sense of existential dread.

Imagine you’re at a party, and everyone is trying to play a game where they need to form pairs. But here’s the catch: only pairs with the same score can play together. So, if you have a bunch of friends with different scores, you might end up with a few lonely souls sitting on the sidelines. This problem is all about maximizing those pairs!

Solution Links

Code Solution


class Solution {
  public int maxOperations(int[] nums) {
    int ans = 1;
    int sum = nums[0] + nums[1];

    for (int i = 2; i + 1 < nums.length; i += 2) {
      if (nums[i] + nums[i + 1] == sum)
        ++ans;
      else
        break;
    }

    return ans;
  }
}

Approach Explanation

The code starts by initializing a counter ans to 1, assuming that the first two numbers can form a pair. It then calculates the sum of the first two numbers. The loop iterates through the array in pairs, checking if the sum of each pair matches the initial sum. If it does, it increments the counter; if not, it breaks out of the loop.

In simpler terms, it’s like checking if your friends can still play the game after the first round. If they can, great! If not, it’s time to call it a night.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n)
Space Complexity	O(1)

Real-World Example

Let’s say you’re organizing a dance-off. You have a group of dancers, and they can only pair up if they have the same dance score. You start with two dancers who have a score of 10. If the next pair also has a score of 10, you can keep pairing them up until someone shows up with a score of 5. At that point, the dance-off is over for that round, and you can only count the pairs that danced together.