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Maximum Number of Operations With the Same Score II

Difficulty: Medium

Problem Description

Given an array of integers called nums, you can perform any of the following operations while nums contains at least 2 elements:

  • Choose the first two elements of nums and delete them.
  • Choose the last two elements of nums and delete them.
  • Choose the first and the last elements of nums and delete them.

The score of the operation is the sum of the deleted elements. Your task is to find the maximum number of operations that can be performed, such that all operations have the same score. Return the maximum number of operations possible that satisfy this condition.

Key Insights

  • The operations can be performed in three different ways, all contributing to a score based on the sum of deleted elements.
  • To maximize the number of operations, we need to identify scores that can be achieved repeatedly through different operations.
  • Using a hashmap (or dictionary) can help track how many times each score can be achieved.
  • The problem can be approached by iterating through possible pairs of elements to compute scores and count the frequency of each score.

Space and Time Complexity

Time Complexity: O(n^2) - We iterate through pairs of elements to compute scores. Space Complexity: O(n) - To store the frequency of each score in a hashmap.

Solution

To solve this problem, we can use a hashmap to track the frequency of each possible score that can be obtained through the defined operations. We will iterate through the array to compute scores for:

  • The first two elements.
  • The last two elements.
  • The first and last elements.

After calculating the scores, we will determine which score has the highest frequency and return that count as the maximum number of operations.

Code Solutions

from collections import defaultdict

def maxOperations(nums):
    score_count = defaultdict(int)
    n = len(nums)
    
    for i in range(n):
        for j in range(i + 1, n):
            # Calculate scores for the different operations
            if i + 1 < n:  # First two elements
                score_count[nums[i] + nums[i + 1]] += 1
            if j - 1 >= 0:  # Last two elements
                score_count[nums[j] + nums[j - 1]] += 1
            if i < n and j < n:  # First and last elements
                score_count[nums[i] + nums[j]] += 1
    
    return max(score_count.values(), default=0)
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