How do you solve Maximum Number of Operations With the Same Score I on LeetCode?

Maximum Number of Operations With the Same Score I (LeetCode #3038) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: The score of each operation must match the previous score for it to count.

What is the brute force solution for Maximum Number of Operations With the Same Score I?

The brute force approach for Maximum Number of Operations With the Same Score I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating the operations by continuously removing the first two elements and checking if their sum matches the previous score. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Number of Operations With the Same Score I?

Maximum Number of Operations With the Same Score I uses the following concepts: Array, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Operations With the Same Score I?

Always explain your thought process while coding. Consider edge cases and test your solution with them. Discuss time and space complexity of your approach.

What are common mistakes when solving Maximum Number of Operations With the Same Score I?

Not considering all pairs when calculating scores. Failing to handle edge cases where the array length is minimal.

#3038

Maximum Number of Operations With the Same Score I

Easy

ArraySimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal approach involves using a frequency map to count occurrences of each possible score. This allows us to determine how many pairs can produce the same score efficiently without simulating each operation.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count how many times each possible score can be formed by summing pairs of elements.
2Step 2: For each unique score in the frequency map, calculate how many pairs can be formed and keep track of the maximum number of operations.
3Step 3: Return the maximum count of operations found.

solution.py10 lines

1from collections import defaultdict
2
3def maxOperations(nums):
4    score_count = defaultdict(int)
5    n = len(nums)
6    for i in range(n):
7        for j in range(i + 1, n):
8            score = nums[i] + nums[j]
9            score_count[score] += 1
10    return max(score_count.values(), default=0)

ℹ

Complexity note: While the time complexity remains O(n²) due to the nested loops, the space complexity improves to O(n) because we store the scores in a frequency map instead of modifying the original array.

1The score of each operation must match the previous score for it to count.
2Using a frequency map allows us to efficiently track possible scores.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.