What is the time complexity of Maximum Number of Operations With the Same Score II?

The optimal solution for Maximum Number of Operations With the Same Score II runs in O(n²) time and uses O(n²) space. The complexity is O(n²) due to the nested loops for filling the DP table, but we use O(n²) space to store results for subarrays.

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

The brute force approach for Maximum Number of Operations With the Same Score II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of operations on the array to see how many can yield the same score. This is straightforward but inefficient as it explores all sequences of operations. 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 II?

Maximum Number of Operations With the Same Score II uses the following concepts: Array, Dynamic Programming, Memoization. The recommended patterns to study are: Dynamic Programming, Two Pointers.

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

Always clarify the problem statement and constraints before diving into coding. Think aloud to explain your thought process; it helps interviewers understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

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

Failing to consider all possible pairs for scores can lead to incorrect results. Not properly managing the state of the array during operations can cause errors.

#3040

Maximum Number of Operations With the Same Score II

Medium

ArrayDynamic ProgrammingMemoizationDynamic ProgrammingTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to track the maximum number of operations that can be performed for each possible score. This reduces redundant calculations and efficiently narrows down the possibilities.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[l][r] represents the maximum operations possible for the subarray nums[l..r].
2Step 2: Iterate through all possible pairs of indices to calculate scores and update the DP table accordingly.
3Step 3: Use previously computed results to build up the solution for larger subarrays.

solution.py18 lines

1# Full working Python code
2
3def maxOperations(nums):
4    n = len(nums)
5    dp = [[0] * n for _ in range(n)]
6    for length in range(2, n + 1):
7        for l in range(n - length + 1):
8            r = l + length - 1
9            score1 = nums[l] + nums[l + 1]
10            score2 = nums[r] + nums[r - 1]
11            score3 = nums[l] + nums[r]
12            if score1 == score2:
13                dp[l][r] = max(dp[l][r], dp[l + 2][r] + 1)
14            if score1 == score3:
15                dp[l][r] = max(dp[l][r], dp[l + 1][r - 1] + 1)
16            if score2 == score3:
17                dp[l][r] = max(dp[l][r], dp[l][r - 2] + 1)
18    return dp[0][n - 1]

ℹ

Complexity note: The complexity is O(n²) due to the nested loops for filling the DP table, but we use O(n²) space to store results for subarrays.

1Understanding how to break down the problem into subproblems is crucial for dynamic programming.
2Recognizing patterns in operations can lead to more efficient solutions.

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