How do you solve Maximize Total Cost of Alternating Subarrays on LeetCode?

Maximize Total Cost of Alternating Subarrays (LeetCode #3196) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding alternating sums is crucial.

What is the brute force solution for Maximize Total Cost of Alternating Subarrays?

The brute force approach for Maximize Total Cost of Alternating Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calculating the cost of every possible subarray and then trying all possible ways to split the array into subarrays. This is simple but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximize Total Cost of Alternating Subarrays?

Maximize Total Cost of Alternating Subarrays uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Maximize Total Cost of Alternating Subarrays?

Explain your thought process clearly while coding. Consider edge cases and clarify constraints with the interviewer. Discuss time and space complexities as you develop your solution.

What are common mistakes when solving Maximize Total Cost of Alternating Subarrays?

Not considering the need to split into at least two subarrays. Forgetting to handle edge cases like single-element arrays.

#3196

Maximize Total Cost of Alternating Subarrays

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to keep track of the maximum cost at each index while considering the alternating sum. This avoids recalculating costs for overlapping subarrays.

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Algorithm

3 steps

1Step 1: Initialize a DP array where dp[i] represents the maximum cost ending at index i.
2Step 2: Iterate through the array and calculate the cost for each index based on the previous index's cost.
3Step 3: Update the maximum total cost when a new subarray starts.

solution.py17 lines

1# Full working Python code
2
3def maxCost(nums):
4    n = len(nums)
5    if n == 1:
6        return 0
7    dp = [0] * n
8    dp[0] = nums[0]
9    max_cost = 0
10    for i in range(1, n):
11        dp[i] = dp[i - 1] + (nums[i] if i % 2 == 1 else -nums[i])
12        if i > 1:
13            max_cost = max(max_cost, dp[i])
14    return max_cost
15
16# Example usage
17print(maxCost([1, -2, 3, 4]))  # Output: 10

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the array once, and the space complexity is O(n) due to the DP array used to store intermediate results.

1Understanding alternating sums is crucial.
2Dynamic programming helps to optimize overlapping subproblems.

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