How do you solve Max Chunks To Make Sorted II on LeetCode?

Max Chunks To Make Sorted II (LeetCode #768) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding how to compare the maximum values helps in determining valid chunks.

What is the time complexity of Max Chunks To Make Sorted II?

The optimal solution for Max Chunks To Make Sorted II runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the array, while the space complexity is O(n) for storing the sorted array.

What is the brute force solution for Max Chunks To Make Sorted II?

The brute force approach for Max Chunks To Make Sorted II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible chunks and checking if sorting each chunk results in the entire array being sorted. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Max Chunks To Make Sorted II?

Max Chunks To Make Sorted II uses the following concepts: Array, Stack, Greedy, Sorting, Monotonic Stack. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Max Chunks To Make Sorted II?

Always consider simpler approaches first before optimizing. Explain your thought process clearly when discussing your approach. Practice dry runs to ensure you understand how your algorithm works on different inputs.

What are common mistakes when solving Max Chunks To Make Sorted II?

Failing to correctly compare the maximum value with the sorted array. Overcomplicating the chunk generation process instead of using a greedy approach.

#768

Max Chunks To Make Sorted II

Hard

ArrayStackGreedySortingMonotonic StackGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses a greedy strategy to determine where to split the array by comparing the maximum value seen so far with the sorted version of the array. This allows us to find valid chunks efficiently.

⚙️

Algorithm

4 steps

1Step 1: Create a sorted version of the array.
2Step 2: Initialize variables for the current maximum and chunk count.
3Step 3: Iterate through the array, updating the current maximum and comparing it with the sorted version at each index.
4Step 4: If the current maximum equals the index, increment the chunk count.

solution.py9 lines

1def maxChunksToSorted(arr):
2    sorted_arr = sorted(arr)
3    max_chunks = 0
4    current_max = 0
5    for i in range(len(arr)):
6        current_max = max(current_max, arr[i])
7        if current_max == sorted_arr[i]:
8            max_chunks += 1
9    return max_chunks

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the array, while the space complexity is O(n) for storing the sorted array.

1Understanding how to compare the maximum values helps in determining valid chunks.
2Sorting the array is crucial to find the correct positions for the chunks.

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