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

Max Chunks To Make Sorted (LeetCode #769) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum value encountered at each index helps determine valid chunks.

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

The optimal solution for Max Chunks To Make Sorted runs in O(n) time and uses O(1) space. This complexity is linear because we only make a single pass through the array, updating the maximum value and counting chunks.

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

The brute force approach for Max Chunks To Make Sorted is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible way to split the array into chunks and sorting each chunk to see if the concatenated result matches the sorted array. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Always consider simpler approaches first; they can lead to insights for optimization. Be clear about the properties of the input (e.g., permutations) and how they can simplify the problem. Practice explaining your thought process clearly, as communication is key in interviews.

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

Overcomplicating the problem by trying to sort chunks instead of tracking maximum values. Not recognizing that the problem is about the relative positions of elements.

#769

Max Chunks To Make Sorted

Medium

ArrayStackGreedySortingMonotonic StackGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the property of permutations. By keeping track of the maximum value encountered so far, we can determine how many chunks we can form without needing to sort.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum value seen as we iterate through the array.
2Step 2: For each element in the array, update the maximum value.
3Step 3: If the current index matches the maximum value, it indicates a valid chunk, increment the chunk count.

solution.py9 lines

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

ℹ

Complexity note: This complexity is linear because we only make a single pass through the array, updating the maximum value and counting chunks.

1The maximum value encountered at each index helps determine valid chunks.
2The problem can be solved in linear time by leveraging properties of permutations.

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