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Max Chunks To Make Sorted

You are given an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1].

We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.

Return the largest number of chunks we can make to sort the array.

 

Example 1:

Input: arr = [4,3,2,1,0]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.

Example 2:

Input: arr = [1,0,2,3,4]
Output: 4
Explanation:
We can split into two chunks, such as [1, 0], [2, 3, 4].
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.

 

Constraints:

  • n == arr.length
  • 1 <= n <= 10
  • 0 <= arr[i] < n
  • All the elements of arr are unique.

Problem: Max Chunks To Make Sorted

Given an integer array arr representing a permutation of integers in the range [0, n-1], we need to split arr into chunks such that when each chunk is individually sorted and concatenated, the result equals the sorted array. The goal is to find the maximum number of chunks possible.

Solution Explanation

There are two main approaches to solve this problem:

Approach 1: Greedy Approach

This approach iterates through the array, keeping track of the maximum value encountered so far (mx). If the current index i equals mx, it means all elements up to index i are within the range [0, i], forming a chunk that can be sorted independently. Therefore, we increment the chunk count (ans).

Time Complexity: O(n), where n is the length of the array. We iterate through the array once. Space Complexity: O(1), constant extra space is used.

Approach 2: Monotonic Stack Approach

This approach utilizes a monotonic stack to track the maximum values encountered. The stack maintains a decreasing order of maximum values. When a value v is encountered, we check if it's greater than or equal to the top of the stack. If it is, we push v onto the stack. Otherwise, we pop elements from the stack until we find an element smaller than or equal to v, indicating that we've encountered a value that requires merging with previous chunks. The final size of the stack represents the maximum number of chunks.

Time Complexity: O(n) in the average case. While the inner while loop could potentially lead to O(n^2) in the worst case, for a permutation, this is unlikely. The number of times elements are popped is less than the total number of pushes which is at most n. Space Complexity: O(n) in the worst case, as the stack might store up to n elements. In the average case it's much smaller.

Code in Different Languages

The code implementations for both approaches are provided in the original response. The code is well-structured and follows best practices. I will not repeat the code here to avoid redundancy. However, I would like to highlight a couple of things:

  • Approach 1 (Greedy): This approach is significantly more efficient both in time and space complexity.
  • Approach 2 (Monotonic Stack): While more complex, it provides a different perspective and illustrates the use of a monotonic stack for solving this kind of problem. It's a good example of a problem where a less efficient solution approach might help one to learn about stack data structure.

It's recommended to use the greedy approach (Approach 1) for its superior efficiency in most cases. Understanding the monotonic stack approach is valuable for its general applicability in other problems. The provided codes in the different languages are all correct and illustrate both approaches efficiently. The comments in the code further aid in understanding the logic of each step.