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Max Chunks To Make Sorted

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ArraysGreedy Algorithms

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.

Try coding it in LeetCode

Max Chunks To Make Sorted

Medium

Asked by:

27 views

Topics:

ArraysGreedy Algorithms

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array?
Can the input array be empty or null?
Are duplicate numbers allowed in the array, and if so, how should they be handled when determining chunks?
If the array is already sorted, what should the return value be?
Is the input array guaranteed to be a permutation of numbers from 0 to n-1, where n is the length of the array?

Brute Force Solution

Approach

The brute force approach for this problem is all about trying every possible way to split up the numbers we're given. We will check if each possible split creates sections that, when sorted individually and then combined, result in the fully sorted version of the original numbers.

Here's how the algorithm would work step-by-step:

Consider every possible place you could make a split in the numbers.
For each possible set of splits, imagine cutting the original group of numbers into smaller groups at those split points.
Individually sort each of those smaller groups of numbers.
Now, put the sorted smaller groups back together in their original order.
Check if the resulting combined group of numbers is completely sorted.
If combining the sorted smaller groups produces a completely sorted result, count that split arrangement as a valid way to break up the numbers.
After checking every possible split arrangement, the answer is the maximum number of groups from all valid arrangements.

Code Implementation

def max_chunks_to_sorted_brute_force(numbers):
    number_length = len(numbers)
    max_chunks = 0

    for i in range(1 << (number_length - 1)):
        # Iterate through all possible split combinations
        splits = []
        for j in range(number_length - 1):
            if (i >> j) & 1:
                splits.append(j + 1)

        chunks = []
        start_index = 0

        # Create chunks based on the split points
        for split_index in splits:
            chunks.append(numbers[start_index:split_index])
            start_index = split_index
        chunks.append(numbers[start_index:])

        sorted_chunks = [sorted(chunk) for chunk in chunks]

        merged_array = []
        for chunk in sorted_chunks:
            merged_array.extend(chunk)

        # If the merged array is sorted, increment max_chunks
        if merged_array == sorted(numbers):
            max_chunks = max(max_chunks, len(chunks))

    # Handle the no-split case separately
    if sorted(numbers) == sorted(numbers):
        max_chunks = max(max_chunks, 1)
    
    return max_chunks

Big(O) Analysis

Time Complexity

O(n! * n log n) – The brute force approach considers every possible split of the array, leading to O(2^(n-1)) potential partitions. For each partition, we must sort the resulting chunks, which takes O(n log n) time overall, considering the total number of elements remains n. Checking if the combined, sorted chunks are sorted takes O(n) time. Since 2^(n-1) can be expressed as O(n!), the dominant factor is the enumeration of all possible splits combined with sorting. Therefore, the overall time complexity is O(n! * n log n).

Space Complexity

O(N) – The brute force approach, as described, generates all possible splits of the input array. For each split, it sorts sub-arrays, requiring temporary storage. The most space-intensive scenario arises when the entire input array needs to be copied into a temporary array for sorting within a particular split configuration. Therefore, the auxiliary space used is proportional to the input size N, leading to a space complexity of O(N).

Optimal Solution

Approach

The key idea is to figure out where we can cut the input sequence into independent pieces. We track the largest value we've seen so far and split the sequence whenever that maximum value equals the position we are at.

Here's how the algorithm would work step-by-step:

Imagine each number in the list as being in its 'correct' spot if the list were sorted.
As we move through the list, keep track of the largest number we've encountered up to that point.
If the largest number we've seen is the same as our current position, it means everything we've seen so far can form a sorted chunk. We can then split the list at that point.
Each time you've split the list, count it as a 'chunk'.
Continue moving through the list and applying the same process. The final count of 'chunks' is your answer.

Code Implementation

def max_chunks_to_sorted(arr):
    chunk_count = 0
    max_value_seen = 0

    for i in range(len(arr)):
        max_value_seen = max(max_value_seen, arr[i])

        # If max so far equals index, it's a sorted chunk.
        if max_value_seen == i:
            # Increment chunk count because we found one.
            chunk_count += 1

    return chunk_count

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once. In each iteration, it performs a constant number of operations: updating the maximum value seen so far and comparing the maximum value to the current index. Therefore, the time complexity is directly proportional to the size of the input array, resulting in O(n) time complexity.

Space Complexity

O(1) – The algorithm maintains a 'max' variable to track the largest number encountered so far and a 'chunks' variable to count the number of chunks. These are the only additional variables needed beyond the input array itself. The amount of memory required for these variables does not depend on the size of the input array (N), and remains constant. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no chunks are possible without an array.
Array with a single element	Return 1 since a single element is always sorted.
Array already sorted in ascending order	Return 1, as the entire array is one chunk.
Array sorted in descending order	Return the length of the array, as each element must be a separate chunk.
Array with duplicate values	The algorithm should correctly handle duplicates by tracking the maximum value seen so far.
Large array with integer overflow potential when calculating sums or maximums	Use appropriate data types (e.g., long) or check for overflow during calculations.
Array with a wide range of values, including negative numbers	The algorithm should correctly handle both positive and negative values when determining chunk boundaries.
Maximum array size leading to memory issues	Consider using an in-place algorithm or a streaming approach if memory is severely limited.

Null or empty input array

How to Handle:

Return 0 since no chunks are possible without an array.

Array with a single element

How to Handle:

Return 1 since a single element is always sorted.

Array already sorted in ascending order

How to Handle:

Return 1, as the entire array is one chunk.

Array sorted in descending order

How to Handle:

Return the length of the array, as each element must be a separate chunk.

Array with duplicate values

How to Handle:

The algorithm should correctly handle duplicates by tracking the maximum value seen so far.

Large array with integer overflow potential when calculating sums or maximums

How to Handle:

Use appropriate data types (e.g., long) or check for overflow during calculations.

Array with a wide range of values, including negative numbers

How to Handle:

The algorithm should correctly handle both positive and negative values when determining chunk boundaries.

Maximum array size leading to memory issues

How to Handle:

Consider using an in-place algorithm or a streaming approach if memory is severely limited.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the input array?
Can the input array be empty or null?
Are duplicate numbers allowed in the array, and if so, how should they be handled when determining chunks?
If the array is already sorted, what should the return value be?
Is the input array guaranteed to be a permutation of numbers from 0 to n-1, where n is the length of the array?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider every possible place you could make a split in the numbers.
For each possible set of splits, imagine cutting the original group of numbers into smaller groups at those split points.
Individually sort each of those smaller groups of numbers.
Now, put the sorted smaller groups back together in their original order.
Check if the resulting combined group of numbers is completely sorted.
If combining the sorted smaller groups produces a completely sorted result, count that split arrangement as a valid way to break up the numbers.
After checking every possible split arrangement, the answer is the maximum number of groups from all valid arrangements.

Code Implementation

def max_chunks_to_sorted_brute_force(numbers):
    number_length = len(numbers)
    max_chunks = 0

    for i in range(1 << (number_length - 1)):
        # Iterate through all possible split combinations
        splits = []
        for j in range(number_length - 1):
            if (i >> j) & 1:
                splits.append(j + 1)

        chunks = []
        start_index = 0

        # Create chunks based on the split points
        for split_index in splits:
            chunks.append(numbers[start_index:split_index])
            start_index = split_index
        chunks.append(numbers[start_index:])

        sorted_chunks = [sorted(chunk) for chunk in chunks]

        merged_array = []
        for chunk in sorted_chunks:
            merged_array.extend(chunk)

        # If the merged array is sorted, increment max_chunks
        if merged_array == sorted(numbers):
            max_chunks = max(max_chunks, len(chunks))

    # Handle the no-split case separately
    if sorted(numbers) == sorted(numbers):
        max_chunks = max(max_chunks, 1)
    
    return max_chunks

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Imagine each number in the list as being in its 'correct' spot if the list were sorted.
As we move through the list, keep track of the largest number we've encountered up to that point.
If the largest number we've seen is the same as our current position, it means everything we've seen so far can form a sorted chunk. We can then split the list at that point.
Each time you've split the list, count it as a 'chunk'.
Continue moving through the list and applying the same process. The final count of 'chunks' is your answer.

Code Implementation

def max_chunks_to_sorted(arr):
    chunk_count = 0
    max_value_seen = 0

    for i in range(len(arr)):
        max_value_seen = max(max_value_seen, arr[i])

        # If max so far equals index, it's a sorted chunk.
        if max_value_seen == i:
            # Increment chunk count because we found one.
            chunk_count += 1

    return chunk_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 since no chunks are possible without an array.
Array with a single element	Return 1 since a single element is always sorted.
Array already sorted in ascending order	Return 1, as the entire array is one chunk.
Array sorted in descending order	Return the length of the array, as each element must be a separate chunk.
Array with duplicate values	The algorithm should correctly handle duplicates by tracking the maximum value seen so far.
Large array with integer overflow potential when calculating sums or maximums	Use appropriate data types (e.g., long) or check for overflow during calculations.
Array with a wide range of values, including negative numbers	The algorithm should correctly handle both positive and negative values when determining chunk boundaries.
Maximum array size leading to memory issues	Consider using an in-place algorithm or a streaming approach if memory is severely limited.

Null or empty input array

How to Handle:

Return 0 since no chunks are possible without an array.

Array with a single element

How to Handle:

Return 1 since a single element is always sorted.

Array already sorted in ascending order

How to Handle:

Return 1, as the entire array is one chunk.

Array sorted in descending order

How to Handle:

Return the length of the array, as each element must be a separate chunk.

Array with duplicate values

How to Handle:

The algorithm should correctly handle duplicates by tracking the maximum value seen so far.

Large array with integer overflow potential when calculating sums or maximums

How to Handle:

Use appropriate data types (e.g., long) or check for overflow during calculations.

Array with a wide range of values, including negative numbers

How to Handle:

The algorithm should correctly handle both positive and negative values when determining chunk boundaries.

Maximum array size leading to memory issues

How to Handle:

Consider using an in-place algorithm or a streaming approach if memory is severely limited.