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Maximum Length of Semi-Decreasing Subarrays

Medium

Asked by:

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Topics:

Arrays

You are given a 0-indexed integer array nums.

A subarray of nums is called semi-decreasing if its first element is greater than or equal to its last element.

Return the length of the longest semi-decreasing subarray of nums.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [3,4,2,1,5]
Output: 2
Explanation: The semi-decreasing subarrays of this array are [3], [4], [2], [1], [5], [3, 4], [2, 1], [4, 2, 1], and [1, 5].
The longest semi-decreasing subarray is [4, 2, 1] so we return 3.

Example 2:

Input: nums = [5,4,3,2,1]
Output: 5
Explanation: Every subarray of nums is semi-decreasing. There fore, the longest one is nums itself, so we return 5.

Example 3:

Input: nums = [1,2,3,4,5]
Output: 1
Explanation: The semi-decreasing subarrays of this array are [1], [2], [3], [4], and [5].
There fore, the longest one has length 1, so we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

Maximum Length of Semi-Decreasing Subarrays

Medium

Asked by:

14 views

Topics:

Arrays

You are given a 0-indexed integer array nums.

A subarray of nums is called semi-decreasing if its first element is greater than or equal to its last element.

Return the length of the longest semi-decreasing subarray of nums.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [3,4,2,1,5]
Output: 2
Explanation: The semi-decreasing subarrays of this array are [3], [4], [2], [1], [5], [3, 4], [2, 1], [4, 2, 1], and [1, 5].
The longest semi-decreasing subarray is [4, 2, 1] so we return 3.

Example 2:

Input: nums = [5,4,3,2,1]
Output: 5
Explanation: Every subarray of nums is semi-decreasing. There fore, the longest one is nums itself, so we return 5.

Example 3:

Input: nums = [1,2,3,4,5]
Output: 1
Explanation: The semi-decreasing subarrays of this array are [1], [2], [3], [4], and [5].
There fore, the longest one has length 1, so we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

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 integer values within the input array, and can I expect negative numbers?
If the input array is empty or null, what should the function return?
Are duplicate values allowed in the array, and if so, how should they be handled when determining the length of a semi-decreasing subarray?
Could you define 'semi-decreasing'? Specifically, if `arr[i] >= arr[i+1]` for all elements within a subarray, is that considered semi-decreasing?
If there are multiple semi-decreasing subarrays with the same maximum length, is it acceptable to return the length of any one of them?

Brute Force Solution

Approach

The brute force approach to finding the longest semi-decreasing part in a list is like trying every possible slice of the list and checking if it follows the rule. We'll look at all combinations, no matter how inefficient.

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

Start by considering the first number in the list as the beginning of a potential semi-decreasing part.
Then, look at the first and second numbers together. Check if they are semi-decreasing (each number must be less than or equal to the number two positions before it).
Next, look at the first three numbers, then the first four, and so on, each time checking if that slice is semi-decreasing.
Repeat the process, starting now with the second number in the list. Take the second number alone, then the second and third numbers, then the second, third, and fourth numbers, and so on, again checking the semi-decreasing rule for each slice.
Keep doing this, moving the starting point further down the list each time, until you've tried every possible slice.
Keep track of the length of each semi-decreasing slice you find.
Finally, compare all the lengths you've recorded and choose the longest one. That's your answer.

Code Implementation

def find_maximum_length_of_semi_decreasing_subarrays(numbers):
    maximum_length = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            sub_array = numbers[start_index:end_index + 1]

            #  A sub array of length 0 or 1 is semi-decreasing
            if len(sub_array) <= 1:
                maximum_length = max(maximum_length, len(sub_array))

            else:
                is_semi_decreasing = True

                # Iterate through sub array to check if it is semi-decreasing
                for index in range(2, len(sub_array)):

                    # This conditional checks if the current number
                    # is greater than the number two indices before it
                    if sub_array[index] > sub_array[index - 2]:
                        is_semi_decreasing = False
                        break

                #  If sub array is semi-decreasing, compare its length
                #  to the existing maximum length.
                if is_semi_decreasing:
                    maximum_length = max(maximum_length, len(sub_array))

    return maximum_length

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible subarrays of the input array of size n. For each starting index i, it considers all subarrays starting at i. The number of such subarrays starting at index i is n-i. Summing this up for all i from 0 to n-1 gives us approximately n + (n-1) + (n-2) + ... + 1, which is a sum of an arithmetic series. This sum can be expressed as n*(n+1)/2. Therefore, the total number of operations is proportional to n², leading to a time complexity of O(n²).

Space Complexity

O(1) – The brute force approach iterates through all possible subarrays but does not create any auxiliary data structures dependent on the input size N (the length of the input list). It only uses a few constant space variables to store the current subarray's length and the maximum length found so far. Thus, the algorithm's space complexity is constant and independent of the input size.

Optimal Solution

Approach

The most efficient way to solve this problem is to move through the data once, tracking the potential 'good' sections as we go. We'll keep track of the longest good section seen so far and update it whenever we find a bigger one.

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

Start at the beginning of the data.
Consider the current spot as the potential start of a 'good' section.
As we move forward, check if the numbers are decreasing or staying the same. If they are, the 'good' section is still valid.
If we find a number that is bigger than the one before it, the 'good' section ends.
When a 'good' section ends, calculate its length. If this length is bigger than the longest 'good' section we've seen so far, remember this new length.
If a 'good' section ends, the next number could be the start of a new 'good' section, so consider that.
Keep going until we've checked all the numbers. The longest 'good' section we remembered is the answer.

Code Implementation

def find_maximum_length_of_semi_decreasing_subarrays(data):
    maximum_length = 0
    current_subarray_start_index = 0
    
    for current_index in range(1, len(data)):

        # Check for the end of the subarray
        if data[current_index] > data[current_index - 1]:

            current_subarray_length = current_index - current_subarray_start_index
            maximum_length = max(maximum_length, current_subarray_length)

            # Start a new subarray from the current index.
            current_subarray_start_index = current_index

    # Process the last subarray if it exists
    current_subarray_length = len(data) - current_subarray_start_index
    maximum_length = max(maximum_length, current_subarray_length)

    return maximum_length

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once. In each iteration, it performs constant-time operations like comparing the current element with the previous one, updating the current subarray length, and updating the maximum subarray length seen so far. Therefore, the total number of operations is directly proportional to the size of the input, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It keeps track of the current 'good' section's start and calculates its length, and it stores the maximum length seen so far. These are all done using a few integer variables and do not depend on the input size N, where N is the number of elements in the input array. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no subarray can be formed.
Array with a single element	Return 1 as a single element is trivially semi-decreasing.
Array with two elements in decreasing order	Return 2 as it's a semi-decreasing subarray.
Array with two elements in increasing order	Return 1 as the longest semi-decreasing subarray has length 1 (either element).
Array with all identical elements	Return 1 as only single elements are semi-decreasing.
Array with large numbers that might cause integer overflow during calculations (if relevant)	Use appropriate data types (e.g., long) to prevent overflow.
Array is already sorted in strictly decreasing order	The entire array is a semi-decreasing subarray, so return the array's length.
Array contains negative numbers or zero	The algorithm should handle negative numbers and zero correctly as they can be part of a semi-decreasing sequence.

Empty input array

How to Handle:

Return 0 since no subarray can be formed.

Array with a single element

How to Handle:

Return 1 as a single element is trivially semi-decreasing.

Array with two elements in decreasing order

How to Handle:

Return 2 as it's a semi-decreasing subarray.

Array with two elements in increasing order

How to Handle:

Return 1 as the longest semi-decreasing subarray has length 1 (either element).

Array with all identical elements

How to Handle:

Return 1 as only single elements are semi-decreasing.

Array with large numbers that might cause integer overflow during calculations (if relevant)

How to Handle:

Use appropriate data types (e.g., long) to prevent overflow.

Array is already sorted in strictly decreasing order

How to Handle:

The entire array is a semi-decreasing subarray, so return the array's length.

Array contains negative numbers or zero

How to Handle:

The algorithm should handle negative numbers and zero correctly as they can be part of a semi-decreasing sequence.

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 integer values within the input array, and can I expect negative numbers?
If the input array is empty or null, what should the function return?
Are duplicate values allowed in the array, and if so, how should they be handled when determining the length of a semi-decreasing subarray?
Could you define 'semi-decreasing'? Specifically, if `arr[i] >= arr[i+1]` for all elements within a subarray, is that considered semi-decreasing?
If there are multiple semi-decreasing subarrays with the same maximum length, is it acceptable to return the length of any one of them?

Brute Force Solution

Approach

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

Start by considering the first number in the list as the beginning of a potential semi-decreasing part.
Then, look at the first and second numbers together. Check if they are semi-decreasing (each number must be less than or equal to the number two positions before it).
Next, look at the first three numbers, then the first four, and so on, each time checking if that slice is semi-decreasing.
Repeat the process, starting now with the second number in the list. Take the second number alone, then the second and third numbers, then the second, third, and fourth numbers, and so on, again checking the semi-decreasing rule for each slice.
Keep doing this, moving the starting point further down the list each time, until you've tried every possible slice.
Keep track of the length of each semi-decreasing slice you find.
Finally, compare all the lengths you've recorded and choose the longest one. That's your answer.

Code Implementation

def find_maximum_length_of_semi_decreasing_subarrays(numbers):
    maximum_length = 0

    for start_index in range(len(numbers)):
        for end_index in range(start_index, len(numbers)):
            sub_array = numbers[start_index:end_index + 1]

            #  A sub array of length 0 or 1 is semi-decreasing
            if len(sub_array) <= 1:
                maximum_length = max(maximum_length, len(sub_array))

            else:
                is_semi_decreasing = True

                # Iterate through sub array to check if it is semi-decreasing
                for index in range(2, len(sub_array)):

                    # This conditional checks if the current number
                    # is greater than the number two indices before it
                    if sub_array[index] > sub_array[index - 2]:
                        is_semi_decreasing = False
                        break

                #  If sub array is semi-decreasing, compare its length
                #  to the existing maximum length.
                if is_semi_decreasing:
                    maximum_length = max(maximum_length, len(sub_array))

    return maximum_length

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the beginning of the data.
Consider the current spot as the potential start of a 'good' section.
As we move forward, check if the numbers are decreasing or staying the same. If they are, the 'good' section is still valid.
If we find a number that is bigger than the one before it, the 'good' section ends.
When a 'good' section ends, calculate its length. If this length is bigger than the longest 'good' section we've seen so far, remember this new length.
If a 'good' section ends, the next number could be the start of a new 'good' section, so consider that.
Keep going until we've checked all the numbers. The longest 'good' section we remembered is the answer.

Code Implementation

def find_maximum_length_of_semi_decreasing_subarrays(data):
    maximum_length = 0
    current_subarray_start_index = 0
    
    for current_index in range(1, len(data)):

        # Check for the end of the subarray
        if data[current_index] > data[current_index - 1]:

            current_subarray_length = current_index - current_subarray_start_index
            maximum_length = max(maximum_length, current_subarray_length)

            # Start a new subarray from the current index.
            current_subarray_start_index = current_index

    # Process the last subarray if it exists
    current_subarray_length = len(data) - current_subarray_start_index
    maximum_length = max(maximum_length, current_subarray_length)

    return maximum_length

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no subarray can be formed.
Array with a single element	Return 1 as a single element is trivially semi-decreasing.
Array with two elements in decreasing order	Return 2 as it's a semi-decreasing subarray.
Array with two elements in increasing order	Return 1 as the longest semi-decreasing subarray has length 1 (either element).
Array with all identical elements	Return 1 as only single elements are semi-decreasing.
Array with large numbers that might cause integer overflow during calculations (if relevant)	Use appropriate data types (e.g., long) to prevent overflow.
Array is already sorted in strictly decreasing order	The entire array is a semi-decreasing subarray, so return the array's length.
Array contains negative numbers or zero	The algorithm should handle negative numbers and zero correctly as they can be part of a semi-decreasing sequence.

Empty input array

How to Handle:

Return 0 since no subarray can be formed.

Array with a single element

How to Handle:

Return 1 as a single element is trivially semi-decreasing.

Array with two elements in decreasing order

How to Handle:

Return 2 as it's a semi-decreasing subarray.

Array with two elements in increasing order

How to Handle:

Return 1 as the longest semi-decreasing subarray has length 1 (either element).

Array with all identical elements

How to Handle:

Return 1 as only single elements are semi-decreasing.

Array with large numbers that might cause integer overflow during calculations (if relevant)

How to Handle:

Use appropriate data types (e.g., long) to prevent overflow.

Array is already sorted in strictly decreasing order

How to Handle:

The entire array is a semi-decreasing subarray, so return the array's length.

Array contains negative numbers or zero

How to Handle:

The algorithm should handle negative numbers and zero correctly as they can be part of a semi-decreasing sequence.