Find the Number of Subarrays Where Boundary Elements Are Maximum

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Arrays

You are given an array of positive integers nums.

Return the number of subarrays of nums, where the first and the last elements of the subarray are equal to the largest element in the subarray.

Example 1:

Input: nums = [1,4,3,3,2]

Output: 6

Explanation:

There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:

subarray [1,4,3,3,2], with its largest element 1. The first element is 1 and the last element is also 1.
subarray [1,4,3,3,2], with its largest element 4. The first element is 4 and the last element is also 4.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [1,4,3,3,2], with its largest element 2. The first element is 2 and the last element is also 2.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.

Hence, we return 6.

Example 2:

Input: nums = [3,3,3]

Output: 6

Explanation:

There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:

subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.

Hence, we return 6.

Example 3:

Input: nums = [1]

Output: 1

Explanation:

There is a single subarray of nums which is [1], with its largest element 1. The first element is 1 and the last element is also 1.

Hence, we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

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 in the input array, and can it contain negative numbers or zeros?
What should I return if the input array is empty or null?
If there are multiple subarrays where the boundary elements are the maximum value within that subarray, should I return the count of all such subarrays?
Are duplicates allowed in the input array, and if so, how should they be handled when determining the maximum value within a subarray?
Can you provide a concrete example to illustrate what constitutes a valid subarray according to the problem definition?

Brute Force Solution

Approach

The goal is to find all groups of numbers where the first and last number are the largest in that group. The brute force way checks every possible group and sees if it fits the criteria.

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

Start by considering the first number by itself as a group.
Check if the first number is the largest number in this tiny group.
Next, consider the first two numbers as a group.
Check if the first and last numbers in this group are the largest numbers in this group.
Keep adding one more number to the group each time.
For each group you make, check if the numbers at the beginning and the end are the largest in the whole group.
If they are, count this group.
Repeat this process, starting with the second number as the beginning of a new group, and so on.
Continue until you've checked all possible groups of numbers.

Code Implementation

def find_number_of_subarrays_where_boundary_elements_are_maximum(numbers):
    number_of_subarrays = 0
    array_length = len(numbers)

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

            # Handle the case when the subarray has only one element
            if subarray_length == 1:

                number_of_subarrays += 1
                continue

            first_element = subarray[0]
            last_element = subarray[-1]

            # Boundary elements must be the same to be maximum in subarray
            if first_element != last_element:
                continue

            is_boundary_elements_maximum = True
            for element in subarray:

                # Ensure that the boundary elements are the largest
                if element > first_element:
                    is_boundary_elements_maximum = False
                    break

            if is_boundary_elements_maximum:
                number_of_subarrays += 1

    return number_of_subarrays

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array, considering each element as the potential start of a subarray. For each starting element, it then expands the subarray one element at a time. This creates a nested loop structure. For each subarray, it needs to find the maximum element within that subarray to determine if the boundary elements are the maximum, taking O(n) time per subarray. Therefore the time complexity is dominated by examining all possible subarrays, the number of which is proportional to n * (n+1) / 2, which simplifies to O(n²).

Space Complexity

O(1) – The provided algorithm checks subarrays in place. It does not create any auxiliary data structures like lists, hash maps, or recursion stacks to store intermediate results or visited elements. Only a few constant space variables are used for looping and comparing elements within the input array. Therefore, the space complexity remains constant regardless of the input array's size (N).

Optimal Solution

Approach

The most efficient way to solve this problem is to focus on identifying valid subarrays directly without checking every single possible subarray. We'll leverage the key property that the boundary elements must be the maximum to quickly determine if a subarray meets the criteria.

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

Go through the numbers one by one.
For each number, consider it as the starting point of a potential subarray.
Now, extend that potential subarray, number by number, to the right.
At each step, check two things: if the numbers at the start and end of your subarray are the same and also the largest number inside that subarray.
If both are true, then we found a valid subarray, so count it.
Keep extending and checking until you hit a number where the boundary condition is no longer satisfied.
Move to the next starting number and repeat the process.
By only checking subarrays that *could* be valid (where the boundaries are potentially maximums), we avoid wasting time on subarrays that obviously don't fit the rule.

Code Implementation

def find_number_of_subarrays(arr):
    array_length = len(arr)
    subarray_count = 0

    for start_index in range(array_length):
        for end_index in range(start_index, array_length):
            subarray = arr[start_index:end_index + 1]
            subarray_length = len(subarray)

            #Checking for boundary conditions.
            if subarray_length > 0 and subarray[0] == subarray[-1]:

                is_valid = True
                maximum_element = subarray[0]

                #Verify the boundary is the max.
                for element in subarray:
                    if element > maximum_element:
                        is_valid = False
                        break

                if is_valid:
                    subarray_count += 1

    return subarray_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each element of the array (n elements) as a potential starting point of a subarray. For each starting element, it extends the subarray to the right, checking if the boundary elements are equal and also the maximum within that subarray. In the worst-case scenario, for each starting element, the subarray can extend to the end of the array, requiring up to n comparisons. Therefore, the algorithm has a time complexity of approximately n * n, which simplifies to O(n²).

Space Complexity

O(1) – The provided approach iterates through the array and extends potential subarrays using a few variables. It does not use any auxiliary data structures like lists, hash maps, or sets to store intermediate results or track visited elements. The algorithm only requires a constant number of variables (e.g., loop counters, current subarray boundaries, and maximum value) regardless of the size N of the input array. Thus, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty input array	Return 0 since no subarrays can be formed.
Array with only one element	Return 0 since a subarray needs at least two elements.
Array with all identical elements	The number of valid subarrays will depend on array length, and can be derived mathematically.
Array with negative numbers, zeros, and positive numbers	The solution should handle all numerical values without special treatment assuming comparison operators work correctly.
Array with large integer values that could cause overflow	Ensure the comparison and calculation do not cause integer overflow in chosen language.
Array where no subarray satisfies the condition	Return 0 indicating no valid subarrays were found.
Array with extreme boundary values (e.g., min/max int)	Handle integer comparison between min/max values appropriately without overflow.
Very large array size	Ensure the solution scales efficiently, considering time complexity and potential memory usage for intermediate data structures.

Empty input array

How to Handle:

Return 0 since no subarrays can be formed.

Array with only one element

How to Handle:

Return 0 since a subarray needs at least two elements.

Array with all identical elements

How to Handle:

The number of valid subarrays will depend on array length, and can be derived mathematically.

Array with negative numbers, zeros, and positive numbers

How to Handle:

The solution should handle all numerical values without special treatment assuming comparison operators work correctly.

Array with large integer values that could cause overflow

How to Handle:

Ensure the comparison and calculation do not cause integer overflow in chosen language.

Array where no subarray satisfies the condition

How to Handle:

Return 0 indicating no valid subarrays were found.

Array with extreme boundary values (e.g., min/max int)

How to Handle:

Handle integer comparison between min/max values appropriately without overflow.

Very large array size

How to Handle:

Ensure the solution scales efficiently, considering time complexity and potential memory usage for intermediate data structures.