What is the time complexity of Find the Number of Subarrays Where Boundary Elements Are Maximum?

The optimal solution for Find the Number of Subarrays Where Boundary Elements Are Maximum runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the storage of left and right indices.

What is the brute force solution for Find the Number of Subarrays Where Boundary Elements Are Maximum?

The brute force approach for Find the Number of Subarrays Where Boundary Elements Are Maximum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if it meets the criteria. This is straightforward but inefficient, as it involves checking all pairs of indices. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Number of Subarrays Where Boundary Elements Are Maximum?

Find the Number of Subarrays Where Boundary Elements Are Maximum uses the following concepts: Array, Binary Search, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Two Pointers, Sliding Window.

What are the interview tips for Find the Number of Subarrays Where Boundary Elements Are Maximum?

Explain your thought process clearly while coding. Discuss the time and space complexity of your approach. Practice explaining your code to someone else.

What are common mistakes when solving Find the Number of Subarrays Where Boundary Elements Are Maximum?

Not considering edge cases like all elements being the same. Failing to optimize the search for maximum elements.

#3113

Find the Number of Subarrays Where Boundary Elements Are Maximum

Hard

ArrayBinary SearchStackMonotonic StackMonotonic StackTwo PointersSliding Window

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Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses a monotonic stack to efficiently find the nearest smaller elements to the left and right of each element. This allows us to calculate valid subarrays in linear time.

⚙️

Algorithm

7 steps

1Step 1: Initialize a stack to keep track of indices and a counter for valid subarrays.
2Step 2: Create an array to store the nearest smaller element indices to the left for each element.
3Step 3: Traverse the array and fill the left nearest smaller indices using the stack.
4Step 4: Create another array for the nearest smaller element indices to the right.
5Step 5: Traverse the array in reverse to fill the right nearest smaller indices using the stack.
6Step 6: For each element, calculate the number of valid subarrays using the left and right indices.
7Step 7: Return the total count of valid subarrays.

solution.py20 lines

1def countSubarrays(nums):
2    n = len(nums)
3    left = [0] * n
4    right = [0] * n
5    stack = []
6    for i in range(n):
7        while stack and nums[stack[-1]] < nums[i]:
8            stack.pop()
9        left[i] = stack[-1] if stack else -1
10        stack.append(i)
11    stack = []
12    for i in range(n-1, -1, -1):
13        while stack and nums[stack[-1]] <= nums[i]:
14            stack.pop()
15        right[i] = stack[-1] if stack else n
16        stack.append(i)
17    count = 0
18    for i in range(n):
19        count += (i - left[i]) * (right[i] - i)
20    return count

ℹ

Complexity note: The time complexity is O(n) because we traverse the array a constant number of times, and the space complexity is O(n) due to the storage of left and right indices.

1Using a monotonic stack can significantly reduce time complexity.
2Understanding boundaries of subarrays is crucial for counting valid ones.

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