How do you solve Count Good Subarrays on LeetCode?

Count Good Subarrays (LeetCode #3878) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: A good subarray's OR equals its maximum element.

What is the time complexity of Count Good Subarrays?

The optimal solution for Count Good Subarrays runs in O(n) time and uses O(n) space. The linear complexity comes from processing each element a constant number of times using the stack.

What is the brute force solution for Count Good Subarrays?

The brute force approach for Count Good Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subarrays and check if the bitwise OR equals the maximum element in each subarray. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Good Subarrays?

Count Good Subarrays uses the following concepts: Array, Stack, Bit Manipulation, Monotonic Stack. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Good Subarrays?

Explain your thought process clearly. Practice similar problems with stacks. Understand bitwise operations thoroughly.

What are common mistakes when solving Count Good Subarrays?

Forgetting to handle equal elements in the stack. Not considering all subarray combinations.

#3878

Count Good Subarrays

Hard

ArrayStackBit ManipulationMonotonic StackHash MapArray

LeetCode ↗

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)

Use a monotonic stack to efficiently find the range where each element is the maximum. This reduces the number of checks needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a stack to keep track of indices and arrays to store left and right bounds for each element.
2Step 2: For each element, use the stack to find the nearest greater element to the left and right.
3Step 3: Calculate the number of good subarrays using the bounds where the element is the maximum.

solution.py20 lines

1def countGoodSubarrays(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.clear()
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 linear complexity comes from processing each element a constant number of times using the stack.

1A good subarray's OR equals its maximum element.
2Use a monotonic stack to efficiently find bounds.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.