How do you solve Count Stable Subarrays on LeetCode?

Count Stable Subarrays (LeetCode #3748) 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: Maximal non-decreasing segments help in counting stable subarrays efficiently.

What is the brute force solution for Count Stable Subarrays?

The brute force approach for Count Stable Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible subarray to see if it's stable. This involves examining all pairs of indices to count inversions. The optimal Optimal Solution approach improves this to O(n).

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

Count Stable Subarrays uses the following concepts: Array, Binary Search, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Stable Subarrays?

Explain your thought process clearly. Consider edge cases like empty arrays or single-element arrays. Practice identifying patterns in sequences.

What are common mistakes when solving Count Stable Subarrays?

Overlooking single-element subarrays as stable. Not resetting the segment length correctly when encountering a decrease.

#3748

Count Stable Subarrays

Hard

ArrayBinary SearchPrefix SumHash 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)

Identify maximal non-decreasing segments in the array. Each segment contributes stable subarrays based on its length.

⚙️

Algorithm

3 steps

1Step 1: Traverse the array to identify lengths of maximal non-decreasing segments.
2Step 2: For each segment of length L, calculate stable subarrays as L * (L + 1) / 2.
3Step 3: Build a prefix sum array to quickly answer each query based on precomputed stable subarray counts.

solution.py14 lines

1def countStableSubarrays(nums, queries):
2    n = len(nums)
3    stable_counts = [0] * n
4    length = 1
5    for i in range(1, n):
6        if nums[i] >= nums[i - 1]: length += 1
7        else:
8            for j in range(length): stable_counts[i - j - 1] += length - j
9            length = 1
10    for j in range(length): stable_counts[n - j - 1] += length - j
11    prefix_sum = [0] * n
12    prefix_sum[0] = stable_counts[0]
13    for i in range(1, n): prefix_sum[i] = prefix_sum[i - 1] + stable_counts[i]
14    return [prefix_sum[r] - (prefix_sum[l - 1] if l > 0 else 0) for l, r in queries]

ℹ

Complexity note: This complexity is achieved by processing the array in a single pass to identify segments and using a prefix sum for quick query responses.

1Maximal non-decreasing segments help in counting stable subarrays efficiently.
2Stable subarrays can be derived from segment lengths using combinatorial counting.

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