Is Count Strictly Increasing Subarrays easy or hard?

The problem is typically rated Medium because recognizing the increasing run pattern is the key insight. The implementation itself is straightforward once you realize that each index contributes the length of the current increasing streak.

How to solve Count Strictly Increasing Subarrays in O(n)?

Maintain a variable representing the length of the current strictly increasing subarray ending at index i. If nums[i] > nums[i-1], increase the streak length; otherwise reset it to 1. Add the streak length to the total at each step. This counts all valid subarrays ending at every index in a single pass.

What is the best approach for Count Strictly Increasing Subarrays?

The optimal approach scans the array once and tracks the length of the current strictly increasing run. Each element contributes the length of the run ending at that index to the answer. This method runs in O(n) time and uses O(1) extra space, making it the expected interview solution.

What data structure is used in Count Strictly Increasing Subarrays?

The solution primarily uses basic array traversal. No advanced data structures are required. The algorithm relies on tracking a running counter for the current increasing streak and accumulating results during the pass.

What is the time complexity of Count Strictly Increasing Subarrays?

The optimal solution runs in O(n) time because the array is traversed exactly once while maintaining the length of the current increasing streak. A naive enumeration approach can take O(n²) time if it checks every possible starting position and extends the subarray forward.

Count Strictly Increasing Subarrays Python or Java solution approach?

In both Python and Java, the logic is identical: iterate through the array, maintain a streak length for increasing elements, and add it to the result each iteration. This produces an O(n) time and O(1) space implementation in any language.

Is Count Strictly Increasing Subarrays asked at Google, Amazon, or Meta?

Problems involving counting increasing subarrays or sequences frequently appear in interviews at companies like Amazon and Google because they test array traversal, pattern recognition, and simple dynamic programming optimizations.

Description Hints 2 Solution Video 5:31

2393. Count Strictly Increasing Subarrays

Medium70.8% acceptance1 companyPremium elsewhere, free here

You are given an array nums consisting of positive integers.

Return the number of subarrays of nums that are in strictly increasing order.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,3,5,4,4,6]
Output: 10
Explanation: The strictly increasing subarrays are the following:
- Subarrays of length 1: [1], [3], [5], [4], [4], [6].
- Subarrays of length 2: [1,3], [3,5], [4,6].
- Subarrays of length 3: [1,3,5].
The total number of subarrays is 6 + 3 + 1 = 10.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 15
Explanation: Every subarray is strictly increasing. There are 15 possible subarrays that we can take.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Asked at 1 company

Bridgewater Associates

Topics

Hints open one at a time

Hint 1

Find the number of strictly increasing subarrays that end at a specific index. Can you calculate that for each index from 0 to n - 1?

Hint 2

The answer will be the sum of the number of subarrays that end at each index.

5:31

Still stuck?Read the written solution, or watch a walkthrough.

Open solution