Count Strictly Increasing Subarrays

Given an integer array arr[], count the number of subarrays in arr[] that are strictly increasing and have a size of at least 2. A subarray is a contiguous sequence of elements from arr[]. A subarray is strictly increasing if each element is greater than its previous element.

Examples:

Input: arr[] = [1, 4, 5, 3, 7, 9]
Output: 6
Explanation: The strictly increasing subarrays are: [1, 4], [1, 4, 5], [4, 5], [3, 7], [3, 7, 9], [7, 9]

Input: arr[] = [1, 3, 3, 2, 3, 5]
Output: 4
Explanation: The strictly increasing subarrays are: [1, 3], [2, 3], [2, 3, 5], [3, 5]

Input: arr[] = [2, 2, 2, 2]
Output: 0
Explanation: No strictly increasing subarray exists.

[Naive Approach] Check All Subarrays - O(n^2) Time and O(1) Space

Count all strictly increasing subarrays by starting from each index i and expanding forward. For every starting point, we use another pointer j to extend the subarray as long as the elements keep increasing. As soon as the increasing order breaks, we stop extending from that starting point and move to the next index.

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[Expected Approach] Using Subarray Count Formula - O(n) Time and O(1) Space

We can do only with a single pass. Instead of checking every subarray explicitly, we track the length of increasing segments using len. When a decreasing element is encountered, we use the formula (len * (len - 1)) / 2 to count subarrays formed by the segment and then reset len. Finally, we add the remaining subarrays after the loop ends.

Steps to implement the above idea:

• Initialize count to store the number of strictly increasing subarrays and len to track the length of increasing sequences.
• Iterate through the array starting from index 1, comparing each element with its previous element to check for increasing order.
• If the current element is greater than the previous, increment len as it extends the increasing subarray.
• If the current element breaks the increasing sequence, update count using the formula (len*(len-1))/2 and reset len to 1.
• Continue iterating until the end of the array, applying the same logic for each increasing and non-increasing sequence.
• After the loop, add the remaining subarrays count using (len * (len - 1)) / 2 to include the last segment.
• Finally, return count, which holds the total number of strictly increasing subarrays in the given array.

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