Count Strictly Increasing Subarrays

class Solution {
public:
    long long countIncreasingSubsequences(vector<int>& elements) {
        long long totalSubsequences = 0;

        for (int index = 0; index < elements.size(); index++) {
            long lengthOfSubsequence = 1;

            while (index + 1 < elements.size() && elements[index] < elements[index + 1]) {
                lengthOfSubsequence++;
                index++;
            }

            totalSubsequences += (lengthOfSubsequence * (lengthOfSubsequence + 1)) / 2;
        }

        return totalSubsequences;
    }
};

The given C++ solution provides a method to count the number of strictly increasing subarrays within an array of integers. This solution is encapsulated within a class named Solution, which features a public method countIncreasingSubsequences taking a vector of integers as its argument.

The method implements an efficient approach using a single scan of the input array:

• Initialize a long long variable totalSubsequences to store the result.
• Loop through each index index of the array elements.
• For each element, calculate the length of the strictly increasing subsequence starting from this index.
• As long as the next element is greater than the current, increment the length and the index.
• The number of subsequences that can be formed with a strictly increasing subsequence of length n is given by (n * (n + 1)) / 2. This formula is derived from the sum of the first n natural numbers.
• Add the computed number of subsequences to totalSubsequences.

The function finally returns totalSubsequences, representing the total count of strictly increasing subsequences within the input array. Remember, the return value captures possible large results using a long long integer to prevent overflow issues associated with large input sizes.

class Solution {
    public long countSubarrays(int[] array) {
        long totalSubarrays = 0;
        
        for (int index = 0; index < array.length; index++) {
            long currentLength = 1;
            
            while (index + 1 < array.length && array[index] < array[index + 1]) {
                currentLength++;
                index++;
            }
            
            totalSubarrays += (currentLength * (currentLength + 1)) / 2;
        }
        
        return totalSubarrays;
    }
}

This solution aims to count the number of strictly increasing subarrays within a given array in Java. The primary approach taken involves iterating through the array while continuously checking for sequences where subsequent elements are strictly greater than their predecessors.

Here's how the solution works:

• Start by initializing a variable totalSubarrays to zero, which will track the count of all valid subarrays.
• Iterate through each element of the array using a for loop.
• For each element, initialize a currentLength variable to 1 to represent the minimum length of any subarray (an element itself).
• Use a nested while loop to check if the current element is less than its next element. If it is:
  • Increment the currentLength which tracks the length of the current strictly increasing sequence.
  • Also increment the index to avoid rechecking the same elements.
• After exiting the while loop, calculate the number of subarrays that the identified sequence can form using the formula ( \text{currentLength} \times (\text{currentLength} + 1) / 2 ), and add this value to totalSubarrays.
• Return the totalSubarrays as the result which represents the total count of strict increasing subarrays.

Key considerations:

• The use of a single loop with a nested while ensures that each element is part of only one strictly increasing sequence calculation which optimizes the performance.
• The formula ( \text{currentLength} \times (\text{currentLength} + 1) / 2 ) efficiently computes the sum of the first n natural numbers, which corresponds to the number of subarrays formed by 'n' consecutive strictly increasing numbers.
• Make sure to handle edge cases such as arrays with no consecutive increasing numbers, very large arrays, or arrays of uniform elements effectively within this framework.

This implementation provides a clear and efficient method to count increasing subarrays, ensuring an optimal solution with a time complexity nearly linear relative to the size of the input array, i.e., ( O(n) ), where n is the number of elements in the array.