Number of Valid Subarrays

class Solution {
public:
    int countValidSubarrays(vector<int>& elements) {
        int total = 0;
            
        stack<int> indices;
        for (int index = 0; index < elements.size(); index++) {
            while (!indices.empty() && elements[index] < elements[indices.top()]) {
                total += (index - indices.top());
                indices.pop();
            }
            indices.push(index);
        }
            
        while (!indices.empty()) {
            total += (elements.size() - indices.top());
            indices.pop();
        }
            
        return total;
    }
};

The provided C++ code aims to solve the problem of counting valid subarrays within a given vector of integers. The technique used involves a stack to store indices of array elements under certain conditions, leveraging the Monotonic Stack method that helps in maintaining element order for quick comparison and validity checks.

To clarify the approach:

• Initialize a counter total to zero which will store the result, i.e., the total number of valid subarrays.
• Create a stack<int> named indices that will maintain the indices of array elements that fulfill a condition during iteration.
• Loop over the elements of the input vector:
  • Within the loop, use a while loop to pop elements from the stack while the current element is smaller than the element at the top index of the stack. For each element popped, update total by adding the difference between the current index and the popped index.
  • Push the current index onto the stack afterwards.
• After completing the primary loop, any remaining indices in the stack are processed in a final while loop. For each of these, add to total the difference between the size of the vector and the index at the top of the stack, then pop the stack.
• Return the computed total as the solution.

This solution effectively counts and returns the total number of subarrays that meet a specific ordering criterion, doing so efficiently with the use of a stack to manage and compare indices dynamically.

class Solution {
    public int countValidSubarrays(int[] elements) {
        int count = 0;
            
        Stack<Integer> indexStack = new Stack<>();
        for (int index = 0; index < elements.length; index++) {
            while(!indexStack.isEmpty() && elements[index] < elements[indexStack.peek()]) {
                count += (index - indexStack.peek());
                indexStack.pop();
            }
            indexStack.push(index);
        }
            
        while (!indexStack.isEmpty()) {
            count += (elements.length - indexStack.peek());
            indexStack.pop();
        }
            
        return count;
    }
}

The provided Java solution efficiently counts the number of valid subarrays in a given integer array. Here’s a breakdown of how the program tackles this task:

1. Initialize a counter variable, count, to zero to store the total count of valid subarrays.

2. Use a Stack<Integer> called indexStack to store indices of array elements. This stack helps keep track of the indices while iterating over the array to find valid subarrays.

3. Loop through each element in the elements array using a for-loop:

   • Inside the loop, use a while-loop to check the conditions that determine if a subarray is valid:
     • If indexStack is not empty and the current element (elements[index]) is less than the element at the index stored at the top of the stack (elements[indexStack.peek()]), then calculate and add the difference between the current index and the index popped from the stack to count.
   • Push the current index to indexStack.

4. After looping through all elements, proceed to empty indexStack if it still contains any indices:

   • While indexStack is not empty, pop the top index and add the difference between the length of the array and the popped index to count.

The final output, returned by int countValidSubarrays(int[] elements), is the total count of valid subarrays that meet the criteria where all sub-arrays (elements[i], elements[i+1], ..., elements[j]) should satisfy the condition that elements[i] is less than or equal to all other elements in the sub-array.