Solving the Maximum Index Problem Using Stack

In the realm of competitive programming and algorithmic challenges, optimizing solutions to find maximum index differences in arrays is a common task. This problem entails finding the maximum value of j - i, where j and i are indices of elements in the array, subject to the constraint that the element at index i is less than the element at index j.

Problem Statement:

Given an array a of n positive integers, the task is to find the maximum of j - i subject to the constraint of a[i] < a[j] and i < j.

Example:

Let's consider an example with the following array:

n = 9
a[] = {34, 8, 10, 3, 2, 80, 30, 33, 1}        

The expected output is 6, as the maximum difference occurs between indices 1 and 7 (i = 1, j = 7), where a[i] < a[j].

Approach:

We can solve this problem efficiently using a stack data structure. Here's a step-by-step explanation of the provided code:

1. We iterate through the array in reverse order (from n-1 to 0) and maintain a stack of pairs. Each pair contains the index and value of the element.
2. We check if the stack is empty or if the value at the current index is greater than the value at the top of the stack. If either condition is true, we push the current index and value onto the stack.
3. Next, we iterate through the array from index 0 to n-1.
4. While the stack is not empty and the value at the top of the stack is greater than or equal to the value at the current index, we compute the difference between the top of the stack's index and the current index and update the maximum difference (ans).
5. Finally, we return the maximum difference.

Code:

class Solution
{
public:
    int maxIndexDiff(int a[], int n)
    {
        stack<pair<int, int>> st; // Create a stack of pairs to store index and value

        // Traverse the array from the end
        for (int i = n - 1; i > -1; i--)
        {
            // If the stack is empty or the top element's value is less than the current element's value
            // Push the current index and value onto the stack
            if (st.empty() || st.top().second < a[i])
                st.push({i, a[i]});
        }

        int ans = 0; // Initialize the answer to 0

        // Traverse the array
        for (int i = 0; i < n; i++)
        {
            // While the stack is not empty and the top element's value is greater than or equal to the current element's value
            // Pop elements from the stack and update the answer by taking the maximum difference between the current index and the popped index
            while (!st.empty() && st.top().second >= a[i])
            {
                ans = max(ans, st.top().first - i);
                st.pop();
            }
        }

        return ans; // Return the maximum index difference
    }
};
        

Complexity Analysis:

• Time Complexity: The algorithm has a time complexity of O(N), where N is the number of elements in the array. This is because we iterate through the array only twice, and each iteration has a linear time complexity.
• Space Complexity: The space complexity is O(N) due to the stack that can potentially hold all elements of the array.

Conclusion:

Efficiently solving problems like maximizing index differences in arrays requires a solid understanding of data structures and algorithmic techniques. By utilizing a stack-based approach, we can achieve optimal solutions with a time complexity of O(N). This approach demonstrates the importance of choosing the right data structure and algorithm for solving different types of problems efficiently.