Understanding the Maximum Index Problem: A Two-Pointer Approach

When working with arrays, a common type of problem is finding the maximum distance between two elements that satisfy a certain condition. One such problem is the Maximum Index problem, where we are given an array of integers and need to find the maximum value of j - i such that A[i] <= A[j] and i < j. This problem can be efficiently solved using a two-pointer approach, which we will explore in this article.

Problem Statement

Given an array A of N integers, find the maximum value of j - i subjected to the constraint of A[i] <= A[j], where i and j are the indices of the array.

Example

Consider the array A = [3, 5, 4, 1]. The maximum index difference that satisfies A[i] <= A[j] is 2, which occurs for the pair (A[0], A[2]) or (3, 4).

The Two-Pointer Approach

The two-pointer approach is a technique that uses two references (pointers) to traverse the array, usually in opposite directions or in a way that maintains a certain relationship between the pointers. For the Maximum Index problem, we will use two arrays, LMin and RMax, and two pointers to find the solution.

Step 1: Preprocessing

We create two auxiliary arrays:

• LMin: For each index i, it stores the minimum value from the start of the array up to i.
• RMax: For each index j, it stores the maximum value from j to the end of the array.

Step 2: Traversing with Two Pointers

We initialize two pointers, i and j, to traverse LMin and RMax. The pointer i starts at the beginning of LMin, and j starts at the beginning of RMax. We compare the elements at these pointers and move them according to the following rules:

• If LMin[i] <= RMax[j], we have a valid pair, and we check if j - i is greater than the current maximum difference. If it is, we update the maximum difference. Then, we increment j to look for a larger difference.
• If LMin[i] > RMax[j], we increment i to find a smaller LMin[i].

We continue this process until either i or j reaches the end of the array.

Step 3: Returning the Result

The maximum value of j - i found during the traversal is the answer to the problem.

Example Walkthrough

Let's apply the two-pointer approach to the example array A = [3, 5, 4, 1].

• Preprocessing:

We create LMin and RMax: LMin = [3, 3, 3, 1] RMax = [5, 5, 4, 1]

• Traversing with Two Pointers:

We start with i = 0 and j = 0 and compare LMin[i] with RMax[j]. Since 3 <= 5, we have a valid pair and maxDiff is updated to 0. We increment j to 1.

Continuing this process, we find that the maximum difference is 2 when i = 0 and j = 2.

• Result:

The maximum index difference is 2, corresponding to the pair (3, 4) in the original array.

def maxDiff(a, n):
    # Create two arrays LMin and RMax
    LMin = [0] * n
    RMax = [0] * n

    # Initialize first element of LMin
    LMin[0] = a[0]
    for i in range(1, n):
        LMin[i] = min(a[i], LMin[i - 1])

    # Initialize last element of RMax
    RMax[n - 1] = a[n - 1]
    for j in range(n - 2, -1, -1):
        RMax[j] = max(a[j], RMax[j + 1])

    # Traverse both arrays to find the maximum j - i
    i, j, maxDiff = 0, 0, -1
    while j < n and i < n:
        if LMin[i] <= RMax[j]:
            maxDiff = max(maxDiff, j - i)
            j += 1
        else:
            i += 1

    return maxDiff