GATE-CS-2000 Problem 8

This is a problem from the GATE (Graduate Aptitude Test in Engineering) for Computer Science in the year 2000. The problem is focused on understanding the fundamentals of programming and data structures.

Problem Statement

We are given an array A of n integers, which may be positive or negative. We need to find a subarray of A whose sum is maximum among all possible subarrays. The problem statement asks to solve this problem using the Divide and Conquer algorithm.

Solution Approach

The approach to solve this problem using Divide and Conquer algorithm is based on the following observations:

1. If the subarray is of length 1, then the maximum subarray sum is equal to the element itself.
2. If the subarray is of length 2, then the maximum subarray sum is either the first element or the second element or the sum of both.
3. If the subarray is of length n, then we can divide it into two subarrays of equal length and recursively find the maximum subarray sum in each subarray. Then we need to find the maximum subarray sum that crosses the midpoint of the original array. This can be done by finding the maximum suffix sum in the left subarray and the maximum prefix sum in the right subarray and adding them.

Using these observations, we can write a recursive function that computes the maximum subarray sum for a given subarray. Here is the pseudocode for the function:

function find_maximum_subarray_sum(A, low, high):
    if low == high:
        return A[low]
    else:
        mid = (low + high) / 2
        left_sum = find_maximum_subarray_sum(A, low, mid)
        right_sum = find_maximum_subarray_sum(A, mid+1, high)
        cross_sum = find_maximum_crossing_sum(A, low, mid, high)
        return max(left_sum, right_sum, cross_sum)

The find_maximum_crossing_sum function computes the maximum subarray sum that crosses the midpoint of the subarray A[low..high]. Here is the pseudocode for the function:

function find_maximum_crossing_sum(A, low, mid, high):
    left_sum = -infinity
    sum = 0
    for i in range(mid, low-1, -1):
        sum = sum + A[i]
        if sum > left_sum:
            left_sum = sum

    right_sum = -infinity
    sum = 0
    for i in range(mid+1, high+1):
        sum = sum + A[i]
        if sum > right_sum:
            right_sum = sum

    return left_sum + right_sum

Conclusion

In this problem, we have seen how to solve the Maximum Subarray Sum problem using the Divide and Conquer algorithm. This approach has a time complexity of O(n log n), which is better than the brute-force approach.