Algorithms – Maximum Subarray

Introduction

Welcome to our blog post, where we unravel the inner workings of the “Maximum Subarray” problem. In this post, we will explore a specific implementation that efficiently finds the maximum sum of a contiguous subarray within a given array. Join us as we dissect each step and understand the logic behind it. Let’s get started!

Step 1 – Handling Empty or Null Arrays

To ensure a valid input, we begin by checking if the given array nums is null or empty. If it is, we return 0 as there are no elements to process.

Step 2 – Initializing Variables

Next, we initialize an auxiliary array s with the same length as nums. This array will store the maximum subarray sum ending at each index. We also set the first element of s and max to the first element of nums, as it is the only element considered at the start.

Step 3 – Calculating Maximum Subarray Sum

Using a loop starting from index 1, we calculate the maximum subarray sum ending at each index i. The value at s[i] is determined based on whether the sum of the current element nums[i] and the previous maximum subarray sum s[i-1] is positive or negative. If it’s positive, we add it to the current element, otherwise, we consider only the current element itself. This allows us to keep track of the maximum subarray sum ending at each index.

Step 4 – Updating Maximum Sum

Within the loop, we also update the max variable to keep track of the overall maximum subarray sum encountered so far. We compare the current maximum subarray sum s[i] with the previous maximum max and update max accordingly.

Step 5 – Returning the Result

After the loop finishes, we have calculated the maximum subarray sum for each index. We return the value of max as the final result, representing the maximum sum of a contiguous subarray within the given array.

Java Implementation

public int maxSubArray(int[] nums) {
    if (nums == null || nums.length == 0)
        return 0;

    int[] s = new int[nums.length];
    s[0] = nums[0];
    int max = nums[0];

    for (int i = 1; i < nums.length; i++) {
        s[i] = s[i - 1] > 0 ? (nums[i] + s[i - 1]) : nums[i];
        max = Math.max(max, s[i]);
    }

    return max;
}

Time and Space complexity

The time complexity of the maxSubArray algorithm is O(n), where n is the length of the input array nums. This is because we iterate through the array once in a single loop to calculate the maximum subarray sum ending at each index.

The space complexity of the algorithm is O(n) as well. This is because we create an auxiliary array s with the same length as nums to store the maximum subarray sum ending at each index. Therefore, the space required is proportional to the size of the input array.

Overall, the maxSubArray algorithm has a linear time complexity and linear space complexity, making it an efficient solution for finding the maximum subarray sum.

Conclusion

In this blog post, we explored the step-by-step process of finding the maximum subarray sum using an efficient implementation. By handling empty or null arrays, initializing variables, calculating the maximum subarray sum, and updating the maximum sum, we can accurately solve the “Maximum Subarray” problem. Now, let’s dive into the implementation and witness these steps in action.

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