Introduction to the Maximum Subarray Problem

Imagine you're given a list of integers, where each number represents your profit or loss for the day. Your task is to find the period during which you would have made the most money if you only had the foresight to start and end trading on specific days.

This is essentially the Maximum Subarray problem: given an integer array nums, find the subarray that has the largest sum and return that sum. This is LeetCode 53: Maximum Subarray, a medium typed problem.

For example, consider nums = [-2,1,-3,4,-1,2,1,-5,4]. The subarray [4,-1,2,1] has the largest sum, 6. It's a classic problem that tests your ability to navigate arrays and optimize solutions, making it a popular question in software engineering interviews.

Given an integer array nums, find the subarray with the largest sum, and return its sum.

Subarray - A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The subarray [4,-1,2,1] has the largest sum 6.

Example 2:

Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.

Understanding the Solution

The most elegant solution to this problem leverages Kadane's Algorithm. The crux of Kadane's Algorithm is to examine each number in the array and decide whether to add it to the current subarray sum (which could be negative) or start a new subarray with the current number. The goal is to maintain the largest sum we've seen as we iterate through the array. This can be solved with either a variable or a whole array.

Array Logic

Computes the sum of subarrays ending at each index. Use an array memory to keep track of these sums. For each index, decide whether to add the current element to the sum of the previous subarray (if that sum is positive) or start a new subarray from the current element.

This decision is encoded in the expression nums[i] + (memory[i - 1] if memory[i - 1] > 0 else 0). This keeps track of the maximum subarray sum found so far.

Variable Logic

Maintain a "local" maximum sum of the subarray ending at the current index and a "global" maximum to keep track of the highest sum encountered so far. For each element, decide whether to start a new subarray from the current element or to extend the existing subarray to include the current element.

This decision is based on whether adding the current element to the existing subarray sum (local + nums[i]) is better than just the current element (nums[i]).

Array vs. Variable Approach

Both approaches have a time complexity of O(n) since they require a single pass through the array.

Solution in Python Using an Array

# Additional comments: The memory array stores the maximum sum subarray ending at each index.
# Allowing easy tracking of the overall maximum.
def maxSubArray(nums):
    # Edge case: if the array is empty
    if not nums:
        return 0

    # Initialize the memory array with the first element
    memory = [nums[0]]
    for i in range(1, len(nums)):
        # Calculate the maximum sum up to the current index
        memory.append(max(nums[i], memory[i-1] + nums[i]))
    # The largest sum is the maximum element in the memory array
    return max(memory)

Solution in Python Using a Variable

# Additional comments: This implementation directly applies Kadane's Algorithm.
# Uses only two variables to keep track of the current and maximum sums.
def maxSubArray(nums):
    currentSum = maxSum = nums[0]
    for num in nums[1:]:
        currentSum = max(num, currentSum + num)
        maxSum = max(maxSum, currentSum)
    return maxSum

Solution in TypeScript

// Additional comments: Similar to the Python variable approach, but in TypeScript
// Showcases how Kadane's Algorithm transcends language specifics.
function maxSubArray(nums: number[]): number {
    let currentSum = nums[0];
    let maxSum = currentSum;

    for (let i = 1; i < nums.length; i++) {
        currentSum = Math.max(nums[i], currentSum + nums[i]);
        maxSum = Math.max(maxSum, currentSum);
    }

    return maxSum;
}

Solution in Java

// Additional comments: Demonstrates Kadane's Algorithm in Java.
// Highlighting the use of Math.max for clarity and simplicity.
public int maxSubArray(int[] nums) {
    int currentSum = nums[0];
    int maxSum = currentSum;

    for (int i = 1; i < nums.length; i++) {
        currentSum = Math.max(nums[i], currentSum + nums[i]);
        maxSum = Math.max(maxSum, currentSum);
    }

    return maxSum;
}

Conclusion

Understanding and implementing Kadane's Algorithm for the Maximum Subarray problem is a fundamental skill for any software engineer.

Whether you're an experienced developer or new to engineering interviews, mastering this problem not only sharpens your coding skills but also prepares you for tackling a variety of dynamic programming questions. Through practice and application of these solutions, you'll be well on your way to acing your next coding interview.

Happy coding, and I genuinely hope this guide aids you in your interview preparation journey. Thank you for taking the time to walk through these solutions with me!