Given an input array that may contain both positive and negative integers, find the sum of continuous subarray of numbers which has the largest sum.
Example:
Input: nums = [-2, -5, 6, -2, -3, 1, 5, -6]
Output: 7
Explanation: sum([6, -2, -3, 1, 5]) = 7
Your algorithm should run in O(n) time and use at most O(n) extra space.
The core challenge of this problem is to find the subarray (a contiguous part of the array) that has the maximum sum. This problem is significant in various fields such as finance (to find the best time to buy and sell stocks) and computer science (for optimization problems).
Potential pitfalls include misunderstanding the requirement for the subarray to be contiguous and not considering negative numbers correctly.
To solve this problem, we can use Kadane's Algorithm, which is an efficient way to find the maximum sum subarray in O(n) time complexity. The naive approach would involve checking all possible subarrays, which would be inefficient with a time complexity of O(n^2).
The naive approach involves iterating through all possible subarrays and calculating their sums, which is not optimal due to its O(n^2) time complexity.
Kadane's Algorithm improves upon the naive approach by maintaining a running sum of the current subarray and updating the maximum sum found so far. The key idea is to decide at each element whether to add it to the current subarray or start a new subarray.
Here is a step-by-step breakdown of Kadane's Algorithm:
maxSoFar
to store the maximum sum found so far and maxEndingHere
to store the maximum sum of the subarray ending at the current position.maxEndingHere
to be the maximum of the current element alone or the current element plus maxEndingHere
.maxSoFar
to be the maximum of maxSoFar
and maxEndingHere
.maxSoFar
as the result.public class MaximumSumSubarray {
public static int maxSubArray(int[] nums) {
// Initialize variables to store the maximum sum so far and the maximum sum ending at the current position
int maxSoFar = nums[0];
int maxEndingHere = nums[0];
// Iterate through the array starting from the second element
for (int i = 1; i < nums.length; i++) {
// Update maxEndingHere to be the maximum of the current element alone or the current element plus maxEndingHere
maxEndingHere = Math.max(nums[i], maxEndingHere + nums[i]);
// Update maxSoFar to be the maximum of maxSoFar and maxEndingHere
maxSoFar = Math.max(maxSoFar, maxEndingHere);
}
// Return the maximum sum found
return maxSoFar;
}
public static void main(String[] args) {
int[] nums = {-2, -5, 6, -2, -3, 1, 5, -6};
System.out.println("Maximum sum subarray: " + maxSubArray(nums)); // Output: 7
}
}
The time complexity of Kadane's Algorithm is O(n) because it involves a single pass through the array. The space complexity is O(1) as it uses a constant amount of extra space.
Potential edge cases include:
Examples:
Input: [-1, -2, -3]
Output: -1
Input: [1]
Output: 1
To test the solution comprehensively, consider a variety of test cases:
When approaching such problems, it is helpful to:
In this blog post, we discussed how to solve the Maximum Sum Subarray problem using Kadane's Algorithm. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
For further reading and practice, consider the following resources: