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 maximum sum of a continuous subarray within a given array of integers, which may include both positive and negative numbers. This problem is significant in various applications such as financial analysis, where one might want to find the period with the maximum profit or minimum loss.
Potential pitfalls include misunderstanding the requirement for the subarray to be continuous and not considering subarrays that start or end with negative numbers.
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 algorithm works by iterating through the array while keeping track of the maximum sum subarray ending at the current position and the overall maximum sum found so far.
A naive solution would involve checking all possible subarrays and calculating their sums, which would result in a time complexity of O(n^2). This is not optimal for large arrays.
Kadane's Algorithm improves upon the naive solution by maintaining two variables: current_max
(the maximum sum of the subarray ending at the current position) and global_max
(the maximum sum found so far). As we iterate through the array, we update these variables based on the current element.
current_max
and global_max
to the first element of the array.current_max
to be the maximum of the current element and the sum of current_max
and the current element.global_max
to be the maximum of global_max
and current_max
.global_max
as the result.#include <iostream>
#include <vector>
#include <algorithm> // for std::max
// Function to find the maximum sum subarray
int maxSubArray(const std::vector<int>& nums) {
// Initialize current_max and global_max to the first element
int current_max = nums[0];
int global_max = nums[0];
// Iterate through the array starting from the second element
for (size_t i = 1; i < nums.size(); ++i) {
// Update current_max
current_max = std::max(nums[i], current_max + nums[i]);
// Update global_max
global_max = std::max(global_max, current_max);
}
return global_max;
}
int main() {
std::vector<int> nums = {-2, -5, 6, -2, -3, 1, 5, -6};
std::cout << "Maximum sum subarray: " << maxSubArray(nums) << std::endl;
return 0;
}
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:[5]
Output: 5
To test the solution comprehensively, consider a variety of test cases:
Using a testing framework like Google Test can help automate and manage these tests.
When approaching such problems, it's essential to:
Understanding and solving the maximum sum subarray problem is crucial for developing efficient algorithms. Kadane's Algorithm provides an optimal solution with O(n) time complexity, making it suitable for large datasets. Practicing such problems helps improve algorithmic thinking and problem-solving skills.
For further reading and practice, consider the following resources: