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 O(1) extra space.
The problem requires finding the maximum sum of a continuous subarray within a given array of integers, which may include both positive and negative numbers.
nums
.Input: nums = [-2, -5, 6, -2, -3, 1, 5, -6] Output: 7 Explanation: sum([6, -2, -3, 1, 5]) = 7
The core challenge is to find the subarray with the maximum sum in linear time. This problem is significant in various fields such as finance (for maximum profit calculation) and computer science (for optimization problems).
Potential pitfalls include misunderstanding the requirement for a continuous subarray and not accounting for 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.
A naive solution would involve checking all possible subarrays and calculating their sums, which would result in O(n^2) time complexity. This is not optimal for large arrays.
Kadane's Algorithm improves upon the naive solution by maintaining a running sum of the current subarray and updating the maximum sum found so far. The key idea is to decide whether to add the current element to the existing subarray or start a new subarray.
Here is a step-by-step breakdown of Kadane's Algorithm:
max_current
and max_global
to the first element of the array.max_current
to be the maximum of the current element and the sum of max_current
and the current element.max_current
is greater than max_global
, update max_global
.max_global
as the result.#include <vector>
#include <algorithm>
#include <iostream>
int maxSubArray(const std::vector<int>& nums) {
// Initialize max_current and max_global to the first element
int max_current = nums[0];
int max_global = nums[0];
// Iterate through the array starting from the second element
for (size_t i = 1; i < nums.size(); ++i) {
// Update max_current to be the maximum of the current element and the sum of max_current and the current element
max_current = std::max(nums[i], max_current + nums[i]);
// Update max_global if max_current is greater
if (max_current > max_global) {
max_global = max_current;
}
}
// Return the maximum sum found
return max_global;
}
int main() {
std::vector<int> nums = {-2, -5, 6, -2, -3, 1, 5, -6};
std::cout << "Maximum sum of continuous 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:
Example edge cases:
Input: nums = [-1, -2, -3] Output: -1 Input: nums = [5] Output: 5
To test the solution comprehensively, consider the following test cases:
Testing frameworks such as Google Test can be used for automated testing.
When approaching such problems, consider the following tips:
Understanding and solving the maximum sum subarray problem using Kadane's Algorithm is crucial for optimizing performance in various applications. Practice and familiarity with such algorithms can significantly enhance problem-solving skills.
For further reading and practice, consider the following resources: