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^2) time and use O(1) extra space.
The core challenge of this problem is to find the subarray with the maximum sum in an array that contains both positive and negative integers. 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 definition of a subarray (which must be contiguous) and not considering negative numbers correctly.
To solve this problem, we can start with a naive approach and then optimize it:
The naive approach involves checking all possible subarrays and calculating their sums. This can be done using two nested loops:
maxSum = nums[0]
for each i : 0 -> n - 1:
for each j : i -> n - 1:
maxSum = max(maxSum, sum(nums[i...j]))
return maxSum
However, this approach is not optimal as it has a time complexity of O(n^3) due to the nested loops and the sum calculation inside the inner loop.
We can optimize the naive approach by calculating the sum of subarrays in O(1) time using a running sum:
maxSum = nums[0]
for each i : 0 -> n - 1:
sum = 0
for each j : i -> n - 1:
sum += nums[j]
maxSum = max(maxSum, sum)
return maxSum
This approach reduces the time complexity to O(n^2) while maintaining O(1) space complexity.
Here is a step-by-step breakdown of the optimized algorithm:
maxSum
to the first element of the array.maxSum
if the current subarray sum is greater than maxSum
.maxSum
after all iterations.#include <iostream>
#include <vector>
#include <algorithm> // for std::max
int maxSubArraySum(const std::vector<int>& nums) {
int maxSum = nums[0]; // Initialize maxSum to the first element
int n = nums.size();
for (int i = 0; i < n; ++i) {
int sum = 0; // Initialize sum for the current subarray
for (int j = i; j < n; ++j) {
sum += nums[j]; // Add the current element to the sum
maxSum = std::max(maxSum, sum); // Update maxSum if needed
}
}
return maxSum; // Return the maximum subarray sum
}
int main() {
std::vector<int> nums = {-2, -5, 6, -2, -3, 1, 5, -6};
std::cout << "Maximum Sum Subarray: " << maxSubArraySum(nums) << std::endl;
return 0;
}
The time complexity of the optimized approach is O(n^2) due to the nested loops. The space complexity is O(1) as we are using only a few extra variables.
Consider the following edge cases:
To test the solution comprehensively, consider the following test cases:
Test Case 1: Input: [1, 2, 3, 4, 5] Expected Output: 15 Test Case 2: Input: [-1, -2, -3, -4, -5] Expected Output: -1 Test Case 3: Input: [3, -2, 5, -1] Expected Output: 6 Test Case 4: Input: [1] Expected Output: 1
When approaching such problems, consider the following tips:
In this blog post, we discussed the problem of finding the maximum sum subarray in an array containing both positive and negative integers. We explored a naive approach and an optimized approach, provided a detailed algorithm, and implemented the solution in C++. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for coding interviews.
For further reading and practice, consider the following resources: