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 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 identify the subarray with the maximum sum. This problem is significant in various applications such as financial analysis, where one might want to find the period with the maximum profit.
Potential pitfalls include misunderstanding the requirement for a continuous subarray and not considering negative numbers correctly.
To solve this problem, we can start with a brute force approach and then optimize it.
The naive solution involves checking all possible subarrays and calculating their sums. This can be achieved using two nested loops:
maxSum = nums[0]
for i in range(len(nums)):
for j in range(i, len(nums)):
currentSum = sum(nums[i:j+1])
maxSum = max(maxSum, currentSum)
return maxSum
While this approach is straightforward, it is not optimal as it has a time complexity of O(n^3) due to the sum calculation inside the inner loop.
We can optimize the naive solution by maintaining a running sum of the current subarray. This reduces the time complexity to O(n^2):
maxSum = nums[0]
for i in range(len(nums)):
currentSum = 0
for j in range(i, len(nums)):
currentSum += nums[j]
maxSum = max(maxSum, currentSum)
return maxSum
This approach ensures that we only traverse the array twice, making it more efficient.
Here is a step-by-step breakdown of the optimized algorithm:
maxSum
with the first element of the array.maxSum
if the current subarray sum is greater.maxSum
after all iterations.def max_subarray_sum(nums):
# Initialize maxSum with the first element of the array
maxSum = nums[0]
# Iterate over all possible subarrays
for i in range(len(nums)):
currentSum = 0
for j in range(i, len(nums)):
# Add the current element to the running sum
currentSum += nums[j]
# Update maxSum if the current sum is greater
maxSum = max(maxSum, currentSum)
return maxSum
# Example usage
nums = [-2, -5, 6, -2, -3, 1, 5, -6]
print(max_subarray_sum(nums)) # Output: 7
The time complexity of the optimized solution is O(n^2) because of the two 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:
def test_max_subarray_sum():
assert max_subarray_sum([-2, -5, 6, -2, -3, 1, 5, -6]) == 7
assert max_subarray_sum([-1, -2, -3, -4]) == -1
assert max_subarray_sum([1, 2, 3, 4]) == 10
assert max_subarray_sum([1]) == 1
assert max_subarray_sum([-1, 2, 3, -4, 5, -6]) == 6
test_max_subarray_sum()
When approaching such problems, consider the following tips:
Understanding and solving the maximum sum subarray problem is crucial for developing strong problem-solving skills. Practice and exploration of different approaches can help in mastering such problems.
For further reading and practice, consider the following resources: