# Max Sum Contiguous Subarray

This is a maximum sum contiguous problem from interviewbit.com

Problem : Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example: Given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the largest sum = 6.

This is my solution :

def max_sub_array(array):
""" Finds msub-array with maximum sum and returns the maximum sum and sub-array"""
max_start, max_stop = 0, 1 # Start, Stop of maximum sub-array
curr = 0                   # pointer to current array
max_sum = array[0]         # Sum of maximum array
current_sum = 0            # sum of current array

for i, elem in enumerate(array):
current_sum +=  elem

if max_sum < current_sum:
max_sum = current_sum
max_start = curr
max_stop = i  + 1
if current_sum < 0:
current_sum = 0
curr = i + 1

return  max_sum , array[max_start:max_stop]


checking test cases:

assert max_sub_array([-4,-2,-3,-4,-5]) == (-2,[-2]), "Wrong evaluation"
assert max_sub_array([-1]) == (-1,[-1]), "Wrong evaluation"
assert max_sub_array([-5, 1, -3, 7, -1, 2, 1, -4, 6]) == (11,[7, -1, 2, 1, -4, 6]), "Wrong evaluation"
assert max_sub_array([-5, 1, -3, 7, -1, 2, 1, -6, 5]) == (9, [7, -1, 2, 1]), "Wrong evaluation"
assert max_sub_array( [6, -3, -2, 7, -5, 2, 1, -7, 6]) == (8,[6, -3, -2, 7]), "Wrong evaluation"
assert max_sub_array([-5, -2, -1, -4, -7]) == (-1,[-1]), "Wrong evaluation"
assert max_sub_array( [4, 1, 1, 4, -4, 10, -4, 10, 3, -3, -9, -8, 2, -6, -6, -5, -1, -7, 7, 8]) == (25,[4, 1, 1, 4, -4, 10, -4, 10, 3]), "Wrong evaluation"
assert max_sub_array([4, -5, -1, 0, -2, 20, -4, -3, -2, 8, -1, 10, -1, -1 ]) ==  (28, [20, -4, -3, -2, 8, -1, 10]), "Wrong evaluation"


How can this code be optimised ?

• It is a known problem in computer science as @kaushal says. Kadane's algorithm is $O(n)$ (linear run time complexity) and is probably as good as it gets. Commented Jul 10, 2018 at 2:27

Expanding upon my comment:

def max_subarray(arr):
max_ending = max_current = arr[0]

for i in arr[1:]:
max_ending = max(i, max_ending + i)
max_current = max(max_current, max_ending)

return max_current

print(max_subarray([-4, -2, -3, -4, 5])) # 5

print(max_subarray([4, -5, -1, 0, -2, 20, -4, -3, -2, 8, -1, 10, -1, -1])) # 28


Like your algorithm, it is $O(n)$. However, it does fewer operations while looping. You could then alter it to return the array which gives that sum, which shouldn't be too hard.

You have written to start a new subarray,

    if current_sum < 0:
current_sum = 0
curr = i +


It is not necessary that current_sum < 0 to start a new subarray. We should be creating a new subarray only when it is less than the current element, i.e. current_sum < array[i].

I have implemented this in C++:

void maximum_sum_subarray(int n, int arr[])
{
// variables for global maximum_sum_subarray
int maxSoFar = arr[0], maxSoFarLeft = 0, maxSoFarRight = 0;
// variables for current subarray
int thisSubarraySum = arr[0], thsSubarrayleft = 0, thisSubarrayRight = 0;

for (int i = 1; i < n; i++)
{
int sumInThisSubArray = arr[i] + thisSubarraySum;

// if sum is less then current elements
// means that we can start a new suarray
if (arr[i] > sumInThisSubArray)
{
thisSubarraySum = arr[i];
thsSubarrayleft = i;
thisSubarrayRight = i;

// if this subarray has sum than
//  previous largest subarray
// then reassign max_subarray diamensions
if (maxSoFar < thisSubarraySum)
{
maxSoFar = thisSubarraySum;
maxSoFarLeft = thsSubarrayleft;
maxSoFarRight = thisSubarrayRight;
}
}
// else continue this subarray
else
{
thisSubarraySum = sumInThisSubArray;
thisSubarrayRight = i;
}
}
cout << maxSoFarLeft << " " << maxSoFarRight << " " << maxSoFar << endl;
}