Python3 Program for Size of The Subarray With Maximum Sum
Last Updated :
27 Jul, 2023
An array is given, find length of the subarray having maximum sum.
Examples :
Input : a[] = {1, -2, 1, 1, -2, 1}
Output : Length of the subarray is 2
Explanation: Subarray with consecutive elements
and maximum sum will be {1, 1}. So length is 2
Input : ar[] = { -2, -3, 4, -1, -2, 1, 5, -3 }
Output : Length of the subarray is 5
Explanation: Subarray with consecutive elements
and maximum sum will be {4, -1, -2, 1, 5}.
This problem is mainly a variation of Largest Sum Contiguous Subarray Problem.
The idea is to update starting index whenever sum ending here becomes less than 0.
Python3
from sys import maxsize
def maxSubArraySum(a,size):
max_so_far = - maxsize - 1
max_ending_here = 0
start = 0
end = 0
s = 0
for i in range ( 0 ,size):
max_ending_here + = a[i]
if max_so_far < max_ending_here:
max_so_far = max_ending_here
start = s
end = i
if max_ending_here < 0 :
max_ending_here = 0
s = i + 1
return (end - start + 1 )
a = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
print (maxSubArraySum(a, len (a)))
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Time Complexity: O(N) where N is the size of the input array. This is because a for loop is executed from 1 to the size of the array.
Auxiliary Space: O(1) as no extra space has been taken.
Approach#2: Using Kadane’s algorithm
This approach implements Kadane’s algorithm to find the maximum subarray sum and returns the size of the subarray with the maximum sum.
Algorithm:
- Initialize max_sum, current_sum, start, end, max_start, and max_end to the first element of the array.
- Iterate through the array from the second element.
- If the current element is greater than the sum of the current element and current_sum, update start to the current index.
- Update current_sum as the maximum of the current element and the sum of current element and current_sum.
- If current_sum is greater than max_sum, update max_sum, end to the current index, and max_start and max_end to start and end respectively.
- Return max_end – max_start + 1 as the size of the subarray with maximum sum.
Below is the implementation of the approach:
Python3
def max_subarray_sum(a):
n = len (a)
max_sum = a[ 0 ]
current_sum = a[ 0 ]
start = 0
end = 0
max_start = 0
max_end = 0
for i in range ( 1 , n):
if a[i] > current_sum + a[i]:
start = i
current_sum = max (a[i], current_sum + a[i])
if current_sum > max_sum:
max_sum = current_sum
end = i
max_start = start
max_end = end
return max_end - max_start + 1
if __name__ = = "__main__" :
a = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
print (max_subarray_sum(a))
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Time Complexity: O(n), where n is length of array
Auxiliary Space: O(1)
Note: The above code assumes that there is at least one positive element in the array. If all the elements are negative, the code needs to be modified to return the maximum element in the array.
Please refer complete article on Size of The Subarray With Maximum Sum for more details!
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