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Python3 Program for Size of The Subarray With Maximum Sum

Last Updated : 27 Jul, 2023
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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




# Python3 program to print largest contiguous array sum
  
from sys import maxsize
  
# Function to find the maximum contiguous subarray
# and print its starting and end index
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)
  
# Driver program to test maxSubArraySum
a = [-2, -3, 4, -1, -2, 1, 5, -3]
print(maxSubArraySum(a,len(a)))


Output : 

5

 

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:

  1.  Initialize max_sum, current_sum, start, end, max_start, and max_end to the first element of the array.
  2.  Iterate through the array from the second element.
  3.  If the current element is greater than the sum of the current element and current_sum, update start to the current index.
  4.  Update current_sum as the maximum of the current element and the sum of current element and current_sum.
  5.  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. 
  6. Return max_end – max_start + 1 as the size of the subarray with maximum sum.

Below is the implementation of the approach:

Python3




# Python3 implementation of the approach
  
# Function to find the maximum subarray sum
  
  
def max_subarray_sum(a):
  
    n = len(a)
  
    # Initializing initial and final sum to 0
    max_sum = a[0]
    current_sum = a[0]
    start = 0
    end = 0
    max_start = 0
    max_end = 0
  
    # Traverse the array
    for i in range(1, n):
        # If the current element is greater than the sum so
        # far plus the current element, then update the
        # start index to the current index
        if a[i] > current_sum + a[i]:
            start = i
  
    # Update the current sum to be either the current
    # element or the sum so far plus the current
    # element
        current_sum = max(a[i], current_sum + a[i])
  
    # If the current sum is greater than the maximum
    # sum so far, then update the maximum sum and its
    # start and end indices
        if current_sum > max_sum:
            max_sum = current_sum
            end = i
            max_start = start
            max_end = end
  
# Return the length of the maximum subarray
    return max_end - max_start + 1
  
  
# Initializing array
if __name__ == "__main__":
    a = [-2, -3, 4, -1, -2, 1, 5, -3]
  
    # Function call
    print(max_subarray_sum(a))


Output

5

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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