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Python Program for Largest Sum Contiguous Subarray

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  • Last Updated : 06 Jan, 2022
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Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum. 

kadane-algorithm

 

Kadane’s Algorithm:

Initialize:
    max_so_far = INT_MIN
    max_ending_here = 0

Loop for each element of the array
  (a) max_ending_here = max_ending_here + a[i]
  (b) if(max_so_far < max_ending_here)
            max_so_far = max_ending_here
  (c) if(max_ending_here < 0)
            max_ending_here = 0
return max_so_far

Explanation: 
The simple idea of Kadane’s algorithm is to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive-sum compare it with max_so_far and update max_so_far if it is greater than max_so_far 

    Lets take the example:
    {-2, -3, 4, -1, -2, 1, 5, -3}

    max_so_far = max_ending_here = 0

    for i=0,  a[0] =  -2
    max_ending_here = max_ending_here + (-2)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=1,  a[1] =  -3
    max_ending_here = max_ending_here + (-3)
    Set max_ending_here = 0 because max_ending_here < 0

    for i=2,  a[2] =  4
    max_ending_here = max_ending_here + (4)
    max_ending_here = 4
    max_so_far is updated to 4 because max_ending_here greater 
    than max_so_far which was 0 till now

    for i=3,  a[3] =  -1
    max_ending_here = max_ending_here + (-1)
    max_ending_here = 3

    for i=4,  a[4] =  -2
    max_ending_here = max_ending_here + (-2)
    max_ending_here = 1

    for i=5,  a[5] =  1
    max_ending_here = max_ending_here + (1)
    max_ending_here = 2

    for i=6,  a[6] =  5
    max_ending_here = max_ending_here + (5)
    max_ending_here = 7
    max_so_far is updated to 7 because max_ending_here is 
    greater than max_so_far

    for i=7,  a[7] =  -3
    max_ending_here = max_ending_here + (-3)
    max_ending_here = 4

Program: 

Python3




# Python program to find maximum contiguous subarray
  
# Function to find the maximum contiguous subarray
from math import inf
maxint=inf
def maxSubArraySum(a,size):
      
    max_so_far = -maxint - 1
    max_ending_here = 0
      
    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if (max_so_far < max_ending_here):
            max_so_far = max_ending_here
 
        if max_ending_here < 0:
            max_ending_here = 0  
    return max_so_far
  
# Driver function to check the above function
a = [-13, -3, -25, -20, -3, -16, -23, -12, -5, -22, -15, -4, -7]
print ("Maximum contiguous sum is", maxSubArraySum(a,len(a)))
  
#This code is contributed by _Devesh Agrawal_

Output:

Maximum contiguous sum is 7

Another approach:

 

Python3




def maxSubArraySum(a,size):
     
    max_so_far = a[0]
    max_ending_here = 0
     
    for i in range(0, size):
        max_ending_here = max_ending_here + a[i]
        if max_ending_here < 0:
            max_ending_here = 0
         
        # Do not compare for all elements. Compare only  
        # when  max_ending_here > 0
        elif (max_so_far < max_ending_here):
            max_so_far = max_ending_here
             
    return max_so_far

Time Complexity: O(n) 

Algorithmic Paradigm: Dynamic Programming
Following is another simple implementation suggested by Mohit Kumar. The implementation handles the case when all numbers in the array are negative. 

Python3




# Python program to find maximum contiguous subarray
 
def maxSubArraySum(a,size):
     
    max_so_far =a[0]
    curr_max = a[0]
     
    for i in range(1,size):
        curr_max = max(a[i], curr_max + a[i])
        max_so_far = max(max_so_far,curr_max)
         
    return max_so_far
 
# Driver function to check the above function
a = [-2, -3, 4, -1, -2, 1, 5, -3]
print("Maximum contiguous sum is" , maxSubArraySum(a,len(a)))
 
#This code is contributed by _Devesh Agrawal_

Output: 

Maximum contiguous sum is 7

To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.  

Python3




# Python 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
 
    print ("Maximum contiguous sum is %d"%(max_so_far))
    print ("Starting Index %d"%(start))
    print ("Ending Index %d"%(end))
 
# Driver program to test maxSubArraySum
a = [-2, -3, 4, -1, -2, 1, 5, -3]
maxSubArraySum(a,len(a))

Output: 

Maximum contiguous sum is 7
Starting index 2
Ending index 6

Kadane’s Algorithm can be viewed both as a greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 (Greedy Part). This is because continuing with a negative sum is way more worse than restarting with a new range. Now it can also be viewed as a DP, at each stage we have 2 choices: Either take the current element and continue with previous sum OR restart a new range. These both choices are being taken care of in the implementation. 

Time Complexity: O(n)

Auxiliary Space: O(1)

Now try the below question 
Given an array of integers (possibly some elements negative), write a C program to find out the *maximum product* possible by multiplying ‘n’ consecutive integers in the array where n ≤ ARRAY_SIZE. Also, print the starting point of the maximum product subarray.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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