Python Program for Largest Sum Contiguous Subarray - GeeksforGeeks

06 Jan, 2022
Write an efficient program to find the sum of contiguous subarray within a one-dimensional array of numbers that has the largest sum. 
 
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 subarrayfrom math import infmaxint=infdef 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 functiona = [-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 ProgrammingFollowing 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 functiona = [-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 indexdef 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 maxSubArraySuma = [-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.
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