Python Program for Largest Sum Contiguous Subarray
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
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
a = [ - 13 , - 3 , - 25 , - 20 , - 3 , - 16 , - 23 , - 12 , - 5 , - 22 , - 15 , - 4 , - 7 ]
print ( "Maximum contiguous sum is" , maxSubArraySum(a, len (a)))
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Output:
Maximum contiguous sum is 7
Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.
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
elif (max_so_far < max_ending_here):
max_so_far = max_ending_here
return max_so_far
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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
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
a = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
print ( "Maximum contiguous sum is" , maxSubArraySum(a, len (a)))
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Output:
Maximum contiguous sum is 7
To print the subarray with the maximum sum, we maintain indices whenever we get the maximum sum.
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
print ( "Maximum contiguous sum is %d" % (max_so_far))
print ( "Starting Index %d" % (start))
print ( "Ending Index %d" % (end))
a = [ - 2 , - 3 , 4 , - 1 , - 2 , 1 , 5 , - 3 ]
maxSubArraySum(a, len (a))
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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.
Last Updated :
14 Mar, 2023
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