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:
# 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
Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.
Another approach:
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.
# 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.
# 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.