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
# 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:
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.
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