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Maximum Subarray Sum after inverting at most two elements | Function to return the maximum required sub - array sum.
Creating one based indexing.
2d array to contain solution for each step.
Case 1 : Choosing current or ( current + previous ) whichever is smaller.
Case 2 : ( a ) Altering sign and add to previous case 1 or value 0.
Case 2 : ( b ) Adding current element with previous case 2 and updating the maximum.
Case 3 : ( a ) Altering sign and add to previous case 2.
Case 3 : ( b ) Adding current element with previous case 3.
Updating the maximum value of variable ans.
Return the final solution.
Driver code"
How can the above be solved in Python?
 ... What is a code to maximum Subarray Sum after inverting at most two elements and the other steps required?
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#Here is code to maximum Subarray Sum after inverting at most two elements and the other steps required.
 def maxSum ( a , n ) :
	ans = 0
	arr = [ 0 ] * ( n + 1 )
	for i in range ( 1 , n + 1 ) :
		arr [ i ] = a [ i - 1 ]
	dp = [ [ 0 for i in range ( 3 ) ] for j in range ( n + 1 ) ]
	for i in range ( 0 , n + 1 ) :
		dp [ i ] [ 0 ] = max ( arr [ i ] , dp [ i - 1 ] [ 0 ] + arr [ i ] )
		dp [ i ] [ 1 ] = max ( 0 , dp [ i - 1 ] [ 0 ] ) - arr [ i ]
		if i >= 2 :
			dp [ i ] [ 1 ] = max ( dp [ i ] [ 1 ] , dp [ i - 1 ] [ 1 ] + arr [ i ] )
		if i >= 2 :
			dp [ i ] [ 2 ] = dp [ i - 1 ] [ 1 ] - arr [ i ]
		if i >= 3 :
			dp [ i ] [ 2 ] = max ( dp [ i ] [ 2 ] , dp [ i - 1 ] [ 2 ] + arr [ i ] )
		ans = max ( ans , dp [ i ] [ 0 ] )
		ans = max ( ans , dp [ i ] [ 1 ] )
		ans = max ( ans , dp [ i ] [ 2 ] )
	return ans
if __name__ == " _ _ main _ _ " :
	arr = [ - 5 , 3 , 2 , 7 , - 8 , 3 , 7 , - 9 , 10 , 12 , - 6 ]
	n = len ( arr )
	print ( maxSum ( arr , n ) )