human
Give me a code to flip minimum signs of array elements to get minimum sum of positive elements possible and the other steps required given the following:
Flip minimum signs of array elements to get minimum sum of positive elements possible | Function to return the minimum number of elements whose sign must be flipped to get the positive sum of array elements as close to 0 as possible.
boolean variable used for toggling between maps.
Calculate the sum of all elements of the array.
Initializing first map ( dp [ 0 ] ) with INT_MAX because for i = 0 , there are no elements in the array to flip.
Base Case.
For toggling.
Required sum is minimum non - negative So , we iterate from i = 0 to sum and find the first i where dp [ flag ^ 1 ] [ i ] != INT_MAX.
In worst case we will flip max n - 1 elements.
Driver code"
How can the above be solved in Python?
assistant
#Here is code to flip minimum signs of array elements to get minimum sum of positive elements possible and the other steps required.
 def solve ( A , n ) :
	dp = [ [ 0 for i in range ( 2000 ) ] for i in range ( 2000 ) ]
	flag = 1
	sum = 0
	for i in range ( n ) :
		sum += A [ i ]
	for i in range ( - sum , sum + 1 ) :
		dp [ 0 ] [ i ] = 10 ** 9
	dp [ 0 ] [ 0 ] = 0
	for i in range ( 1 , n + 1 ) :
		for j in range ( - sum , sum + 1 ) :
			dp [ flag ] [ j ] = 10 ** 9
			if ( j - A [ i - 1 ] <= sum and j - A [ i - 1 ] >= - sum ) :
				dp [ flag ] [ j ] = dp [ flag ^ 1 ] [ j - A [ i - 1 ] ]
			if ( j + A [ i - 1 ] <= sum and j + A [ i - 1 ] >= - sum and dp [ flag ^ 1 ] [ j + A [ i - 1 ] ] != 10 ** 9 ) :
				dp [ flag ] [ j ] = min ( dp [ flag ] [ j ] , dp [ flag ^ 1 ] [ j + A [ i - 1 ] ] + 1 )
		flag = flag ^ 1
	for i in range ( sum + 1 ) :
		if ( dp [ flag ^ 1 ] [ i ] != 10 ** 9 ) :
			return dp [ flag ^ 1 ] [ i ]
	return n - 1
arr = [ 10 , 22 , 9 , 33 , 21 , 50 , 41 , 60 ]
n = len ( arr )
print ( solve ( arr , n ) )