human
Smallest indexed array element required to be flipped to make sum of array equal to 0 | Function to find the smallest indexed array element required to be flipped to make sum of the given array equal to 0.
Stores the required index.
Stores the sum of the array.
Traverse the given array.
Update ArrSum.
Traverse the given array.
If sum of array elements double the value of the current element.
Update pos.
Driver Code"
Solution in Python:
assistant
#Here is code to smallest indexed array element required to be flipped to make sum of array equal to 0 and the other steps required.
 def smallestIndexArrayElementsFlip ( arr , N ) :
	pos = - 1
	ArrSum = 0
	for i in range ( N ) :
		ArrSum += arr [ i ]
	for i in range ( N ) :
		if ( 2 * arr [ i ] == ArrSum ) :
			pos = i
			break
	return pos
arr = [ 1 , 3 , - 5 , 3 , 4 ]
N = len ( arr )
print ( smallestIndexArrayElementsFlip ( arr , N ) )