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Count of non decreasing Arrays with ith element in range [ A [ i ] , B [ i ] ] | Function to count the total number of possible valid arrays.
Make a 2D DP table.
Make a 2D prefix sum table.
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Base Case.
Initialize the prefix values.
Iterate over the range and update the dp table accordingly.
Add the dp values to the prefix sum.
Update the prefix sum table.
Find the result count of arrays formed.
Return the total count of arrays.
Driver Code"
Solution in Python:

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What is a code to count of non decreasing Arrays with ith element in range [ A [ i ] , B [ i ] ] and the other steps required?
assistant
#Here is code to count of non decreasing Arrays with ith element in range [ A [ i ] , B [ i ] ] and the other steps required.
 def totalValidArrays ( a , b , N ) :
	dp = [ [ 0 for _ in range ( b [ N - 1 ] + 1 ) ] for _ in range ( N + 1 ) ]
	pref = [ [ 0 for _ in range ( b [ N - 1 ] + 1 ) ] for _ in range ( N + 1 ) ]
/ * Initialize all values to 0 * /
	dp [ 0 ] [ 0 ] = 1
	for i in range ( 0 , b [ N - 1 ] + 1 ) :
		pref [ 0 ] [ i ] = 1
	for i in range ( 1 , N + 1 ) :
		for j in range ( a [ i - 1 ] , b [ i - 1 ] + 1 ) :
			dp [ i ] [ j ] += pref [ i - 1 ] [ j ]
			pref [ i ] [ j ] += dp [ i ] [ j ]
		for j in range ( 0 , b [ N - 1 ] + 1 ) :
			if ( j > 0 ) :
				pref [ i ] [ j ] += pref [ i ] [ j - 1 ]
	ans = 0
	for i in range ( a [ N - 1 ] , b [ N - 1 ] + 1 ) :
		ans += dp [ N ] [ i ]
	return ans
if __name__ == " _ _ main _ _ " :
	A = [ 1 , 1 ]
	B = [ 2 , 3 ]
	N = len ( A )
	print ( totalValidArrays ( A , B , N ) )