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Smallest subarray with sum greater than a given value | Returns length of smallest subarray with sum greater than x . If there is no subarray with given sum , then returns n + 1.
Initialize current sum and minimum length.
Initialize starting and ending indexes.
Keep adding array elements while current sum is smaller than or equal to x.
If current sum becomes greater than x ..
Update minimum length if needed.
remove starting elements.
Driver program"
Solution in Python:
 ... What is a code to smallest subarray with sum greater than a given value and the other steps required?
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#Here is code to smallest subarray with sum greater than a given value and the other steps required.
 def smallestSubWithSum ( arr , n , x ) :
	curr_sum = 0
	min_len = n + 1
	start = 0
	end = 0
	while ( end < n ) :
		while ( curr_sum <= x and end < n ) :
			curr_sum += arr [ end ]
			end += 1
		while ( curr_sum > x and start < n ) :
			if ( end - start < min_len ) :
				min_len = end - start
			curr_sum -= arr [ start ]
			start += 1
	return min_len
arr1 = [ 1 , 4 , 45 , 6 , 10 , 19 ]
x = 51
n1 = len ( arr1 )
res1 = smallestSubWithSum ( arr1 , n1 , x )
print ( " Not ▁ possible " ) if ( res1 == n1 + 1 ) else print ( res1 )
arr2 = [ 1 , 10 , 5 , 2 , 7 ]
n2 = len ( arr2 )
x = 9
res2 = smallestSubWithSum ( arr2 , n2 , x )
print ( " Not ▁ possible " ) if ( res2 == n2 + 1 ) else print ( res2 )
arr3 = [ 1 , 11 , 100 , 1 , 0 , 200 , 3 , 2 , 1 , 250 ]
n3 = len ( arr3 )
x = 280
res3 = smallestSubWithSum ( arr3 , n3 , x )
print ( " Not ▁ possible " ) if ( res3 == n3 + 1 ) else print ( res3 )