Ceil of a number using binary search

ceil.py

# http://www.geeksforgeeks.org/search-floor-and-ceil-in-a-sorted-array/


def binarySearch(arr, l, r, x):

    if x < arr[l]:
    	return arr[l]

    if x > arr[r]:
    	return 	arr[r]

    while l <= r:
        mid = l + (r - l) / 2

        # Check if x is present at mid
        if arr[mid] == x:
            return arr[mid]

        # the two if cond at the top prevents out of bound exception
        if arr[mid] < x and arr[mid + 1] > x:
            return arr[mid + 1]

        # If x is greater, ignore left half
        if arr[mid] < x:
            l = mid + 1

        # If x is smaller, ignore right half
        else:
            r = mid - 1

    # If we reach here, then the element was not present
    return -1


# Test array
arr = [2, 3, 5, 10, 40]
x = 1

# Function call
result = binarySearch(arr, 0, len(arr) - 1, x)

if result != -1:
    print "Ceil Is %d" % result
else:
    print "None"