Python3 Program for Ceiling in a sorted array

Given a sorted array and a value x, the ceiling of x is the smallest element in array greater than or equal to x, and the floor is the greatest element smaller than or equal to x. Assume that the array is sorted in non-decreasing order. Write efficient functions to find floor and ceiling of x. 

Examples :

For example, let the input array be {1, 2, 8, 10, 10, 12, 19}
For x = 0:    floor doesn't exist in array,  ceil  = 1
For x = 1:    floor  = 1,  ceil  = 1
For x = 5:    floor  = 2,  ceil  = 8
For x = 20:   floor  = 19,  ceil doesn't exist in array

In below methods, we have implemented only ceiling search functions. Floor search can be implemented in the same way.
Method 1 (Linear Search)
Algorithm to search ceiling of x: 
1) If x is smaller than or equal to the first element in array then return 0(index of first element) 
2) Else Linearly search for an index i such that x lies between arr[i] and arr[i+1]. 
3) If we do not find an index i in step 2, then return -1 

Below is the implementation of the above approach:

Python3

# Function to get index of ceiling of x in arr[low..high] */
def ceilSearch(arr, low, high, x):

    # If x is smaller than or equal to first element,
    # then return the first element */
    if x <= arr[low]:
        return low

    # Otherwise, linearly search for ceil value */
    i = low
    for i in range(high):
        if arr[i] == x:
            return i

        # if x lies between arr[i] and arr[i+1] including
        # arr[i+1], then return arr[i+1] */
        if arr[i] < x and arr[i+1] >= x:
            return i+1

    # If we reach here then x is greater than the last element
    # of the array,  return -1 in this case */
    return -1

# Driver program to check above functions */
arr = [1, 2, 8, 10, 10, 12, 19]
n = len(arr)
x = 3
index = ceilSearch(arr, 0, n-1, x);

if index == -1:
    print ("Ceiling of %d doesn't exist in array "% x)
else:
    print ("ceiling of %d is %d"%(x, arr[index]))

# This code is contributed by Shreyanshi Arun

Output

ceiling of 3 is 8

Time Complexity : O(n)
Auxiliary Space: O(1)

Method 2 (Binary Search)
Instead of using linear search, binary search is used here to find out the index. Binary search reduces time complexity to O(Logn). 

Python3

# Function to get index of ceiling of x in arr[low..high]*/
def ceilSearch(arr, low, high, x):

    # If x is smaller than or equal to the first element,
    # then return the first element */
    if x <= arr[low]:
        return low

    # If x is greater than the last element, then return -1 */
    if x > arr[high]:
        return -1

    # get the index of middle element of arr[low..high]*/
    mid = (low + high)/2;  # low + (high - low)/2 */

    # If x is same as middle element, then return mid */
    if arr[mid] == x:
        return mid

    # If x is greater than arr[mid], then either arr[mid + 1]
    # is ceiling of x or ceiling lies in arr[mid+1...high] */
    elif arr[mid] < x:
        if mid + 1 <= high and x <= arr[mid+1]:
            return mid + 1
        else:
            return ceilSearch(arr, mid+1, high, x)

    # If x is smaller than arr[mid], then either arr[mid]
    # is ceiling of x or ceiling lies in arr[low...mid-1] */
    else:
        if mid - 1 >= low and x > arr[mid-1]:
            return mid
        else:
            return ceilSearch(arr, low, mid - 1, x)

# Driver program to check above functions */
arr = [1, 2, 8, 10, 10, 12, 19]
n = len(arr)
x = 20
index = ceilSearch(arr, 0, n-1, x);

if index == -1:
    print ("Ceiling of %d doesn't exist in array "% x)
else:
    print ("ceiling of %d is %d"%(x, arr[index]))

# This code is contributed by Shreyanshi Arun

Output

Ceiling of 20 doesn't exist in array 

Time Complexity: O(log n)
Auxiliary Space: O(1)

Please write comments if you find any of the above codes/algorithms incorrect, or find better ways to solve the same problem, or want to share code for floor implementation.

Please refer complete article on Ceiling in a sorted array for more details!