Javascript 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 than 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 

Javascript

<script>

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

    let i;

    /* 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 */
    for(i = low; i < high; i++)
    {
        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 && 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 code

    let arr = [1, 2, 8, 10, 10, 12, 19];
    let n = arr.length;
    let x = 3;
    let index = ceilSearch(arr, 0, n-1, x);
    if(index == -1)
        document.write("Ceiling of " + x + " doesn't exist in array ");
    else
        document.write ("ceiling of " + x + " is " + arr[index]);


</script>

Output :

ceiling of 3 is 8

Time Complexity : O(n)

Auxiliary Space: O(1)

As constant extra space is used.
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). 

Javascript

<script>
// Javascript Program for Ceiling in
// a sorted array

// Function to get index of ceiling
// of x in arr[low..high]
function ceilSearch(arr, low, high, x)
{
    let mid;

    /* 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] */
    // low + (high - low)/2
    mid = (low + high)/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] */
    else if(arr[mid] < x)
    {
        if(mid + 1 <= high && 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 && x > arr[mid - 1])
          return mid;
        else
         return ceilSearch(arr, low, mid - 1, x);
    }
}

// Driver Code
let arr = [1, 2, 8, 10, 10, 12, 19];
let n = arr.length;
let x = 20;
let index = ceilSearch(arr, 0, n - 1, x);

if(index == -1){
    document.write(`Ceiling of ${x} doesn't exist in array `);
}else{
    document.write(`ceiling of ${x} is ${arr[index]}`);
}

// This code is contributed by _saurabh_jaiswal.

</script>

Output :

Ceiling of 20 doesn't exist in array 

Time Complexity: O(Logn)

Auxiliary Space: O(Logn)

The extra space is used in recursive call stack.

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!