PHP 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:
- If x is smaller than or equal to the first element in array then return 0(index of first element)
- Else Linearly search for an index i such that x lies between arr[i] and arr[i+1].
- If we do not find an index i in step 2, then return -1
Below is the implementation of the above approach:
<?php
// Function to get index of
// ceiling of x in arr[low..high]
function 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 $low;
// Otherwise, linearly search
// for ceil value
for($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
$arr = array(1, 2, 8, 10, 10, 12, 19);
$n = sizeof($arr);
$x = 3;
$index = ceilSearch($arr, 0, $n - 1, $x);
if($index == -1)
echo("Ceiling of " . $x .
" doesn't exist in array ");
else
echo("ceiling of " . $x . " is " .
$arr[$index]);
// This code is contributed by Ajit.
?>
Output
ceiling of 3 is 8
Complexity Analysis:
- 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).
<?php
// PHP 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)
{
$mid;
/* If x is smaller than or
equal to the first element,
then return the first element */
if($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
$arr = array(1, 2, 8, 10, 10, 12, 19);
$n = sizeof($arr);
$x = 20;
$index = ceilSearch($arr, 0, $n - 1, $x);
if($index == -1)
echo("Ceiling of $x doesn't exist in array ");
else
echo("ceiling of $x is");
echo(isset($arr[$index]));
// This code is contributed by nitin mittal.
?>
Output
Ceiling of 20 doesn't exist in array
Complexity Analysis:
- Time Complexity: O(Logn)
- Auxiliary Space: O(Logn)
The extra space is used in recursive call stack.
Related Articles:
Foor in a Sorted Array
Find floor and ceil in an unsorted array
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!
