Given a sorted array and a value x, the ceiling of x is the smallest element in an 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 the 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 the 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 the array then return 0(index of the 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:
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
/* Function to get index of ceiling of x in arr[low..high] */ int ceilSearch( int arr[], int low, int high, int x)
{ int 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*/ int main()
{ int arr[] = {1, 2, 8, 10, 10, 12, 19};
int n = sizeof (arr)/ sizeof (arr[0]);
int x = 3;
int index = ceilSearch(arr, 0, n-1, x);
if (index == -1)
cout << "Ceiling of " << x << " doesn't exist in array " ;
else
cout << "ceiling of " << x << " is " << arr[index];
return 0;
} // This is code is contributed by rathbhupendra |
#include<stdio.h> /* Function to get index of ceiling of x in arr[low..high] */ int ceilSearch( int arr[], int low, int high, int x)
{ int 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 program to check above functions */ int main()
{ int arr[] = {1, 2, 8, 10, 10, 12, 19};
int n = sizeof (arr)/ sizeof (arr[0]);
int x = 3;
int index = ceilSearch(arr, 0, n-1, x);
if (index == -1)
printf ( "Ceiling of %d doesn't exist in array " , x);
else
printf ( "ceiling of %d is %d" , x, arr[index]);
getchar ();
return 0;
} |
class Main
{ /* Function to get index of ceiling
of x in arr[low..high] */
static int ceilSearch( int arr[], int low, int high, int x)
{
int 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 program to check above functions */
public static void main (String[] args)
{
int arr[] = { 1 , 2 , 8 , 10 , 10 , 12 , 19 };
int n = arr.length;
int x = 3 ;
int index = ceilSearch(arr, 0 , n- 1 , x);
if (index == - 1 )
System.out.println( "Ceiling of " +x+ " doesn't exist in array" );
else
System.out.println( "ceiling of " +x+ " is " +arr[index]);
}
} |
# 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 |
// C# program to find ceiling // in a sorted array using System;
class GFG {
// Function to get index of ceiling
// of x in arr[low..high]
static int ceilSearch( int [] arr, int low,
int high, int x)
{
int 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
public static void Main()
{
int [] arr = { 1, 2, 8, 10, 10, 12, 19 };
int n = arr.Length;
int x = 3;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
Console.Write( "Ceiling of " + x +
" doesn't exist in array" );
else
Console.Write( "ceiling of " + x +
" is " + arr[index]);
}
} // This code is contributed by Sam007. |
<?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. ?> |
<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> |
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 the time complexity to O(Logn).
#include <bits/stdc++.h> using namespace std;
/* Function to get index of ceiling of x in arr[low..high]*/
int ceilSearch( int arr[], int low, int high, int x)
{ int 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]*/
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] */
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 int main()
{ int arr[] = { 1, 2, 8, 10, 10, 12, 19 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 20;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
cout << "Ceiling of " << x
<< " doesn't exist in array " ;
else
cout << "ceiling of " << x << " is " << arr[index];
return 0;
} // This code is contributed by rathbhupendra |
#include <stdio.h> /* Function to get index of ceiling of x in arr[low..high]*/ int ceilSearch( int arr[], int low, int high, int x)
{ int 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]*/
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] */
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 program to check above functions */ int main()
{ int arr[] = { 1, 2, 8, 10, 10, 12, 19 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 20;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
printf ( "Ceiling of %d doesn't exist in array " , x);
else
printf ( "ceiling of %d is %d" , x, arr[index]);
getchar ();
return 0;
} |
import java.util.Arrays;
public class CeilSearch {
// Function to get index of ceiling of x in arr[low..high]
public static int ceilSearch( int [] arr, int low, int high, int x) {
int 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]*/
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] */
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);
}
}
public static void main(String[] args) {
int [] arr = { 1 , 2 , 8 , 10 , 10 , 12 , 19 };
int n = arr.length;
int x = 20 ;
int index = ceilSearch(arr, 0 , n - 1 , x);
if (index == - 1 )
System.out.println( "Ceiling of " + x + " doesn't exist in array" );
else
System.out.println( "ceiling of " + x + " is " + arr[index]);
}
} |
# 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 |
// C# program to find ceiling // in a sorted array using System;
class GFG {
// Function to get index of ceiling
// of x in arr[low..high]
static int ceilSearch( int [] arr, int low, int high,
int x)
{
int 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]
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]
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
public static void Main()
{
int [] arr = { 1, 2, 8, 10, 10, 12, 19 };
int n = arr.Length;
int x = 20;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
Console.Write( "Ceiling of " + x
+ " doesn't exist in array" );
else
Console.Write( "ceiling of " + x + " is "
+ arr[index]);
}
} // This code is contributed by Sam007. |
<?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. ?> |
<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> |
Ceiling of 20 doesn't exist in array
Time Complexity: O(log(n)),
Auxiliary Space: O(1)
Another Implementation of Method 2 :
As like previous method here also binary search is being used but the code logic is different instead of lots of if else check i will simply return and lets understand through below steps :
Step 1 : { low->1, 2, 8, 10<-mid, 10, 12, 19<-high};
if( x < mid) yes set high = mid -1;
Step 2 : { low ->1, 2 <-mid, 8 <-high, 10, 10, 12, 19};
if( x < mid) no set low = mid + 1;
Step 3 : {1, 2, 8<-high,low,mid, 10, 10, 12, 19};
if( x == mid ) yes return mid if(x < mid ) no low = mid + 1
Step 4 : {1, 2, 8<-high,mid, 10<-low, 10, 12, 19};
check while(low =< high)
condition break and return low which is my ceiling of target.
#include <bits/stdc++.h> using namespace std;
/* Function to get index of ceiling of x in arr[low..high]*/
int ceilSearch( int arr[], int low, int high, int x)
{ // base condition if length of arr == 0 then return -1
if ( sizeof (arr) / sizeof (arr[0]) == 0) {
return -1;
}
int mid;
// this while loop function will run until condition not
// break once condition break loop will return start and
// ans is low which will be next smallest greater than
// target which is ceiling
while (low <= high) {
mid = low + (high - low) / 2;
if (arr[mid] == x)
return mid;
else if (x < arr[mid])
high = mid - 1;
else
low = mid + 1;
}
return low;
} /* step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high}; if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12,
19}; if( x < mid) no set low = mid + 1; step 3 : {1, 2, 8
= high,low,low, 10, 10, 12, 19}; if( x == mid ) yes
return mid if(x < mid ) no low = mid + 1 step 4 : {1, 2,
8 = high,mid, 10 = low, 10, 12, 19}; check while(low < =
high) condition break and return low which will next
greater of target */
/* Driver program to check above functions */ int main()
{ int arr[] = { 1, 2, 8, 10, 10, 12, 19 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 8;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
printf ( "Ceiling of %d does not exist in an array" ,
x);
else
printf ( "Ceiling of %d is %d" , x, arr[index]);
return 0;
} |
#include <stdio.h> // Function to get index of ceiling of x in arr[low..high] int ceilSearch( int arr[], int low, int high, int x)
{ // base condition if length of arr == 0 then return -1
if (x == 0) {
return -1;
}
int mid;
// this while loop function will run until condition not
// break once condition break loop will return start and
// ans is low which will be next smallest greater than
// target which is ceiling
while (low <= high) {
mid = low + (high - low) / 2;
if (arr[mid] == x)
return mid;
else if (x < arr[mid])
high = mid - 1;
else
low = mid + 1;
}
return low;
} /* step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high}; if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12,
19}; if( x < mid) no set low = mid + 1; step 3 : {1, 2, 8
= high,low,low, 10, 10, 12, 19}; if( x == mid ) yes
return mid if(x < mid ) no low = mid + 1 step 4 : {1, 2,
8 = high,mid, 10 = low, 10, 12, 19}; check while(low < =
high) condition break and return low which will next
greater of target */
/* Driver program to check above functions */ int main()
{ int arr[] = { 1, 2, 8, 10, 10, 12, 19 };
int n = sizeof (arr) / sizeof (arr[0]);
int x = 8;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
printf ( "Ceiling of %d does not exist in an array" , x);
else
printf ( "Ceiling of %d is %d" , x, arr[index]);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
class Main {
/* Function to get index of
ceiling of x in arr[low..high]*/
static int ceilSearch( int arr[], int low, int high,
int x)
{
// base condition if length of arr == 0 then return
// -1
if (x == 0 ) {
return - 1 ;
}
/* this while loop function will run until condition
not break once condition break loop will return
start and ans is low which will be next smallest
greater than target which is ceiling*/
while (low <= high) {
int mid
= low + (high - low) / 2 ; // calculate mid
if (x == arr[mid]) {
return mid;
}
if (x < arr[mid]) {
high = mid - 1 ;
}
else {
low = mid + 1 ;
}
}
return low;
}
/* step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high};
if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12, 19};
if( x < mid) no set low = mid + 1;
step 3 : {1, 2, 8 = high,low,low, 10, 10, 12, 19};
if( x == mid ) yes return mid
if(x < mid ) no low = mid + 1
step 4 : {1, 2, 8 = high,mid, 10 = low, 10, 12, 19};
check while(low < = high)
condition break and return low which will
next greater of target */
/* Driver program to check above functions */
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 8 , 10 , 10 , 12 , 19 };
int n = arr.length;
int x = 8 ;
int index = ceilSearch(arr, 0 , n - 1 , x);
if (index == - 1 )
System.out.println( "Ceiling of " + x
+ " doesn't exist in array" );
else
System.out.println( "ceiling of " + x + " is "
+ arr[index]);
}
} |
# Function to get index of ceiling of x in arr[low..high] def ceilSearch(arr, low, high, x):
# base condition if length of arr == 0 then return -1
if (x = = 0 ):
return - 1
"""this while loop function will run until
condition not break once condition break
loop will return start and ans is low
which will be next smallest greater than target
which is ceiling"""
while (low < = high):
mid = low + (high - low) / 2
mid = int (mid)
if (arr[mid] = = x):
return mid
elif (x < arr[mid]):
high = mid - 1
else :
low = mid + 1
return low
""" step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high}; if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12, 19};
if( x < mid) no set low = mid + 1;
step 3 : {1, 2, 8 = high,low,low, 10, 10, 12, 19};
if( x == mid ) yes return mid
if(x < mid ) no low = mid + 1
step 4 : {1, 2, 8 = high,mid, 10 = low, 10, 12, 19};
check while(low < = high)
condition break and return low which will next greater of target """
# Driver program to check above functions arr = [ 1 , 2 , 8 , 10 , 10 , 12 , 19 ]
n = len (arr)
x = 8
index = ceilSearch(arr, 0 , n - 1 , x)
if (index = = - 1 ):
print ( "Ceiling of" , x, "does not exist in an array" )
else :
print ( "Ceiling of" , x, "is" , arr[index])
|
// C# program for the above approach using System;
class GFG
{ /* Function to get index of
ceiling of x in arr[low..high]*/
static int ceilSearch( int [] arr, int low, int high, int x)
{
// base condition if length of arr == 0 then return -1
if (x == 0)
{
return -1;
}
/* this while loop function will run until condition not break once condition break
loop will return start and ans is low which will be next smallest greater than target
which is ceiling*/
while (low <= high)
{
int mid = low + (high - low) / 2; //calculate mid
if (x == arr[mid])
{
return mid;
}
if (x < arr[mid])
{
high = mid - 1;
}
else
{
low = mid + 1;
}
}
return low;
/* step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high};
if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12, 19};
if( x < mid) no set low = mid + 1;
step 3 : {1, 2, 8 = high,low,low, 10, 10, 12, 19};
if( x == mid ) yes return mid
if(x < mid ) no low = mid + 1
step 4 : {1, 2, 8 = high,mid, 10 = low, 10, 12, 19};
check while(low < = high)
condition break and return low which will next greater of target */
}
/* Driver program to check above functions */
public static void Main()
{
int [] arr = { 1, 2, 8, 10, 10, 12, 19 };
int n = arr.Length;
int x = 8;
int index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
Console.WriteLine( "Ceiling of " + x + " doesn't exist in array" );
else
Console.WriteLine( "ceiling of " + x + " is " + arr[index]);
}
} |
//JS program to implement the approach /* Function to get index of ceiling of x in arr[low..high]*/
function ceilSearch(arr, low, high, x)
{ // base condition if length of arr == 0 then return -1
if (x == 0)
{
return -1;
}
var mid;
/* this while loop function will run until
condition not break once condition break
loop will return start and ans is low
which will be next smallest greater than target
which is ceiling*/
while (low <= high)
{
mid = low + (high - low) / 2;
if (arr[mid] == x)
{
return mid;
}
else if (x < arr[mid])
{
high = mid - 1;
}
else
{
low = mid + 1;
}
}
return low;
} /* step 1 : { low = 1, 2, 8, 10= mid, 10, 12, 19= high}; if( x < mid) yes set high = mid -1;
step 2 : { low = 1, 2 = mid, 8 = high, 10, 10, 12, 19};
if( x < mid) no set low = mid + 1;
step 3 : {1, 2, 8 = high,low,low, 10, 10, 12, 19};
if( x == mid ) yes return mid
if(x < mid ) no low = mid + 1
step 4 : {1, 2, 8 = high,mid, 10 = low, 10, 12, 19};
check while(low < = high)
condition break and return low which will next greater of target */
/* Driver program to check above functions */ var arr = [1, 2, 8, 10, 10, 12, 19];
var n = arr.length;
var x = 8;
var index = ceilSearch(arr, 0, n - 1, x);
if (index == -1)
{
console.log( "Ceiling of " + x + " does not exist in an array" );
}
else
{
console.log( "Ceiling of " + x + " is " + arr[index]);
}
|
Ceiling of 8 is 8
Time Complexity: O(log(n)), where n is the length of the given array,
Auxiliary Space: O(1)
Method Using C++ STL lower_bound
The lower_bound() method in C++ is used to return an iterator pointing to the first element in the range [first, last) which has a value not less than val. This means that the function returns an iterator pointing to the next smallest number just greater than or equal to that number. If there are multiple values that are equal to val, lower_bound() returns the iterator of the first such value.
Simpler and Shorter code :
#include <bits/stdc++.h> using namespace std;
int main()
{ vector< int > arr = { 1, 2, 8, 10, 10, 12, 19 };
int n = arr.size();
int x = 8;
auto itr = lower_bound(arr.begin(), arr.end(),x); // returns iterator
int idx = itr - arr.begin(); // converting to index;
if (idx == n) {
cout << "Ceil Element does not exist " << endl;
}
else {
cout << "Ceil Element of " << x << " is " << arr[idx] << endl;
}
return 0;
} |
import java.util.Arrays;
class GFG {
public static void main(String[] args)
{
int [] arr = { 1 , 2 , 8 , 10 , 10 , 12 , 19 };
int n = arr.length;
int x = 8 ;
// Use binary search to find the index of the
// ceiling element
int idx = Arrays.binarySearch(arr, x);
if (idx < 0 ) {
idx = Math.abs(idx) - 1 ;
}
// Checking if idx is valid
if (idx == n) {
System.out.println(
"Ceiling Element does not exist" );
}
else {
System.out.println( "Ceiling Element of " + x
+ " is " + arr[idx]);
}
}
} // This code is contributed by phasing17 |
from bisect import bisect_left
arr = [ 1 , 2 , 8 , 10 , 10 , 12 , 19 ]
n = len (arr)
x = 8
# Use bisect to get ceiling element idx = bisect_left(arr, x)
# Checking if idx is valid if idx = = n:
print ( "Ceil Element does not exist" )
else :
print (f "Ceil Element of {x} is {arr[idx]}" )
|
using System;
public class GFG {
public static void Main( string [] args)
{
int [] arr = { 1, 2, 8, 10, 10, 12, 19 };
int n = arr.Length;
int x = 8;
// Use Array.BinarySearch to find the index of the
// ceiling element
int idx = Array.BinarySearch(arr, x);
if (idx < 0) {
idx = Math.Abs(idx) - 1;
}
// Checking if idx is valid
if (idx == n) {
Console.WriteLine(
"Ceiling Element does not exist" );
}
else {
Console.WriteLine( "Ceiling Element of " + x
+ " is " + arr[idx]);
}
}
} // This code is contributed by Prasad Kandekar(prasad264) |
const arr = [1, 2, 8, 10, 10, 12, 19]; const n = arr.length; const x = 8; // Use the Array.findIndex() method to find the index of the // first element that is greater than or equal to x let idx = arr.findIndex(val => val >= x); // Checking if idx is valid if (idx === -1) {
console.log( "Ceiling Element does not exist" );
} else {
console.log(`Ceiling Element of ${x} is ${arr[idx]}`);
} // This code is contributed by Prasad Kandekar(prasad264) |
Ceil Element of 8 is 8
Time Complexity: O(log(n)), where n is the length of the given array,
Auxiliary Space: O(1)
https://www.youtube.com/watch?v=Nzm9emAkSCM
Related Articles:
Floor 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, find better ways to solve the same problem, or want to share code for floor implementation.