# 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

## C++

 `// C++ implementation of above approach ` `#include ` `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 `

## C

 `#include ` ` `  `/* 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; ` `} `

## Java

 `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]); ` `    ``}   ` `} `

## 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 `

## C#

 `// C# program to find celing ` `// 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

 `= ``\$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
```

Time Complexity : O(n)

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).

## C

 `#include ` ` `  `/* 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[mid-1...high] */`     `  ``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; ` `} `

## Java

 `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` `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[mid-1...high] */`    `      ``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 */` `    ``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]); ` `    ``}   ` `} `

## 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[mid-1...high] */    ` `    ``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#

 `// C# program to find celing ` `// 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[mid-1...high]  ` `        ``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 = 8; ` `        ``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

 ` ``\$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[mid-1...high] */` `    ``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
```

Time Complexity: O(Logn)

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, or find better ways to solve the same problem, or want to share code for floor implementation.