# Front and Back Search in unsorted array

• Difficulty Level : Easy
• Last Updated : 05 Aug, 2022

Given an unsorted array of integers and an element x, find if x is present in array using Front and Back search.

Examples :

```Input : arr[] = {10, 20, 80, 30, 60, 50,
110, 100, 130, 170}
x = 110;
Output : Yes

Input : arr[] = {10, 20, 80, 30, 60, 50,
110, 100, 130, 170}
x = 175;
Output : No```

A simple solution is to perform linear search. The linear search algorithm always results in worst case complexity of O(n) when element to be searched is last element in the array. Front and Back search algorithm for finding element with value x works the following way:

1. Initialize indexes front and back pointing to first and last element respectively of the array.
2. If front is greater than rear, return false.
3. Check the element x at front and rear index.
4. If element x is found return true.
5. Else increment front and decrement rear and go to step 2.

Key Points:

• The worst case complexity is O(n/2) (equivalent to O(n)) when element is in the middle or not present in the array.
• The best case complexity is O(1) when element is first or last element in the array.

Implementation:

## C++

 `// CPP program to implement front and back``// search``#include``using` `namespace` `std;` `bool` `search(``int` `arr[], ``int` `n, ``int` `x)``{``    ``// Start searching from both ends``    ``int` `front = 0, back = n - 1;` `    ``// Keep searching while two indexes``    ``// do not cross.``    ``while` `(front <= back)``    ``{``        ``if` `(arr[front] == x || arr[back] == x)``          ``return` `true``;``        ``front++;``        ``back--;``    ``}``    ``return` `false``;``}` `int` `main()``{``   ``int` `arr[] = {10, 20, 80, 30, 60, 50,``                     ``110, 100, 130, 170};``   ``int` `x = 130;``   ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``   ``if` `(search(arr, n, x))``      ``cout << ``"Yes"``;``   ``else``      ``cout << ``"No"``;``   ``return` `0;``}`

## Java

 `// Java program to implement front and back``// search``class` `GFG {``    ` `    ``static` `boolean` `search(``int` `arr[], ``int` `n, ``int` `x)``    ``{``        ` `        ``// Start searching from both ends``        ``int` `front = ``0``, back = n - ``1``;``    ` `        ``// Keep searching while two indexes``        ``// do not cross.``        ``while` `(front <= back)``        ``{``            ``if` `(arr[front] == x || arr[back] == x)``                ``return` `true``;``            ``front++;``            ``back--;``        ``}``        ` `        ``return` `false``;``    ``}``    ` `    ``// driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `arr[] = {``10``, ``20``, ``80``, ``30``, ``60``, ``50``,``                        ``110``, ``100``, ``130``, ``170``};``        ``int` `x = ``130``;``        ``int` `n = arr.length;``        ` `        ``if` `(search(arr, n, x))``            ``System.out.print(``"Yes"``);``        ``else``            ``System.out.print(``"No"``);``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to implement``# front and back search` `def` `search(arr, n, x):` `    ``# Start searching from both ends``    ``front ``=` `0``; back ``=` `n ``-` `1``    ` `    ``# Keep searching while two``    ``# indexes do not cross.``    ``while` `(front <``=` `back):``    ` `        ``if` `(arr[front] ``=``=` `x ``or` `arr[back] ``=``=` `x):``            ``return` `True``        ``front ``+``=` `1``        ``back ``-``=` `1``    ` `    ``return` `False` `# Driver code``arr ``=` `[``10``, ``20``, ``80``, ``30``, ``60``,``       ``50``, ``110``, ``100``, ``130``, ``170``]``x ``=` `130``n ``=` `len``(arr)` `if` `(search(arr, n, x)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)``    ` `# This code is contributed by Anant Agarwal.`

## C#

 `// C# program to implement front and back``// search``using` `System;` `public` `class` `GFG {` `    ``static` `bool` `search(``int` `[]arr, ``int` `n, ``int` `x)``    ``{``        ` `        ``// Start searching from both ends``        ``int` `front = 0, back = n - 1;``    ` `        ``// Keep searching while two indexes``        ``// do not cross.``        ``while` `(front <= back)``        ``{``            ``if` `(arr[front] == x || arr[back] == x)``                ``return` `true``;``            ``front++;``            ``back--;``        ``}``        ` `        ``return` `false``;``    ``}` `    ``static` `public` `void` `Main ()``    ``{``        ``int` `[]arr = {10, 20, 80, 30, 60, 50,``                    ``110, 100, 130, 170};``        ``int` `x = 130;``        ``int` `n = arr.Length;``        ` `        ``if` `(search(arr, n, x))``            ``Console.WriteLine(``"Yes"``);``        ``else``            ``Console.WriteLine(``"No"``);``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

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