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Reverse an array upto a given position
• Difficulty Level : Easy
• Last Updated : 20 Apr, 2021

Given an array arr[] and a position in array, k. Write a function name reverse (a[], k) such that it reverses subarray arr[0..k-1]. Extra space used should be O(1) and time complexity should be O(k).
Example:

```Input:
arr[] = {1, 2, 3, 4, 5, 6}
k = 4

Output:
arr[] = {4, 3, 2, 1, 5, 6} ```

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Below is the implementation for the same.

## C++

 `// C++ program to reverse a subarray arr[0..k-1]``#include ``using` `namespace` `std;` `// Reverse subarray a[0..k-1]``void` `reverse(``int` `a[], ``int` `n, ``int` `k)``{``    ``if` `(k > n)``    ``{``        ``cout << ``"Invalid k"``;``        ``return``;``    ``}` `    ``// One by one reverse first and last elements of a[0..k-1]``    ``for` `(``int` `i = 0; i < k/2; i++)``        ``swap(a[i], a[k-i-1]);``}` `// Driver program``int` `main()``{``    ``int` `a[] = {1, 2, 3, 4, 5, 6};``    ``int` `n = ``sizeof``(a) / ``sizeof``(``int``), k = 4;` `    ``reverse(a, n, k);` `    ``for` `(``int` `i = 0; i < n; ++i)``        ``printf``(``"%d "``, a[i]);` `    ``return` `0;``}`

## Java

 `// java program to reverse a``// subarray arr[0..k-1]` `public` `class` `GFG {` `    ``// Reverse subarray a[0..k-1]``    ``static` `void` `reverse(``int` `[]a, ``int` `n, ``int` `k)``    ``{``        ``if` `(k > n)``        ``{``            ``System.out.println( ``"Invalid k"``);``            ``return``;``        ``}``    ` `        ``// One by one reverse first``        ``// and last elements of a[0..k-1]``        ``for` `(``int` `i = ``0``; i < k / ``2``; i++)``        ``{``            ``int` `tempswap = a[i];``                ``a[i] = a[k - i - ``1``];``                ``a[k - i - ``1``] = tempswap;            ``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `[]a = {``1``, ``2``, ``3``, ``4``, ``5``, ``6``};``        ``int` `n = a.length, k = ``4``;``        ``reverse(a, n, k);``        ``for` `(``int` `i = ``0``; i < n; ++i)``            ``System.out.print(a[i] + ``" "``);``    ``}``}` `// This code is contributed by Sam007.`

## Python3

 `# python program to reverse a subarray``# arr[0..k-1]``from` `__future__ ``import` `print_function` `# Reverse subarray a[0..k-1]``def` `reverse(a, n, k):``    ` `    ``if` `(k > n):``        ``print``( ``"Invalid k"``)``        ``return``    ` `    ``# One by one reverse first and``    ``# last elements of a[0..k-1]``    ``for` `i ``in` `range``(``0``, (``int``)(k``/``2``)):``        ``temp ``=` `a[i]``        ``a[i] ``=` `a[k``-``i``-``1``]``        ``a[k``-``i``-``1``] ``=` `temp``        ` `# Driver program``a ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``]``n ``=` `len``(a)``k ``=` `4` `reverse(a, n, k);` `for` `i ``in` `range``(``0``, n):``    ``print``(a[i], end``=``" "``)``    ` `# This code is contributed by Sam007.`

## C#

 `// C# program to reverse a``// subarray arr[0..k-1]``using` `System;` `class` `GFG {``    ` `static` `void` `SwapNum(``ref` `int` `x, ``ref` `int` `y)``{``    ``int` `tempswap = x;``    ``x = y;``    ``y = tempswap;            ``}``    ` `// Reverse subarray a[0..k-1]``static` `void` `reverse(``int` `[]a, ``int` `n,``                             ``int` `k)``{``    ``if` `(k > n)``    ``{``        ``Console.Write( ``"Invalid k"``);``        ``return``;``    ``}` `    ``// One by one reverse first``    ``// and last elements of a[0..k-1]``    ``for` `(``int` `i = 0; i < k / 2; i++)``        ``SwapNum(``ref` `a[i], ``ref` `a[k - i - 1]);``        ` `}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]a = {1, 2, 3, 4, 5, 6};``    ``int` `n = a.Length, k = 4;` `    ``reverse(a, n, k);` `    ``for` `(``int` `i = 0; i < n; ++i)``        ``Console.Write(a[i] + ``" "``);``}``}` `// This code is contributed by Sam007`

## Javascript

 ``

Output:

`4 3 2 1 5 6`

Time complexity: O(k)
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