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# Reverse an array in groups of given size | Set 3 (Single traversal)

• Difficulty Level : Basic
• Last Updated : 13 Aug, 2021

Given an array, reverse every sub-array formed by consecutive k elements.

Examples:

Input: arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3.
Output: [3, 2, 1, 6, 5, 4, 9, 8, 7, 10]
Input: arr = [1, 2, 3, 4, 5, 6, 7], k = 5.
Output: [5, 4, 3, 2, 1, 7, 6]

Approach:

1. We will use two pointers technique to solve this problem.
2. First, we will initialize our 1st pointer d with value k-1 (d = k-1) and a variable m with value 2 (m = 2).
3. Now, we will iterate the array with our 2nd pointer i and check
• If i < d, Swap (arr[i], arr[d]) and decrement d by 1. Otherwise,
• Make d = k * m – 1, i = k * (m – 1) – 1 and m = m + 1.

Below is the implementation of the above approach.

## C++

 `// C++ program to reverse every sub-array``// formed by consecutive k elements``#include``using` `namespace` `std;` `// Function to reverse every sub-array``// formed by consecutive k elements``void` `ReverseInGroup(``int` `arr[], ``int` `n, ``int` `k)``{``    ``if``(n < k)``    ``{``        ``k = n;``    ``}` `    ``// Initialize variables``    ``int` `d = k - 1, m = 2;``    ``int` `i = 0;``        ` `    ``for``(i = 0; i < n; i++)``    ``{``       ` `       ``if` `(i >= d)``       ``{``           ``// Update the variables``           ``d = k * (m);``           ``if``(d >= n)``           ``{``               ``d = n;        ``           ``}    ``           ``i = k * (m - 1) - 1;``           ``m++;``       ``}``       ``else``       ``{``           ``int` `t = arr[i];``           ``arr[i] = arr[d];``           ``arr[d] = t;``       ``}``       ``d = d - 1;``    ``}``    ``return``;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5, 6, 7 };``    ``int` `k = 3;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``ReverseInGroup(arr, n, k);` `    ``for``(``int` `i = 0; i < n; i++)``       ``cout << arr[i] << ``" "``;` `    ``return` `0;``}` `// This code is contributed by Code_Mech`

## C

 `// C program to reverse every sub-array``// formed by consecutive k elements``#include` `// Function to reverse every sub-array``// formed by consecutive k elements``void` `ReverseInGroup(``int` `arr[], ``int` `n, ``int` `k)``{``    ``if``(n= d)``        ``{  ``            ``// Update the variables``            ``d = k * (m);``            ``if``(d>=n)``            ``{``                ``d = n;            ``            ``}          ``            ``i = k * (m - 1)-1;``            ``m++;``            ` `        ``}``        ``else``        ``{``            ``int` `t = arr[i];``            ``arr[i] = arr[d];``            ``arr[d] = t;``        ``}` `     ``d = d - 1;     ``       ` `    ``}``    ``return``;``        ` `}` `// Driver code``int` `main()``{``    ``int` `arr[] = {1, 2, 3, 4, 5, 6, 7};``    ``int` `k = 3;``  ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``  ` `    ``ReverseInGroup(arr, n, k);``  ` `    ``for` `(``int` `i = 0; i < n; i++)``        ``printf``(``"%d "``, arr[i]);``  ` `    ``return` `0;``}`

## Java

 `// Java program to reverse every sub-array``// formed by consecutive k elements``class` `GFG{``    ` `// Function to reverse every sub-array``// formed by consecutive k elements``static` `void` `ReverseInGroup(``int` `arr[],``                           ``int` `n, ``int` `k)``{``    ``if``(n < k)``    ``{``        ``k = n;``    ``}` `    ``// Initialize variables``    ``int` `d = k - ``1``, m = ``2``;``    ``int` `i = ``0``;``        ` `    ``for``(i = ``0``; i < n; i++)``    ``{``       ``if` `(i >= d)``       ``{``           ` `           ``// Update the variables``           ``d = k * (m);``           ``if``(d >= n)``           ``{``               ``d = n;            ``           ``}        ``           ``i = k * (m - ``1``) - ``1``;``           ``m++;``       ``}``       ``else``       ``{``           ``int` `t = arr[i];``           ``arr[i] = arr[d];``           ``arr[d] = t;``       ``}``       ``d = d - ``1``;    ``    ``}``    ``return``;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7` `};``    ``int` `k = ``3``;``    ``int` `n = arr.length;``    ` `    ``ReverseInGroup(arr, n, k);``    ` `    ``for``(``int` `i = ``0``; i < n; i++)``       ``System.out.printf(``"%d "``, arr[i]);``}``}` `// This code is contributed by sapnasingh4991`

## Python3

 `# Python3 program to reverse``# every sub-array formed by``# consecutive k elements` `# Function to reverse every``# sub-array formed by consecutive``# k elements``def` `ReverseInGroup(arr, n, k):` `    ``if``(n < k):``        ``k ``=` `n` `    ``# Initialize variables``    ``d ``=` `k ``-` `1``    ``m ``=` `2``    ``i ``=` `0` `    ``while` `i < n:``        ``if` `(i >``=` `d):` `            ``# Update the``            ``# variables``            ``d ``=` `k ``*` `(m)``            ` `            ``if``(d >``=` `n):``                ``d ``=` `n` `            ``i ``=` `k ``*` `(m ``-` `1``) ``-` `1``            ``m ``+``=` `1` `        ``else``:``            ``arr[i], arr[d] ``=` `(arr[d],``                              ``arr[i])``        ``d ``=` `d ``-` `1``        ``i ``+``=` `1` `    ``return` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``2``, ``3``,``           ``4``, ``5``, ``6``, ``7``]``    ``k ``=` `3``    ``n ``=` `len``(arr)``    ``ReverseInGroup(arr, n, k)` `    ``for` `i ``in` `range``(n):``        ``print``(arr[i],``              ``end ``=` `" "``)` `# This code is contributed by Chitranayal`

## C#

 `// C# program to reverse every sub-array``// formed by consecutive k elements``using` `System;``class` `GFG{``    ` `// Function to reverse every sub-array``// formed by consecutive k elements``static` `void` `ReverseInGroup(``int` `[]arr,``                           ``int` `n, ``int` `k)``{``    ``if``(n < k)``    ``{``        ``k = n;``    ``}` `    ``// Initialize variables``    ``int` `d = k - 1, m = 2;``    ``int` `i = 0;``        ` `    ``for``(i = 0; i < n; i++)``    ``{``        ``if` `(i >= d)``        ``{``                ` `            ``// Update the variables``            ``d = k * (m);``            ``if``(d >= n)``            ``{``                ``d = n;            ``            ``}        ``            ``i = k * (m - 1) - 1;``            ``m++;``        ``}``        ``else``        ``{``            ``int` `t = arr[i];``            ``arr[i] = arr[d];``            ``arr[d] = t;``        ``}``        ``d = d - 1;    ``    ``}``    ``return``;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 1, 2, 3, 4, 5, 6, 7 };``    ``int` `k = 3;``    ``int` `n = arr.Length;``    ` `    ``ReverseInGroup(arr, n, k);``    ` `    ``for``(``int` `i = 0; i < n; i++)``        ``Console.Write(arr[i] + ``" "``);``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``
Output:
`3 2 1 6 5 4 7`

Time Complexity: O(N)
Auxiliary Space: O(1)

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