Reverse an array in groups of given size - GeeksforGeeks

Reverse an array in groups of given size - GeeksforGeeks
Given an array, reverse every sub-array formed by consecutive k elements.

The idea is very simple. We consider every sub-array of size k starting from beginning of the array and reverse it. We need to handle some special cases. If k is not multiple of n where n is size of the array, for last group we will have less than k elements left, we need to reverse all remaining elements. If k = 1, array should remain unchanged. If k >= n, we reverse all elements present in the array.

void reverse(int arr[], int n, int k)

{

    for (int i = 0; i < n; i += k)

    {

        int left = i;

        // to handle case when k is not multiple of n

        int right = min(i + k - 1, n - 1);

        // reverse the sub-array [left, right]

        while (left < right)

            swap(arr[left++], arr[right--]);

    }

}

http://www.geeksforgeeks.org/reverse-an-array-in-groups-of-given-size-2/

Variation 1: Reverse every alternate sub-array formed by consecutive k elements.

void reverse(int arr[], int n, int k)

{

    // increment i in multiples of 2*k

    for (int i = 0; i < n; i += 2*k)

    {

        int left = i;

 

        // to handle case when 2*k is not multiple of n

        int right = min(i + k - 1, n - 1);

 

        // reverse the sub-array [left, right]

        while (left < right)

            swap(arr[left++], arr[right--]);

    }   

}

Variation 2: Reverse every sub-array formed by consecutive k elements at given distance apart.
Examples:

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

void reverse(int arr[], int n, int k, int m)

{

    // increment i in multiples of k + m

    for (int i = 0; i < n; i += k + m)

    {

        int left = i;

 

        // to handle case when k + m is not multiple of n

        int right = min(i + k - 1, n - 1);

 

        // reverse the sub-array [left, right]

        while (left < right)

            swap(arr[left++], arr[right--]);

    }

}

Variation 3: Reverse every sub-array formed by consecutive k elements where k doubles itself with every sub-array.
Examples:

Input: 
arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
k = 1
Output:  
[1], [3, 2], [7, 6, 5, 4], [15, 14, 13, 12, 11, 10, 9, 8]

void reverse(int arr[], int n, int k)

{

    // increment i in multiples of k where value

    // of k is doubled with each iteration

    for (int i = 0; i < n; i += k/2)

    {

        int left = i;

 

        // to handle case when number of elements in

        // last group is less than k

        int right = min(i + k - 1, n - 1);

 

        // reverse the sub-array [left, right]

        while (left < right)

            swap(arr[left++], arr[right--]);

 

        // double value of k with each iteration

        k = k*2;

    }

}

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