Given an array arr[] consisting of 2* N elements in the form of { a1, a2, …, aN, b1, b2, …, bN }, the task is to shuffle the array to {a1, b1, a2, b2, …, an, b1} without using extra space.
Examples :
Input: arr[] = { 1, 3, 5, 2, 4, 6 }
Output: 1 2 3 4 5 6
Explanation:
The output contains the elements in the form of { a1, b1, a2, b2, a3, b3 }.Input: arr[] = {1, 2, 3, -1, -2, -3, }
Output: 1 -1 2 -2 3 -3
Divide and Conquer-based Approach: If the size of an array is a power of 2, then follow the article Shuffle 2n integers in format {a1, b1, a2, b2, a3, b3, ……, an, bn} using divide and conquer technique.
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
Alternate Approach: The above approach will work for all possible values of N by recursively dividing the array such that the length of both halves is even. Follow the steps below to solve the problem:
- Define a recursive function, say shuffle(start, end).
- If array length is divisible by 4, then calculate mid-point of the array, say mid = start + (end – start + 1) / 2.
- Otherwise, mid = start + (end – start + 1) / 2 – 1.
- Calculate mid-points of both subarrays, say mid1 = start + (mid – start)/2, and mid2 = mid + (end – mid + 1) / 2.
- Reverse the subarrays in the ranges [mid1, mid2], [mid1, mid-1], and [mid, mid2 – 1].
- Make recursive calls for subarrays [start, mid – 1] and [mid, end], i.e. shuffle(start, mid – 1) and shuffle(mid, end) respectively.
- Finally, print the array.
Illustration:
Consider an array arr[] = {a1, a2, a3, b1, b2, b3}:
- Split the array into two halves, both of even length, i.e. a1, a2 : a3, b1, b2, b3.
- Reverse the mid of first half to mid of 2nd half, i.e. a1, b1 : a3, a2, b2, b3.
- Now, reverse the mid of first half to mid of subarray [0, 5], a1, b1 : a3, a2, b2, b3.
- Now reverse mid of subarray [0, 5] to mid of 2nd half, a1, b1 : a2, a3, b2, b3.
- Recursively call for arrays {a1, b1}, and {a2, a3, b2, b3}.
- Now the array {a2, a3, b2, b3} modifies to {a2, b2, a3, b3} after applying the above operations.
- Now the arr[] modifies to {a1, b1, a2, b2, a3, b3}.
Below is the implementation of the above approach:
// C program for the above approach #include <stdio.h> // Function to reverse the array from the // position 'start' to position 'end' void reverse( int arr[], int start, int end)
{ // Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for ( int i = 0; i < mid; i++) {
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return ;
} // Utility function to shuffle the given array // in the of form {a1, b1, a2, b2, ....an, bn} void shuffleArrayUtil( int arr[], int start, int end)
{ int i;
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return ;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4) {
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
} // Function to shuffle the given array in // the form of {a1, b1, a2, b2, ....an, bn} void shuffleArray( int arr[], int N,
int start, int end)
{ // Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for ( int i = 0; i < N; i++)
printf ( "%d " , arr[i]);
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
return 0;
} |
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to reverse the array from the // position 'start' to position 'end' void reverse( int arr[], int start, int end)
{ // Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for ( int i = 0; i < mid; i++) {
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return ;
} // Utility function to shuffle the given array // in the of form {a1, b1, a2, b2, ....an, bn} void shuffleArrayUtil( int arr[], int start, int end)
{ int i;
// Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return ;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4) {
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
} // Function to shuffle the given array in // the form of {a1, b1, a2, b2, ....an, bn} void shuffleArray( int arr[], int N, int start, int end)
{ // Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for ( int i = 0; i < N; i++)
cout << arr[i] << " " ;
} // Driver Code int main()
{ // Given array
int arr[] = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = sizeof (arr) / sizeof (arr[0]);
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
return 0;
} // This code is contributed by jainlovely450. |
// Java program for the above approach class GFG{
// Function to reverse the array from the // position 'start' to position 'end' static void reverse( int arr[], int start, int end)
{ // Stores mid of start and end
int mid = (end - start + 1 ) / 2 ;
// Traverse the array in
// the range [start, end]
for ( int i = 0 ; i < mid; i++)
{
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return ;
} // Utility function to shuffle the given array // in the of form {a1, b1, a2, b2, ....an, bn} static void shuffleArrayUtil( int arr[], int start, int end)
{ int i;
// Stores the length of the array
int l = end - start + 1 ;
// If length of the array is 2
if (l == 2 )
return ;
// Stores mid of the { start, end }
int mid = start + l / 2 ;
// Divide array into two
// halves of even length
if (l % 4 > 0 )
{
// Update mid
mid -= 1 ;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2 ;
int mid2 = mid + (end + 1 - mid) / 2 ;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1 );
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1 );
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1 );
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1 );
shuffleArrayUtil(arr, mid, end);
} // Function to shuffle the given array in // the form of {a1, b1, a2, b2, ....an, bn} static void shuffleArray( int arr[], int N,
int start, int end)
{ // Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for ( int i = 0 ; i < N; i++)
System.out.printf( "%d " , arr[i]);
} // Driver Code public static void main(String[] args)
{ // Given array
int arr[] = { 1 , 3 , 5 , 2 , 4 , 6 };
// Size of the array
int N = arr.length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0 , N - 1 );
} } // This code is contributed by 29AjayKumar |
# Python3 program for the above approach # Function to reverse the array from the # position 'start' to position 'end' def reverse(arr, start, end):
# Stores mid of start and end
mid = (end - start + 1 ) / / 2
# Traverse the array in
# the range [start, end]
for i in range (mid):
# Stores arr[start + i]
temp = arr[start + i]
# Update arr[start + i]
arr[start + i] = arr[end - i]
# Update arr[end - i]
arr[end - i] = temp
return arr
# Utility function to shuffle the given array # in the of form {a1, b1, a2, b2, ....an, bn} def shuffleArrayUtil(arr, start, end):
i = 0
# Stores the length of the array
l = end - start + 1
# If length of the array is 2
if (l = = 2 ):
return
# Stores mid of the { start, end }
mid = start + l / / 2
# Divide array into two
# halves of even length
if (l % 4 ):
# Update mid
mid - = 1
# Calculate the mid-points of
# both halves of the array
mid1 = start + (mid - start) / / 2
mid2 = mid + (end + 1 - mid) / / 2
# Reverse the subarray made
# from mid1 to mid2
arr = reverse(arr, mid1, mid2 - 1 )
# Reverse the subarray made
# from mid1 to mid
arr = reverse(arr, mid1, mid - 1 )
# Reverse the subarray made
# from mid to mid2
arr = reverse(arr, mid, mid2 - 1 )
# Recursively calls for both
# the halves of the array
shuffleArrayUtil(arr, start, mid - 1 )
shuffleArrayUtil(arr, mid, end)
# Function to shuffle the given array in # the form of {a1, b1, a2, b2, ....an, bn} def shuffleArray(arr, N, start, end):
# Function Call
shuffleArrayUtil(arr, start, end)
# Print the modified array
for i in arr:
print (i, end = " " )
# Driver Code if __name__ = = '__main__' :
# Given array
arr = [ 1 , 3 , 5 , 2 , 4 , 6 ]
# Size of the array
N = len (arr)
# Shuffles the given array to the
# required permutation
shuffleArray(arr, N, 0 , N - 1 )
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
public class GFG{
// Function to reverse the array from the // position 'start' to position 'end' static void reverse( int [] arr, int start, int end)
{ // Stores mid of start and end
int mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for ( int i = 0; i < mid; i++)
{
// Stores arr[start + i]
int temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return ;
} // Utility function to shuffle the given array // in the of form {a1, b1, a2, b2, ....an, bn} static void shuffleArrayUtil( int [] arr, int start, int end)
{ // Stores the length of the array
int l = end - start + 1;
// If length of the array is 2
if (l == 2)
return ;
// Stores mid of the { start, end }
int mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4 > 0)
{
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
int mid1 = start + (mid - start) / 2;
int mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
} // Function to shuffle the given array in // the form of {a1, b1, a2, b2, ....an, bn} static void shuffleArray( int [] arr, int N,
int start, int end)
{ // Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for ( int i = 0; i < N; i++)
Console.Write(arr[i] + " " );
} // Driver Code static public void Main ()
{ // Given array
int [] arr = { 1, 3, 5, 2, 4, 6 };
// Size of the array
int N = arr.Length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
} } // This code is contributed by Dharanendra L V |
<script> // Javascript program of the above approach // Function to reverse the array from the // position 'start' to position 'end' function reverse(arr, start, end)
{ // Stores mid of start and end
let mid = (end - start + 1) / 2;
// Traverse the array in
// the range [start, end]
for (let i = 0; i < mid; i++)
{
// Stores arr[start + i]
let temp = arr[start + i];
// Update arr[start + i]
arr[start + i] = arr[end - i];
// Update arr[end - i]
arr[end - i] = temp;
}
return ;
} // Utility function to shuffle the given array // in the of form {a1, b1, a2, b2, ....an, bn} function shuffleArrayUtil(arr, start, end)
{ let i;
// Stores the length of the array
let l = end - start + 1;
// If length of the array is 2
if (l == 2)
return ;
// Stores mid of the { start, end }
let mid = start + l / 2;
// Divide array into two
// halves of even length
if (l % 4 > 0)
{
// Update mid
mid -= 1;
}
// Calculate the mid-points of
// both halves of the array
let mid1 = start + (mid - start) / 2;
let mid2 = mid + (end + 1 - mid) / 2;
// Reverse the subarray made
// from mid1 to mid2
reverse(arr, mid1, mid2 - 1);
// Reverse the subarray made
// from mid1 to mid
reverse(arr, mid1, mid - 1);
// Reverse the subarray made
// from mid to mid2
reverse(arr, mid, mid2 - 1);
// Recursively calls for both
// the halves of the array
shuffleArrayUtil(arr, start, mid - 1);
shuffleArrayUtil(arr, mid, end);
} // Function to shuffle the given array in // the form of {a1, b1, a2, b2, ....an, bn} function shuffleArray(arr, N,
start, end)
{ // Function Call
shuffleArrayUtil(arr, start, end);
// Print the modified array
for (let i = 0; i < N; i++)
document.write(arr[i] + " " );
} // Driver Code
// Given array
let arr = [ 1, 3, 5, 2, 4, 6 ];
// Size of the array
let N = arr.length;
// Shuffles the given array to the
// required permutation
shuffleArray(arr, N, 0, N - 1);
</script> |
1 2 3 4 5 6
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)