An in-place algorithm for String Transformation

Given a string, move all even positioned elements to the end of the string. While moving elements, keep the relative order of all even positioned and odd positioned elements the same. For example, if the given string is “a1b2c3d4e5f6g7h8i9j1k2l3m4”, convert it to “abcdefghijklm1234567891234” in-place and in O(n) time complexity.

Below are the steps:
1. Cut out the largest prefix sub-string of the size of the form 3^k + 1. In this step, we find the largest non-negative integer k such that 3^k+1 is smaller than or equal to n (length of the string)
2. Apply cycle leader iteration algorithm ( it has been discussed below ), starting with index 1, 3, 9…… to this sub-string. Cycle leader iteration algorithm moves all the items of this sub-string to their correct positions, i.e. all the alphabets are shifted to the left half of the sub-string and all the digits are shifted to the right half of this sub-string.
3. Process the remaining sub-string recursively using steps#1 and #2.
4. Now, we only need to join the processed sub-strings together. Start from any end ( say from left ), pick two sub-strings, and apply the below steps:
….4.1 Reverse the second half of the first sub-string. 
….4.2 Reverse the first half of the second sub-string. 
….4.3 Reverse the second half of the first sub-string and the first half of second sub-string together.
5. Repeat step#4 until all sub-strings are joined. It is similar to k-way merging where the first sub-string is joined with the second. The resultant is merged with third and so on.

Let us understand it with an example:

Please note that we have used values like 10, 11 12 in the below example. Consider these values as single characters only. These values are used for better readability.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
a 1 b 2 c 3 d 4 e 5 f  6  g  7  h  8  i  9  j  10 k  11 l  12 m  13

After breaking into the size of the form 3^k + 1, two sub-strings are formed of size 10 each. The third sub-string is formed of size 4 and the fourth sub-string is formed of size 2.



0 1 2 3 4 5 6 7 8 9         
a 1 b 2 c 3 d 4 e 5         

10 11 12 13 14 15 16 17 18 19          
f  6  g  7  h  8  i  9  j  10           

20 21 22 23 
k  11 l  12 

24 25
m  13

After applying the cycle leader iteration algorithm to first sub-string:

0 1 2 3 4 5 6 7 8 9          
a b c d e 1 2 3 4 5          

10 11 12 13 14 15 16 17 18 19          
f  6  g  7  h  8  i  9  j  10 

20 21 22 23 
k  11 l  12 

24 25
m  13

After applying cycle leader iteration algorithm to second sub-string:

0 1 2 3 4 5 6 7 8 9          
a b c d e 1 2 3 4 5          

10 11 12 13 14 15 16 17 18 19           
f  g  h  i  j  6  7  8  9  10 

20 21 22 23 
k  11 l  12 

24 25
m 13

After applying cycle leader iteration algorithm to third sub-string:

0 1 2 3 4 5 6 7 8 9          
a b c d e 1 2 3 4 5          

10 11 12 13 14 15 16 17 18 19            
f  g  h  i  j  6  7  8  9  10

20 21 22 23 
k  l  11 12 

24 25
m  13

After applying cycle leader iteration algorithm to fourth sub-string:

0 1 2 3 4 5 6 7 8 9  
a b c d e 1 2 3 4 5  

10 11 12 13 14 15 16 17 18 19             
f  g  h  i  j  6  7  8  9  10   

20 21 22 23 
k  l  11 12 

24 25
m  13

Joining first sub-string and second sub-string: 
1. Second half of first sub-string and the first half of second sub-string reversed.

0 1 2 3 4 5 6 7 8 9          
a b c d e 5 4 3 2 1            <--------- First Sub-string  

10 11 12 13 14 15 16 17 18 19             
j  i  h  g  f  6  7  8  9  10  <--------- Second Sub-string  

20 21 22 23 
k  l  11 12 

24 25
m  13

2. Second half of first sub-string and the first half of second sub-string reversed together( They are merged, i.e. there are only three sub-strings now ).

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
a b c d e f g h i j 1  2  3  4  5  6  7  8  9  10

20 21 22 23 
k  l  11 12 

24 25
m  13

Joining first sub-string and second sub-string: 
1. Second half of first sub-string and the first half of the second sub-string reversed.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
a b c d e f g h i j 10 9  8  7  6  5  4  3  2  1 <--------- First Sub-string  

20 21 22 23 
l  k  11 12                                      <--------- Second Sub-string

24 25
m  13

2. Second half of first sub-string and the first half of second sub-string reversed together( They are merged, i.e. there are only two sub-strings now ).



0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23  
a b c d e f g h i j k  l  1  2  3  4  5  6  7  8  9  10 11 12  

24 25
m  13 

Joining first sub-string and second sub-string: 
1. Second half of first sub-string and the first half of second sub-string reversed.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 
a b c d e f g h i j k  l  12 11 10 9  8  7  6  5  4  3  2  1 <----- First Sub-string

24 25
m  13   <----- Second Sub-string 

2. Second half of first sub-string and the first half of second sub-string reversed together( They are merged, i.e. there is only one sub-string now ).

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
a b c d e f g h i j k  l  m  1  2  3  4  5  6  7  8  9  10 11 12 13

Since all sub-strings have been joined together, we are done.

How does the cycle leader iteration algorithm work?
Let us understand it with an example: 

Input:
0 1 2 3 4 5 6 7 8 9
a 1 b 2 c 3 d 4 e 5

Output:
0 1 2 3 4 5 6 7 8 9 
a b c d e 1 2 3 4 5

Old index    New index
0        0
1        5
2        1
3        6
4        2
5        7
6        3
7        8
8        4
9        9

Let len be the length of the string. If we observe carefully, we find that the new index is given by the below formula: 

if( oldIndex is odd )
    newIndex = len / 2 + oldIndex / 2;
else
        newIndex = oldIndex / 2;

So, the problem reduces to shifting the elements to new indexes based on the above formula.
Cycle leader iteration algorithm will be applied starting from the indices of the form 3^k, starting with k = 0.

Below are the steps:
1. Find new position for item at position i. Before putting this item at new position, keep the back-up of element at new position. Now, put the item at new position.
2. Repeat step#1 for new position until a cycle is completed, i.e. until the procedure comes back to the starting position.
3. Apply cycle leader iteration algorithm to the next index of the form 3^k. Repeat this step until 3^k < len. 
Consider input array of size 28:
The first cycle leader iteration, starting with index 1:
1->14->7->17->22->11->19->23->25->26->13->20->10->5->16->8->4->2->1
The second cycle leader iteration, starting with index 3:
3->15->21->24->12->6->3
The third cycle leader iteration, starting with index 9:
9->18->9

Based on the above algorithm, below is the code:  

C++

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// C++ implimentation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// A utility function to swap characters
void swap ( char* a, char* b )
{
    char t = *a;
    *a = *b;
    *b = t;
}
 
// A utility function to reverse string str[low..high]
void reverse ( char* str, int low, int high )
{
    while ( low < high )
    {
        swap( &str[low], &str[high] );
        ++low;
        --high;
    }
}
 
// Cycle leader algorithm to move all even
//  positioned elements at the end.
void cycleLeader ( char* str, int shift, int len )
{
    int j;
    char item;
 
    for (int i = 1; i < len; i *= 3 )
    {
        j = i;
 
        item = str[j + shift];
        do
        {
            // odd index
            if ( j & 1 )
                j = len / 2 + j / 2;
            // even index
            else
                j /= 2;
 
            // keep the back-up of element at new position
            swap (&str[j + shift], &item);
        }
        while ( j != i );
    }
}
 
// The main function to transform a string. This function 
// mainly uses cycleLeader() to transform
void moveNumberToSecondHalf( char* str )
{
    int k, lenFirst;
 
    int lenRemaining = strlen( str );
    int shift = 0;
 
    while ( lenRemaining )
    {
        k = 0;
 
        // Step 1: Find the largest prefix
        // subarray of the form 3^k + 1
        while ( pow( 3, k ) + 1 <= lenRemaining )
            k++;
        lenFirst = pow( 3, k - 1 ) + 1;
        lenRemaining -= lenFirst;
 
        // Step 2: Apply cycle leader algorithm
        // for the largest subarrau
        cycleLeader ( str, shift, lenFirst );
 
        // Step 4.1: Reverse the second half of first subarray
        reverse ( str, shift / 2, shift - 1 );
 
        // Step 4.2: Reverse the first half of second sub-string.
        reverse ( str, shift, shift + lenFirst / 2 - 1 );
 
        // Step 4.3 Reverse the second half of first sub-string
        // and first half of second sub-string together
        reverse ( str, shift / 2, shift + lenFirst / 2 - 1 );
 
        // Increase the length of first subarray
        shift += lenFirst;
    }
}
 
// Driver program to test above function
int main()
{
    char str[] = "a1b2c3d4e5f6g7";
    moveNumberToSecondHalf( str );
    cout<<str;
    return 0;
}
 
// This is code is contributed by rathbhupendra

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Java

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// Java implementation of above approach
import java.util.*;
 
class GFG{
     
static char []str;
 
// A utility function to reverse
// String str[low..high]
static void reverse(int low, int high)
{
    while (low < high)
    {
        char t = str[low];
        str[low] = str[high];
        str[high] = t;
        ++low;
        --high;
    }
}
 
// Cycle leader algorithm to move all even
// positioned elements at the end.
static void cycleLeader(int shift, int len)
{
    int j;
    char item;
 
    for(int i = 1; i < len; i *= 3)
    {
        j = i;
        item = str[j + shift];
         
        do
        {
             
            // odd index
            if (j % 2 == 1)
                j = len / 2 + j / 2;
                 
            // even index
            else
                j /= 2;
 
            // Keep the back-up of element at
            // new position
            char t = str[j + shift];
            str[j + shift] = item;
            item = t;
        }
        while (j != i);
    }
}
 
// The main function to transform a String.
// This function mainly uses cycleLeader()
// to transform
static void moveNumberToSecondHalf()
{
    int k, lenFirst;
    int lenRemaining = str.length;
    int shift = 0;
 
    while (lenRemaining > 0)
    {
        k = 0;
 
        // Step 1: Find the largest prefix
        // subarray of the form 3^k + 1
        while (Math.pow(3, k) + 1 <= lenRemaining)
            k++;
             
        lenFirst = (int)Math.pow(3, k - 1) + 1;
        lenRemaining -= lenFirst;
 
        // Step 2: Apply cycle leader algorithm
        // for the largest subarrau
        cycleLeader(shift, lenFirst);
 
        // Step 4.1: Reverse the second half
        // of first subarray
        reverse(shift / 2, shift - 1);
 
        // Step 4.2: Reverse the first half
        // of second sub-String.
        reverse(shift, shift + lenFirst / 2 - 1);
 
        // Step 4.3 Reverse the second half
        // of first sub-String and first half
        // of second sub-String together
        reverse(shift / 2, shift + lenFirst / 2 - 1);
 
        // Increase the length of first subarray
        shift += lenFirst;
    }
}
 
// Driver code
public static void main(String[] args)
{
    String st = "a1b2c3d4e5f6g7";
    str = st.toCharArray();
     
    moveNumberToSecondHalf();
     
    System.out.print(str);
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python implimentation of above approach
 
# A utility function to reverse string str[low..high]
def Reverse(string: list, low: int, high: int):
    while low < high:
        string[low], string[high] = string[high], string[low]
        low += 1
        high -= 1
 
# Cycle leader algorithm to move all even
# positioned elements at the end.
def cycleLeader(string: list, shift: int, len: int):
    i = 1
    while i < len:
        j = i
        item = string[j + shift]
 
        while True:
 
            # odd index
            if j & 1:
                j = len // 2 + j // 2
 
            # even index
            else:
                j //= 2
 
            # keep the back-up of element at new position
            string[j + shift], item = item, string[j + shift]
 
            if j == i:
                break
        i *= 3
 
# The main function to transform a string. This function
# mainly uses cycleLeader() to transform
def moveNumberToSecondHalf(string: list):
    k, lenFirst = 0, 0
    lenRemaining = len(string)
    shift = 0
 
    while lenRemaining:
        k = 0
 
        # Step 1: Find the largest prefix
        # subarray of the form 3^k + 1
        while pow(3, k) + 1 <= lenRemaining:
            k += 1
        lenFirst = pow(3, k - 1) + 1
        lenRemaining -= lenFirst
 
        # Step 2: Apply cycle leader algorithm
        # for the largest subarrau
        cycleLeader(string, shift, lenFirst)
 
        # Step 4.1: Reverse the second half of first subarray
        Reverse(string, shift // 2, shift - 1)
 
        # Step 4.2: Reverse the first half of second sub-string
        Reverse(string, shift, shift + lenFirst // 2 - 1)
 
        # Step 4.3 Reverse the second half of first sub-string
        # and first half of second sub-string together
        Reverse(string, shift // 2, shift + lenFirst // 2 - 1)
 
        # Increase the length of first subarray
        shift += lenFirst
 
# Driver Code
if __name__ == "__main__":
 
    string = "a1b2c3d4e5f6g7"
    string = list(string)
    moveNumberToSecondHalf(string)
    print(''.join(string))
 
# This code is contributed by
# sanjeev2552

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C#

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// C# implementation of
// the above approach
using System;
class GFG{
     
static char []str;
 
// A utility function to
// reverse String str[low
// ..high]
static void reverse(int low,
                    int high)
{
  while (low < high)
  {
    char t = str[low];
    str[low] = str[high];
    str[high] = t;
    ++low;
    --high;
  }
}
 
// Cycle leader algorithm to
// move all even positioned
// elements at the end.
static void cycleLeader(int shift,
                        int len)
{
  int j;
  char item;
 
  for(int i = 1;
          i < len; i *= 3)
  {
    j = i;
    item = str[j + shift];
 
    do
    {
      // odd index
      if (j % 2 == 1)
        j = len / 2 + j / 2;
 
      // even index
      else
        j /= 2;
 
      // Keep the back-up of
      // element at new position
      char t = str[j + shift];
      str[j + shift] = item;
      item = t;
    }
    while (j != i);
  }
}
 
// The main function to transform
// a String. This function mainly
// uses cycleLeader() to transform
static void moveNumberToSecondHalf()
{
  int k, lenFirst;
  int lenRemaining = str.Length;
  int shift = 0;
 
  while (lenRemaining > 0)
  {
    k = 0;
 
    // Step 1: Find the largest prefix
    // subarray of the form 3^k + 1
    while (Math.Pow(3, k) +
           1 <= lenRemaining)
      k++;
 
    lenFirst = (int)Math.Pow(3,
                             k - 1) + 1;
    lenRemaining -= lenFirst;
 
    // Step 2: Apply cycle leader
    // algorithm for the largest
    // subarrau
    cycleLeader(shift, lenFirst);
 
    // Step 4.1: Reverse the second
    // half of first subarray
    reverse(shift / 2,
            shift - 1);
 
    // Step 4.2: Reverse the
    // first half of second
    // sub-String.
    reverse(shift, shift +
            lenFirst / 2 - 1);
 
    // Step 4.3 Reverse the second
    // half of first sub-String and
    // first half of second sub-String
    // together
    reverse(shift / 2, shift +
            lenFirst / 2 - 1);
 
    // Increase the length of
    // first subarray
    shift += lenFirst;
  }
}
 
// Driver code
public static void Main(String[] args)
{
  String st = "a1b2c3d4e5f6g7";
  str = st.ToCharArray();
  moveNumberToSecondHalf();
  Console.Write(str);
}
}
 
// This code is contributed by 29AjayKumar

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Output:

abcdefg1234567




Click here to see various test cases.
Notes: 
1. If the array size is already in the form 3^k + 1, We can directly apply cycle leader iteration algorithm. There is no need of joining.
2. Cycle leader iteration algorithm is only applicable to arrays of size of the form 3^k + 1.

How is the time complexity O(n) ? 
Each item in a cycle is shifted at most once. Thus time complexity of the cycle leader algorithm is O(n). The time complexity of the reverse operation is O(n). We will soon update the mathematical proof of the time complexity of the algorithm.

Exercise: 
Given string in the form “abcdefg1234567”, convert it to “a1b2c3d4e5f6g7” in-place and in O(n) time complexity.

References: 
A Simple In-Place Algorithm for In-Shu?e.
__Aashish Barnwal.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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