An in-place algorithm for String Transformation
Last Updated :
20 Jun, 2022
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:
- 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)
- 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.
- Process the remaining sub-string recursively using steps#1 and #2.
- 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.
- 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:
- 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.
- Repeat step#1 for new position until a cycle is completed, i.e. until the procedure comes back to the starting position.
- 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++
#include <bits/stdc++.h>
using namespace std;
void swap ( char* a, char* b )
{
char t = *a;
*a = *b;
*b = t;
}
void reverse ( char* str, int low, int high )
{
while ( low < high )
{
swap( &str[low], &str[high] );
++low;
--high;
}
}
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
{
if ( j & 1 )
j = len / 2 + j / 2;
else
j /= 2;
swap (&str[j + shift], &item);
}
while ( j != i );
}
}
void moveNumberToSecondHalf( char* str )
{
int k, lenFirst;
int lenRemaining = strlen( str );
int shift = 0;
while ( lenRemaining )
{
k = 0;
while ( pow( 3, k ) + 1 <= lenRemaining )
k++;
lenFirst = pow( 3, k - 1 ) + 1;
lenRemaining -= lenFirst;
cycleLeader ( str, shift, lenFirst );
reverse ( str, shift / 2, shift - 1 );
reverse ( str, shift, shift + lenFirst / 2 - 1 );
reverse ( str, shift / 2, shift + lenFirst / 2 - 1 );
shift += lenFirst;
}
}
int main()
{
char str[] = "a1b2c3d4e5f6g7";
moveNumberToSecondHalf( str );
cout<<str;
return 0;
}
|
Java
import java.util.*;
class GFG{
static char []str;
static void reverse(int low, int high)
{
while (low < high)
{
char t = str[low];
str[low] = str[high];
str[high] = t;
++low;
--high;
}
}
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
{
if (j % 2 == 1)
j = len / 2 + j / 2;
else
j /= 2;
char t = str[j + shift];
str[j + shift] = item;
item = t;
}
while (j != i);
}
}
static void moveNumberToSecondHalf()
{
int k, lenFirst;
int lenRemaining = str.length;
int shift = 0;
while (lenRemaining > 0)
{
k = 0;
while (Math.pow(3, k) + 1 <= lenRemaining)
k++;
lenFirst = (int)Math.pow(3, k - 1) + 1;
lenRemaining -= lenFirst;
cycleLeader(shift, lenFirst);
reverse(shift / 2, shift - 1);
reverse(shift, shift + lenFirst / 2 - 1);
reverse(shift / 2, shift + lenFirst / 2 - 1);
shift += lenFirst;
}
}
public static void main(String[] args)
{
String st = "a1b2c3d4e5f6g7";
str = st.toCharArray();
moveNumberToSecondHalf();
System.out.print(str);
}
}
|
Python3
def Reverse(string: list, low: int, high: int):
while low < high:
string[low], string[high] = string[high], string[low]
low += 1
high -= 1
def cycleLeader(string: list, shift: int, len: int):
i = 1
while i < len:
j = i
item = string[j + shift]
while True:
if j & 1:
j = len // 2 + j // 2
else:
j //= 2
string[j + shift], item = item, string[j + shift]
if j == i:
break
i *= 3
def moveNumberToSecondHalf(string: list):
k, lenFirst = 0, 0
lenRemaining = len(string)
shift = 0
while lenRemaining:
k = 0
while pow(3, k) + 1 <= lenRemaining:
k += 1
lenFirst = pow(3, k - 1) + 1
lenRemaining -= lenFirst
cycleLeader(string, shift, lenFirst)
Reverse(string, shift // 2, shift - 1)
Reverse(string, shift, shift + lenFirst // 2 - 1)
Reverse(string, shift // 2, shift + lenFirst // 2 - 1)
shift += lenFirst
if __name__ == "__main__":
string = "a1b2c3d4e5f6g7"
string = list(string)
moveNumberToSecondHalf(string)
print(''.join(string))
|
C#
using System;
class GFG{
static char []str;
static void reverse(int low,
int high)
{
while (low < high)
{
char t = str[low];
str[low] = str[high];
str[high] = t;
++low;
--high;
}
}
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
{
if (j % 2 == 1)
j = len / 2 + j / 2;
else
j /= 2;
char t = str[j + shift];
str[j + shift] = item;
item = t;
}
while (j != i);
}
}
static void moveNumberToSecondHalf()
{
int k, lenFirst;
int lenRemaining = str.Length;
int shift = 0;
while (lenRemaining > 0)
{
k = 0;
while (Math.Pow(3, k) +
1 <= lenRemaining)
k++;
lenFirst = (int)Math.Pow(3,
k - 1) + 1;
lenRemaining -= lenFirst;
cycleLeader(shift, lenFirst);
reverse(shift / 2,
shift - 1);
reverse(shift, shift +
lenFirst / 2 - 1);
reverse(shift / 2, shift +
lenFirst / 2 - 1);
shift += lenFirst;
}
}
public static void Main(String[] args)
{
String st = "a1b2c3d4e5f6g7";
str = st.ToCharArray();
moveNumberToSecondHalf();
Console.Write(str);
}
}
|
Javascript
<script>
let str;
function reverse(low,high)
{
while (low < high)
{
let t = str[low];
str[low] = str[high];
str[high] = t;
++low;
--high;
}
}
function cycleLeader(shift,len)
{
let j;
let item;
for(let i = 1; i < len; i *= 3)
{
j = i;
item = str[j + shift];
do
{
if (j % 2 == 1)
j = Math.floor(len / 2) + Math.floor(j / 2);
else
j =Math.floor(j/2);
let t = str[j + shift];
str[j + shift] = item;
item = t;
}
while (j != i);
}
}
function moveNumberToSecondHalf()
{
let k, lenFirst;
let lenRemaining = str.length;
let shift = 0;
while (lenRemaining > 0)
{
k = 0;
while (Math.pow(3, k) + 1 <= lenRemaining)
k++;
lenFirst = Math.floor(Math.pow(3, k - 1)) + 1;
lenRemaining -= lenFirst;
cycleLeader(shift, lenFirst);
reverse(Math.floor(shift / 2), shift - 1);
reverse(shift, shift + Math.floor(lenFirst / 2) - 1);
reverse(Math.floor(shift / 2), shift + Math.floor(lenFirst / 2) - 1);
shift += lenFirst;
}
}
let st = "a1b2c3d4e5f6g7";
str = st.split("");
moveNumberToSecondHalf();
document.write(str.join(""));
</script>
|
Time complexity: O(n2)
Auxiliary Space: O(1)
Click here to see various test cases.
Notes:
- 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.
- 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.
Similar Reads
Move To Front Data Transform Algorithm
What is the MTF transform? The MTF (Move to Front) is a data transformation algorithm that restructures data in such a way that the transformed message is more compressible and therefore used as an extra step in compression. Technically, it is an invertible transform of a sequence of input character
9 min read
Minimize the String Transformation Cost
Given a string Str and cost array C[] of size n replacing Str[i] with any character will have cost C[i]. The task is to minimize the cost where every odd i, for all i from 1 to n have to match either, S[i] = S[n-i+1] and S[i+1] = S[n-i] or,S[i] = S[n-i] and S[i+1] = S[n-i+1]Examples: Input: N = 4, S
6 min read
Maximum weight transformation of a given string
Given a string consisting of only A's and B's. We can transform the given string to another string by toggling any character. Thus many transformations of the given string are possible. The task is to find Weight of the maximum weight transformation. Weight of a string is calculated using below form
9 min read
Check if it is possible to transform one string to another
Given two strings s1 and s2(all letters in uppercase). Check if it is possible to convert s1 to s2 by performing following operations. Make some lowercase letters uppercase. Delete all the lowercase letters. Examples: Input : s1 = daBcd s2 = ABC Output : yes Explanation : daBcd -> dABCd -> ABC
7 min read
Z algorithm (Linear time pattern searching Algorithm)
This algorithm efficiently locates all instances of a specific pattern within a text in linear time. If the length of the text is "n" and the length of the pattern is "m," then the total time taken is O(m + n), with a linear auxiliary space. It is worth noting that the time and auxiliary space of th
13 min read
Transform an Empty String to a Given String
Given a string s1 and an empty string s2, you have to transform the empty string s2 to given string s1 using the given moves. In one move, you can create a sequence of identical characters on string s2 or overwrite any substring with a new character on string s2. The task is to minimize the number o
7 min read
Remove all occurrences of string t in string s using KMP Algorithm
Given two strings s and t, the task is to remove all occurrences of t in s and return the modified string s, and you have to use the KMP algorithm to solve this. Examples: Input: s = "abcdefgabcabcabdefghabc", t = "abc"Output: "defgdefgh" Input: s = "aaabbbccc", t = "bbb"Output: "aaaccc" Approach: T
8 min read
Finite Automata algorithm for Pattern Searching
Given a text txt[0..n-1] and a pattern pat[0..m-1], write a function search(char pat[], char txt[]) that prints all occurrences of pat[] in txt[]. You may assume that n > m.Examples: Input: txt[] = "THIS IS A TEST TEXT" pat[] = "TEST" Output: Pattern found at index 10 Input: txt[] = "AABAACAADAAB
13 min read
Transform One String to Another using Minimum Number of Given Operation
Given two strings A and B, the task is to convert A to B if possible. The only operation allowed is to put any character from A and insert it at front. Find if it's possible to convert the string. If yes, then output minimum no. of operations required for transformation. Examples: Input: A = "ABD",
15+ min read
Move spaces to front of string in single traversal
Given a string that has set of words and spaces, write a program to move all spaces to front of string, by traversing the string only once. Examples: Input : str = "geeks for geeks" Output : str = " geeksforgeeks" Input : str = "move these spaces to beginning" Output : str = " movethesespacestobegin
6 min read