Swap characters in a String

Read

Discuss

Given a String S of length N, two integers B and C, the task is to traverse characters starting from the beginning, swapping a character with the character after C places from it, i.e. swap characters at position i and (i + C)%N. Repeat this process B times, advancing one position at a time. Your task is to find the final String after B swaps.

Examples:

Input : S = "ABCDEFGH", B = 4, C = 3;
Output:  DEFGBCAH
Explanation:
         after 1st swap: DBCAEFGH
         after 2nd swap: DECABFGH
         after 3rd swap: DEFABCGH
         after 4th swap: DEFGBCAH
Input : S = "ABCDE", B = 10, C = 6;
Output : ADEBC
Explanation:
         after 1st swap: BACDE
         after 2nd swap: BCADE
         after 3rd swap: BCDAE
         after 4th swap: BCDEA
         after 5th swap: ACDEB
         after 6th swap: CADEB
         after 7th swap: CDAEB
         after 8th swap: CDEAB
         after 9th swap: CDEBA
         after 10th swap: ADEBC

Naive Approach:

For large values of B, the naive approach of looping B times, each time swapping ith character with (i + C)%N-th character will result in high CPU time.
The trick to solving this problem is to observe the resultant string after every N iterations, where N is the length of the string S.
Again, if C is greater than or equal to the N, it is effectively equal to the remainder of C divided by N.
Hereon, let’s consider C to be less than N.

Below is the implementation of the approach:

C++

// C++ Program to Swap characters in a String
#include <iostream>
#include <string>
 
using namespace std;
 
string swapCharacters(string s, int B, int C)
{
    int N = s.size();
    // If c is greater than n
    C = C % N;
    // loop to swap ith element with (i + C) % n th element
    for (int i = 0; i < B; i++) {
        swap(s[i], s[(i + C) % N]);
    }
    return s;
}
 
int main()
{
    string s = "ABCDEFGH";
    int B = 4;
    int C = 3;
    s = swapCharacters(s, B, C);
    cout << s << endl;
    return 0;
}
 
// This code is contributed by Susobhan Akhuli

Java

// Java Program to Swap characters in a String
import java.util.*;
 
public class Main {
    public static String swapCharacters(String s, int B,
                                        int C)
    {
        int N = s.length();
        // If c is greater than n
        C = C % N;
        // loop to swap ith element with (i + C) % n th
        // element
        char[] arr = s.toCharArray();
        for (int i = 0; i < B; i++) {
            char temp = arr[i];
            arr[i] = arr[(i + C) % N];
            arr[(i + C) % N] = temp;
        }
        return new String(arr);
    }
 
    public static void main(String[] args)
    {
        String s = "ABCDEFGH";
        int B = 4;
        int C = 3;
        s = swapCharacters(s, B, C);
        System.out.println(s);
    }
}
 
// This code is contributed by Susobhan Akhuli

Python3

# Python Program to Swap characters in a String
def swapCharacters(s, B, C):
    N = len(s)
    # If c is greater than n
    C = C % N
    # loop to swap ith element with (i + C) % n th element
    s = list(s)
    for i in range(B):
        s[i], s[(i + C) % N] = s[(i + C) % N], s[i]
    return ''.join(s)
 
# Driver code
s = "ABCDEFGH"
B = 4
C = 3
s = swapCharacters(s, B, C)
print(s)
 
# This code is contributed by Susobhan Akhuli

C#

using System;
 
class Program
{
    // Function to swap characters in a string
    static string SwapCharacters(string s, int B, int C)
    {
        int N = s.Length;
        // If C is greater than N, take the modulo to ensure it's within bounds
        C = C % N;
        // Loop to swap the ith element with (i + C) % N th element
        for (int i = 0; i < B; i++)
        {
            int j = (i + C) % N; // Calculate the index of the character to swap with
            char temp = s[i]; // Store the character at index i in a temporary variable
            // Replace character at index i with the character at index j
            s = s.Remove(i, 1).Insert(i, s[j].ToString());
            // Replace character at index j with the character stored in temp
            s = s.Remove(j, 1).Insert(j, temp.ToString());
        }
        return s;
    }
 
    static void Main()
    {
        string s = "ABCDEFGH";
        int B = 4;
        int C = 3;
        s = SwapCharacters(s, B, C);
        Console.WriteLine(s);
    }
}

Javascript

// JS Program to Swap characters in a String
function swapCharacters(s, B, C) {
    const N = s.length;
     
    // If c is greater than n
    C = C % N;
     
    // loop to swap ith element with (i + C) % n th element
    for (let i = 0; i < B; i++) {
        const temp = s[i];
        s = s.substring(0, i) + s[(i + C) % N] + s.substring(i + 1);
        s = s.substring(0, (i + C) % N) + temp + s.substring((i + C) % N + 1);
    }
    return s;
}
 
let s = "ABCDEFGH";
const B = 4;
const C = 3;
s = swapCharacters(s, B, C);
console.log(s);

Output

DEFGBCAH

Time Complexity: O(B), to iterate B times.
Auxiliary Space: O(1)

Efficient Approach:

If we observe the string that is formed after every N successive iterations and swaps (let’s call it one full iteration), we can start to get a pattern.
We can find that the string is divided into two parts: the first part of length C comprising of the first C characters of S, and the second part comprising of the rest of the characters.
The two parts are rotated by some places. The first part is rotated right by (N % C) places every full iteration.
The second part is rotated left by C places every full iteration.
We can calculate the number of full iterations f by dividing B by N.
So, the first part will be rotated left by ( N % C ) * f . This value can go beyond C and so, it is effectively ( ( N % C ) * f ) % C, i.e. the first part will be rotated by ( ( N % C ) * f ) % C places left.
The second part will be rotated left by C * f places. Since, this value can go beyond the length of the second part which is ( N – C ), it is effectively ( ( C * f ) % ( N – C ) ), i.e. the second part will be rotated by ( ( C * f ) % ( N – C ) ) places left.
After f full iterations, there may still be some iterations remaining to complete B iterations. This value is B % N which is less than N. We can follow the naive approach on these remaining iterations after f full iterations to get the resultant string.

Example:
s = ABCDEFGHIJK; c = 4;
parts: ABCD EFGHIJK
after 1 full iteration: DABC IJKEFGH
after 2 full iteration: CDAB FGHIJKE
after 3 full iteration: BCDA JKEFGHI
after 4 full iteration: ABCD GHIJKEF
after 5 full iteration: DABC KEFGHIJ
after 6 full iteration: CDAB HIJKEFG
after 7 full iteration: BCDA EFGHIJK
after 8 full iteration: ABCD IJKEFGH

Below is the implementation of the approach:

C++

// C++ program to find new after swapping
// characters at position i and i + c
// b times, each time advancing one
// position ahead
#include <bits/stdc++.h>
using namespace std;
 
string rotateLeft(string s, int p)
{
     
    // Rotating a string p times left is
    // effectively cutting the first p
    // characters and placing them at the end
    return s.substr(p) + s.substr(0, p);
}
 
// Method to find the required string
string swapChars(string s, int c, int b)
{
     
    // Get string length
    int n = s.size();
     
    // If c is larger or equal to the length of
    // the string is effectively the remainder of
    // c divided by the length of the string
    c = c % n;
     
    if (c == 0)
    {
         
        // No change will happen
        return s;
    }
    int f = b / n;
    int r = b % n;
     
    // Rotate first c characters by (n % c)
    // places f times
    string p1 = rotateLeft(s.substr(0, c),
                  ((n % c) * f) % c);
                   
    // Rotate remaining character by
    // (n * f) places
    string p2 = rotateLeft(s.substr(c),
                  ((c * f) % (n - c)));
                   
    // Concatenate the two parts and convert the
    // resultant string formed after f full
    // iterations to a string array
    // (for final swaps)
    string a = p1 + p2;
     
    // Remaining swaps
    for(int i = 0; i < r; i++)
    {
         
        // Swap ith character with
        // (i + c)th character
        char temp = a[i];
        a[i] = a[(i + c) % n];
        a[(i + c) % n] = temp;
    }
     
    // Return final string
    return a;
}
 
// Driver code
int main() 
{
     
    // Given values
    string s1 = "ABCDEFGHIJK";
    int b = 1000;
    int c = 3;
     
    // Get final string print final string
    cout << swapChars(s1, c, b) << endl;
}
 
// This code is contributed by rag2127

Java

// Java Program to find new after swapping
// characters at position i and i + c
// b times, each time advancing one
// position ahead
 
class GFG {
    // Method to find the required string
 
    String swapChars(String s, int c, int b)
    {
        // Get string length
        int n = s.length();
 
        // if c is larger or equal to the length of
        // the string is effectively the remainder of
        // c divided by the length of the string
        c = c % n;
 
        if (c == 0) {
            // No change will happen
            return s;
        }
 
        int f = b / n;
        int r = b % n;
 
        // Rotate first c characters by (n % c)
        // places f times
        String p1 = rotateLeft(s.substring(0, c),
                               ((n % c) * f) % c);
 
        // Rotate remaining character by
        // (n * f) places
        String p2 = rotateLeft(s.substring(c),
                               ((c * f) % (n - c)));
 
        // Concatenate the two parts and convert the
        // resultant string formed after f full
        // iterations to a character array
        // (for final swaps)
        char a[] = (p1 + p2).toCharArray();
 
        // Remaining swaps
        for (int i = 0; i < r; i++) {
 
            // Swap ith character with
            // (i + c)th character
            char temp = a[i];
            a[i] = a[(i + c) % n];
            a[(i + c) % n] = temp;
        }
 
        // Return final string
        return new String(a);
    }
 
    String rotateLeft(String s, int p)
    {
        // Rotating a string p times left is
        // effectively cutting the first p
        // characters and placing them at the end
        return s.substring(p) + s.substring(0, p);
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Given values
        String s1 = "ABCDEFGHIJK";
        int b = 1000;
        int c = 3;
 
        // get final string
        String s2 = new GFG().swapChars(s1, c, b);
 
        // print final string
        System.out.println(s2);
    }
}

Python3

# Python3 program to find new after swapping 
# characters at position i and i + c 
# b times, each time advancing one 
# position ahead 
 
# Method to find the required string 
def swapChars(s, c, b):
     
    # Get string length 
    n = len(s)
     
    # If c is larger or equal to the length of 
    # the string is effectively the remainder of 
    # c divided by the length of the string 
    c = c % n
     
    if (c == 0):
         
        # No change will happen 
        return s
         
    f = int(b / n)
    r = b % n
     
    # Rotate first c characters by (n % c) 
    # places f times 
    p1 = rotateLeft(s[0 : c], ((c * f) % (n - c)))
     
    # Rotate remaining character by 
    # (n * f) places 
    p2 = rotateLeft(s[c:], ((c * f) % (n - c)))
     
    # Concatenate the two parts and convert the 
    # resultant string formed after f full 
    # iterations to a character array 
    # (for final swaps) 
    a = p1 + p2
    a = list(a)
     
    # Remaining swaps 
    for i in range(r):
         
        # Swap ith character with 
        # (i + c)th character 
        temp = a[i]
        a[i] = a[(i + c) % n]
        a[(i + c) % n] = temp
 
    # Return final string 
    return str("".join(a))
 
def rotateLeft(s, p):
     
    # Rotating a string p times left is 
    # effectively cutting the first p 
    # characters and placing them at the end 
    return s[p:] + s[0 : p]
 
# Driver code 
 
# Given values 
s1 = "ABCDEFGHIJK"
b = 1000
c = 3
 
# Get final string 
s2 = swapChars(s1, c, b)
 
# Print final string 
print(s2)
 
# This code is contributed by avanitrachhadiya2155

C#

// C# Program to find new after swapping
// characters at position i and i + c
// b times, each time advancing one
// position ahead
using System;
 
class GFG
{
    // Method to find the required string
    String swapChars(String s, int c, int b)
    {
        // Get string length
        int n = s.Length;
 
        // if c is larger or equal to the length of
        // the string is effectively the remainder of
        // c divided by the length of the string
        c = c % n;
 
        if (c == 0)
        {
            // No change will happen
            return s;
        }
 
        int f = b / n;
        int r = b % n;
 
        // Rotate first c characters by (n % c)
        // places f times
        String p1 = rotateLeft(s.Substring(0, c),
                            ((n % c) * f) % c);
 
        // Rotate remaining character by
        // (n * f) places
        String p2 = rotateLeft(s.Substring(c),
                            ((c * f) % (n - c)));
 
        // Concatenate the two parts and convert the
        // resultant string formed after f full
        // iterations to a character array
        // (for readonly swaps)
        char []a = (p1 + p2).ToCharArray();
 
        // Remaining swaps
        for (int i = 0; i < r; i++) 
        {
 
            // Swap ith character with
            // (i + c)th character
            char temp = a[i];
            a[i] = a[(i + c) % n];
            a[(i + c) % n] = temp;
        }
 
        // Return readonly string
        return new String(a);
    }
 
    String rotateLeft(String s, int p)
    {
         
        // Rotating a string p times left is
        // effectively cutting the first p
        // characters and placing them at the end
        return s.Substring(p) + s.Substring(0, p);
    }
 
    // Driver code
    public static void Main(String []args)
    {
        // Given values
        String s1 = "ABCDEFGHIJK";
        int b = 1000;
        int c = 3;
 
        // get readonly string
        String s2 = new GFG().swapChars(s1, c, b);
 
        // print readonly string
        Console.WriteLine(s2);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript Program to find new after swapping
// characters at position i and i + c
// b times, each time advancing one
// position ahead
 
function swapChars(s, c, b)
{
 
    // Get string length
        let n = s.length;
  
        // if c is larger or equal to the length of
        // the string is effectively the remainder of
        // c divided by the length of the string
        c = c % n;
  
        if (c == 0) {
            // No change will happen
            return s;
        }
  
        let f = Math.floor(b / n);
        let r = b % n;
  
        // Rotate first c characters by (n % c)
        // places f times
        let p1 = rotateLeft(s.substring(0, c),
                               ((n % c) * f) % c);
  
        // Rotate remaining character by
        // (n * f) places
        let p2 = rotateLeft(s.substring(c),
                               ((c * f) % (n - c)));
  
        // Concatenate the two parts and convert the
        // resultant string formed after f full
        // iterations to a character array
        // (for final swaps)
        let a = (p1 + p2).split("");
  
        // Remaining swaps
        for (let i = 0; i < r; i++) {
  
            // Swap ith character with
            // (i + c)th character
            let temp = a[i];
            a[i] = a[(i + c) % n];
            a[(i + c) % n] = temp;
        }
  
        // Return final string
        return (a).join("");
}
 
function rotateLeft(s,p)
{
    // Rotating a string p times left is
        // effectively cutting the first p
        // characters and placing them at the end
        return s.substring(p) + s.substring(0, p);
}
 
 // Driver code
// Given values
let s1 = "ABCDEFGHIJK";
let b = 1000;
let c = 3;
 
// get final string
let s2 = swapChars(s1, c, b);
 
// print final string
document.write(s2);
     
// This code is contributed by ab2127
</script>

Output

CADEFGHIJKB

Time Complexity: O(n)
Space Complexity: O(n)

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 26 Oct, 2023