Swap characters in a String

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++ 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 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

# 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

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);
    }
}

// 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);

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++ 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 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 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# 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

<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>

CADEFGHIJKB

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

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Article Tags :

• Strings
• Java
• DSA
• rotation
• Constructive Algorithms
• Swap-Program

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Practice Tags :

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• Strings