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Swap characters in a String
  • Difficulty Level : Expert
  • Last Updated : 29 Dec, 2020

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.

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


Output: 

CADEFGHIJKB

 

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

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