C++ Program to 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 % N], 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

DEFGBCAH

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

CADEFGHIJKB

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

Please refer complete article on Swap characters in a String for more details!