C++ Program to Swap characters in a String

Last Updated : 18 Aug, 2023

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 % 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
 
  

Output

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++

   // 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
 
  

Output:

CADEFGHIJKB

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

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

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