# C++ Program to Swap characters in a String

• Last Updated : 25 Jan, 2022

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;
Explanation:
after 1st swap: BACDE
after 3rd swap: BCDAE
after 4th swap: BCDEA
after 5th swap: ACDEB
after 7th swap: CDAEB
after 8th swap: CDEAB
after 9th swap: CDEBA

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 ` `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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