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;
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
```

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 thanN.

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 to the left by some places. The first part is rotated by (N % C) places every full iteration.
• The second part is rotated 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:

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); ` `    ``} ` `} `

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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