# Java 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:

## 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);``    ``}``}`
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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