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# Swap characters in a String

• Difficulty Level : Hard
• Last Updated : 05 Jul, 2021

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`

## 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);``    ``}``}`

## Python3

 `# Python3 program to find new after swapping``# characters at position i and i + c``# b times, each time advancing one``# position ahead` `# Method to find the required string``def` `swapChars(s, c, b):``    ` `    ``# Get string length``    ``n ``=` `len``(s)``    ` `    ``# 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``        ` `    ``f ``=` `int``(b ``/` `n)``    ``r ``=` `b ``%` `n``    ` `    ``# Rotate first c characters by (n % c)``    ``# places f times``    ``p1 ``=` `rotateLeft(s[``0` `: c], ((c ``*` `f) ``%` `(n ``-` `c)))``    ` `    ``# Rotate remaining character by``    ``# (n * f) places``    ``p2 ``=` `rotateLeft(s[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)``    ``a ``=` `p1 ``+` `p2``    ``a ``=` `list``(a)``    ` `    ``# Remaining swaps``    ``for` `i ``in` `range``(r):``        ` `        ``# Swap ith character with``        ``# (i + c)th character``        ``temp ``=` `a[i]``        ``a[i] ``=` `a[(i ``+` `c) ``%` `n]``        ``a[(i ``+` `c) ``%` `n] ``=` `temp` `    ``# Return final string``    ``return` `str``("".join(a))` `def` `rotateLeft(s, p):``    ` `    ``# Rotating a string p times left is``    ``# effectively cutting the first p``    ``# characters and placing them at the end``    ``return` `s[p:] ``+` `s[``0` `: p]` `# Driver code` `# Given values``s1 ``=` `"ABCDEFGHIJK"``b ``=` `1000``c ``=` `3` `# Get final string``s2 ``=` `swapChars(s1, c, b)` `# Print final string``print``(s2)` `# This code is contributed by avanitrachhadiya2155`

## 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`

## Javascript

 ``
Output:
`CADEFGHIJKB`

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

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