# Python3 Program for 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:

## 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`

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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