Python3 Program for 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;
Output : ADEBC
Explanation:
         after 1st swap: BACDE
         after 2nd swap: BCADE
         after 3rd swap: BCDAE
         after 4th swap: BCDEA
         after 5th swap: ACDEB
         after 6th swap: CADEB
         after 7th swap: CDAEB
         after 8th swap: CDEAB
         after 9th swap: CDEBA
         after 10th swap: ADEBC

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.

Below is the implementation of the approach:

Python3

# Python Program to Swap characters in a String
 
def swapCharacters(s, B, C): 

    N = len(s) 

    # If c is greater than n 

    C = C % N

    # Converting string to list

    s = list(s)

    # loop to swap ith element with (i + C) % n th element 

    for i in range(B):

        s[i], s[(i + C) % N] = s[(i + C) % N], s[i] 

    s = ''.join(s)

    return s
 
# Driver program 

s = "ABCDEFGH"

B = 4

C = 3

print(swapCharacters(s, B, C)) 
 
# This code is contributed by Susobhan Akhuli

Output

DEFGBCAH

Time Complexity: O(B), to iterate B times.
Auxiliary Space: O(1)

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