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:

# 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

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 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

CADEFGHIJKB

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

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