Python: Circular shift number

Last update on December 21 2024 07:44:12 (UTC/GMT +8 hours)

Circular Shift of Digits

Write a Python program to shift the decimal digits n places to the left, wrapping the extra digits around. If the shift > the number of digits in n, reverse the string.

Input:
n = 12345  and shift = 1
Output:
Result =  23451
Input:
n = 12345  and shift = 2
Output:
Result =  34512
Input:
n = 12345  and shift = 3
Output:
Result =  45123
Input:
n = 12345  and shift = 5
Output:
Result =  12345
Input:
n = 12345  and shift = 6
Output:
Result =  54321

Visual Presentation:

Sample Solution-1:

Python Code:

# License: https://bit.ly/3oLErEI

def test(n, shift):
    # Convert the number to a string
    s = str(n)
    
    # Check if shift is greater than the number of digits in n
    if shift > len(s):
        # If so, reverse the string
        return s[::-1]
    
    # Shift the decimal digits to the left by 'shift' places
    return s[shift:] + s[:shift]

# Display the purpose of the code
print("Shift the decimal digits n places to the left. If shift > the number of digits of n, reverse the string.")

# Example 1
n1 = 12345
shift1 = 1
print("\nn =", n1, " and shift =", shift1)
print("Result =", test(n1, shift1))

# Example 2
n2 = 12345
shift2 = 2
print("\nn =", n2, " and shift =", shift2)
print("Result =", test(n2, shift2))

# Example 3
n3 = 12345
shift3 = 3
print("\nn =", n3, " and shift =", shift3)
print("Result =", test(n3, shift3))

# Example 4
n4 = 12345
shift4 = 5
print("\nn =", n4, " and shift =", shift4)
print("Result =", test(n4, shift4))

# Example 5
n5 = 12345
shift5 = 6
print("\nn =", n5, " and shift =",shift5)
print("Result = ",test(n5, shift))

Sample Output:

Shift the decimal digits n places to the left. If shift > the number of digits of n, reverse the string.:

n = 12345  and shift = 1
Result =  23451

n = 12345  and shift = 2
Result =  34512

n = 12345  and shift = 3
Result =  45123

n = 12345  and shift = 5
Result =  12345

n = 12345  and shift = 6
Result =  54321

Flowchart:

Sample Solution-2:

Python Code:

# License: https://bit.ly/3oLErEI

def test(n, shift):
    # Convert the number to a list of individual digits
    shifted_digits = [int(x) for x in str(n)]
    
    # Shift the digits to the left by 'shift' places using list manipulation
    for i in range(shift):
        shifted_digits.append(shifted_digits.pop(0))
    
    # Check if shift is greater than the number of digits in n
    if shift > len(shifted_digits):
        # If so, reverse the string representation of n
        return str(n)[::-1]
    else:
        # Convert the shifted digits back to a string and join them
        return ''.join(str(x) for x in shifted_digits)

# Display the purpose of the code
print("Shift the decimal digits n places to the left. If shift > the number of digits of n, reverse the string.")

# Example 1
n1 = 12345
shift1 = 1
print("\nn =", n1, " and shift =", shift1)
print("Result =", test(n1, shift1))

# Example 2
n2 = 12345
shift2 = 2
print("\nn =", n2, " and shift =", shift2)
print("Result =", test(n2, shift2))

# Example 3
n3 = 12345
shift3 = 3
print("\nn =", n3, " and shift =", shift3)
print("Result =", test(n3, shift3))

# Example 4
n4 = 12345
shift4 = 5
print("\nn =", n4, " and shift =", shift4)
print("Result =", test(n4, shift4))

# Example 5
n5 = 12345
shift5 = 6
print("\nn =", n5, " and shift =", shift5)
print("Result =", test(n5, shift5))

Sample Output:

Shift the decimal digits n places to the left. If shift > the number of digits of n, reverse the string.:

n = 12345  and shift = 1
Result =  23451

n = 12345  and shift = 2
Result =  34512

n = 12345  and shift = 3
Result =  45123

n = 12345  and shift = 5
Result =  12345

n = 12345  and shift = 6
Result =  54321

Flowchart:

Python Code Editor :

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