Check if the characters of a given string are in alphabetical order in Python

Suppose we have a string s. We have to check whether characters in s are in alphabetical order or not.

So, if the input is like s = "mnnooop", then the output will be True.

Method 1: Using Sorting Approach

To solve this, we will follow these steps ?

• char_arr := a new list from the characters present in s
• sort the list char_arr
• return char_arr is same as list of all characters in s then true otherwise false

Example

def solve(s):
    char_arr = list(s)
    char_arr.sort()
    return char_arr == list(s)

s = "mnnooop"
print(solve(s))

The output of the above code is ?

True

Method 2: Using Direct Comparison

A more efficient approach is to directly compare the string with its sorted version ?

def is_alphabetical_order(s):
    return s == ''.join(sorted(s))

# Test with multiple examples
strings = ["mnnooop", "hello", "abc", "dcba"]

for string in strings:
    result = is_alphabetical_order(string)
    print(f"'{string}' is in alphabetical order: {result}")

The output of the above code is ?

'mnnooop' is in alphabetical order: True
'hello' is in alphabetical order: False
'abc' is in alphabetical order: True
'dcba' is in alphabetical order: False

Method 3: Using Loop Comparison

We can also check by comparing adjacent characters without sorting ?

def check_alphabetical_loop(s):
    for i in range(len(s) - 1):
        if s[i] > s[i + 1]:
            return False
    return True

# Test the function
test_string = "mnnooop"
print(f"'{test_string}' is in alphabetical order: {check_alphabetical_loop(test_string)}")

test_string2 = "hello"
print(f"'{test_string2}' is in alphabetical order: {check_alphabetical_loop(test_string2)}")

The output of the above code is ?

'mnnooop' is in alphabetical order: True
'hello' is in alphabetical order: False

Comparison

Conclusion

Use the loop comparison method for optimal performance with O(n) time complexity. The sorting approaches are simpler to understand but less efficient for large strings.