Python program to check if a string contains all unique characters
To implement an algorithm to determine if a string contains all unique characters.
Examples:
Input : s = “abcd”
Output: True
“abcd” doesn’t contain any duplicates. Hence the output is True.Input : s = “abbd”
Output: False
“abbd” contains duplicates. Hence the output is False.
One solution is to create an array of boolean values, where the flag at the index i indicates whether character i in the alphabet is contained in the string. The second time you see this character you can immediately return false.
You can also return false if the string length exceeds the number of unique characters in the alphabet.
Implementation:
Python
def isUniqueChars(st): # String length cannot be more than # 256. if len (st) > 256 : return False # Initialize occurrences of all characters char_set = [ False ] * 128 # For every character, check if it exists # in char_set for i in range ( 0 , len (st)): # Find ASCII value and check if it # exists in set. val = ord (st[i]) if char_set[val]: return False char_set[val] = True return True # driver code st = "abcd" print (isUniqueChars(st)) |
True
Complexity Analysis:
- Time Complexity: O(N), where N is the length of the string.
- Auxiliary Space: O(1), no extra space required so it is a constant.
Method #2:Using Built-in Python Functions:
- Count the frequencies of characters using Counter() function
- If the keys in the frequency dictionary(which gives the count of distinct characters) is equal to the length of string then print True else False
Implementation:
Python3
from collections import Counter def isUniqueChars(string): # Counting frequency freq = Counter(string) if ( len (freq) = = len (string)): return True else : return False # driver code st = "abcd" print (isUniqueChars(st)) # This code is contributed by vikkycirus |
True
Complexity Analysis:
- Time Complexity: O(N), where N is the length of the string.
- Auxiliary Space: O(26), in total there are 26 letters in alphabet and no extra space required so it is a constant.
Method #3 : Using list() and set() methods
Python3
st = "abcbd" x = list ( set (st)) y = list (st) x.sort() y.sort() if (x = = y): print ( True ) else : print ( False ) |
False
Time Complexity: O(logn)
Auxiliary Space: O(1)
Method #4: Using for loop and membership operators
Python3
st = "abcbd" a = "" for i in st: if i not in a: a + = i if (a = = st): print ( True ) else : print ( False ) |
False
Time Complexity: O(N)
Auxiliary Space: O(N)
Method #5 : Using Operator.countOf() function
Python3
import operator as op def isUniqueChars(string): for i in string: if op.countOf(string, i) > 1 : return False return True # driver code st = "abcd" print (isUniqueChars(st)) |
True
Complexity Analysis:
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1), no extra space required so it is a constant.
Method #6 : Using count() function
Python3
def isUniqueChars(string): for i in string: if string.count(i) > 1 : return False return True # driver code st = "abcd" print (isUniqueChars(st)) |
True
Time Complexity: O(N)
Auxiliary Space: O(1)