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:
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:
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
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
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
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
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)