Python | Check if frequencies of all characters of a string are different

Last Updated : 23 Mar, 2023
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Given a string S consisting only of lowercase letters, the task is to check if the frequency of all characters of the string is unique. 
Examples:

Input : abaccc
Output : Yes
‘a’  occurs two times, ‘b’ occurs once
and ‘c’ occurs three times.

Input : aabbc
Output : No
Frequency of both 'a' and 'b' are same.

Approach: One thing to observe in this problem is that position of characters does not matter here so, simply count the frequency of characters. Go through the string and count the occurrence of all characters. Then sort the frequency array and check if consecutive elements are same and immediately print "No" else print "Yes". 

Python3 
# Python program to check if frequency of
# all characters of a string are unique
  
    # creating a frequency array 
    freq =[0]*26
  
    # Finding length of s 
    n = len(s) 
  
    for i in range(n): 
  
        # counting frequency of all characters 
        freq[ord(s[i])-97] += 1

    # sorting the frequency array
    freq.sort()
    for i in range(25): 
  
        # checking if element is zero
        if (freq[i] == 0):
            continue

        # if element is non-zero
        # checking if frequencies are unique
        if (freq[i] == freq[i + 1]): 
            return False

    return True
  
# Driver code 
s ="abaccc"

if(check(s)): 
    print("Yes") 
else: 
    print("No")
Output:
Yes

Approach : Using count() and len() methods.First find the count of each element, append the count to a list. If the count of first element of list equals to the length of list, then the frequencies are unique.

Python3 
# Python program to check if frequency of
# all characters of a string are unique

s ="aabbc"
v=[]
for i in s:
    if i not in v:
        v.append(s.count(i))
x=[]
for i in v:
  if i not in x:
    x.append(i)
if(len(x)==len(v)):
  print("Yes")
else:
  print("No")

Output
No

Approach: using operator.countOf() method

Python3 
# Python program to check if frequency of
# all characters of a string are unique
import operator as op
s ="aabbc"
v=[]
for i in s:
    if i not in v:
        v.append(op.countOf(s,i))
x=[]
for i in v:
  if i not in x:
    x.append(i)
if(len(x)==len(v)):
  print("Yes")
else:
  print("No")

Output
No

Time Complexity: O(N)

Auxiliary Space : O(N)

Using counter and set()

Algorithm:

  • Import the Counter module from the collections package
  • Convert the input string into a Counter object
  • Extract the frequency of each character and store in a set
  • If the length of the set is equal to the number of unique frequencies, return "Yes"
  • Otherwise, return "No"
Python3 
from collections import Counter

def check_freq(s):
    # Convert string into a Counter object
    freq = Counter(s)
    
    # Extract the frequency of each character and store in a set
    freq_set = set(freq.values())
    
    # Check if the length of the set is equal to the number of unique frequencies
    if len(freq_set) == len(set(s)):
        return "Yes"
    else:
        return "No"
        
# Driver code
s = "abaccc"

print(check_freq(s))

Output
Yes

Complexity analysis:

Time complexity: The Counter function has a time complexity of O(N) where N is the length of the string. The set function also has a time complexity of O(N). Therefore, the overall time complexity of the code is O(N).
Space complexity: The space complexity of the code is O(N) because we are creating a Counter object and a set object, both of which can store up to N elements.


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Count frequencies of all elements in array in Python using collections module

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Pravash Jha
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