Python program to check if two Strings are Anagram or not
Check if The Two Strings are Anagram or not
Anagram program in python is when strings share the same no of characters and also the same characters then strings are called anagrams. The rearrangement of similar characters or letters in a string even if they don’t have the same meanings are anagrams. Only one strict rule is followed if we create anagrams of a string ‘count of every character available in the 1st string should be equal to the count of characters in the 2nd string for the same character.
Anagram Program in Python
Two strings S1 and S2 are anagrams of one another if –
By re-arranging characters of string1 (s1) we can form string2 (s2) and vice versa
Check if two Strings are Anagram or not in Python
Algorithm
- Method 1: Sorts both string and check if the results are same or not
- Method 2: Uses counter method that gives results of individual counts of items present in the list, and compare them
Method 1
- This method first converts both strings into lowercase
- We sort both of the strings
- We then compare both resultant strings
- If they are same then they must be anagrams
# Anagram program in python
# take user input String1 = "Listen" String2 = "Silent" String1 = sorted(String1.lower()) String2 = sorted(String2.lower()) print("String1 after sorting: ", String1) print("String2 after sorting: ", String2) # check if now strings matches if String1 == String2: print('Strings are anagram') else: print('Strings are not anagram')
Output:
String1 after sorting: ['e', 'i', 'l', 'n', 's', 't'] String2 after sorting: ['e', 'i', 'l', 'n', 's', 't'] Strings are anagram
Method 2
We use counter method from the collections library for this.
The counter method can be used as –
- For string: Listen
- Counter(string)
printing this counter string will give a result –
Counter({'l': 1, 'i': 1, 's': 1, 't': 1, 'e': 1, 'n': 1})We can then compare both counters for both the strings.
# Check if two Strings are Anagram or not in Python
from collections import Counter def check(s1, s2): # printing counter print(Counter(s1)) print(Counter(s2)) # implementing counter function if Counter(s1) == Counter(s2): print("The strings are anagrams.") else: print("The strings aren't anagrams.") # driver code s1 = "Listen" s2 = "silent" check(s1.lower(), s2.lower())
Output:
Counter({'l': 1, 'i': 1, 's': 1, 't': 1, 'e': 1, 'n': 1})
Counter({'s': 1, 'i': 1, 'l': 1, 'e': 1, 'n': 1, 't': 1})
The strings are anagrams.
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st1=’silent’
st2=’listn’
def check(st1,st2):
for i in st1:
if i in st2:
st2=st2.replace(i,”)
else:
return False
return True
print(check(st1,st2))
#USING SIMPLE IF ELSE CONDITIONS
a=set(input().lower())
b=set(input().lower())
# flag=True
if(len(a)==len(b)):
for i in a:
if(i in b):
flag=True
else:
flag=False
break
if(flag):
print(“yes”)
else:
print(“no”)
from collections import Counter
def anagram(s):
return Counter(s);
s1,s2 = input().split(“,”);
str1 = anagram(s1);
str2 = anagram(s2);
if str1==str2 :
print(“Anagram”);
else :
print(“Not anagram”);
s1,s2 = input().split(“,”);
str1 = sorted(s1.lower());
str2 = sorted(s2.lower());
if str1 == str2 :
print(“anagram”);
else :
print(“Not anagram”);
s1=input()
s2=input()
l1=[]
l2=[]
for i in s1:
l1.append(i.lower())
l1.sort()
for j in s2:
l2.append(j.lower())
l2.sort()
if(l1==l2):
print(“A”)
else:
print(“n”)
a=input()
b=input()
a=sorted(a.lower())
b=sorted(b.lower())
print(a)
print(b)
if(a==b):
print(“strings are anagram”)
else:
print(“strings are not anagram”)
a = str(input(“enter first string :”))
b = str(input(“enter second string: “))
a =list(a)
b =list(b)
a.sort()
b.sort()
if a ==b:
print(“valid”)
else:
print(“invalid”)
#OUTPUT
enter first string :prepinsta
enter second string: preepinsta
invalid