Problem 8-Anagram Check Analysis Using Python - Competitive Programming

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[krishnaik]

Revise Competitive Programming Playlist

[shivamjindal]

def anagram(str1, str2):

# convert capital to lower
str1 = str1.lower()
str2 = str2.lower()

#cleaning up spaces in string
str1 = str1.replace(" ", "")
str2 = str2.replace(" ", "")

#convert string to list which includes specific letters
list_str1 = list(str1)
list_str2 = list(str2)
anagram_not_exit = 0

for i in list_str1:
if i in list_str2:
pass
else:
anagram_not_exit+=1


for i in list_str2:
if i in list_str1:
pass
else:
anagram_not_exit+=1

if anagram_not_exit>0:
print('Anagram Does Not Exist')
else:
print('Anagram Exist')

[yashwanthvb]

def anagram_check(s1, s2):
s1=s1.replace(' ', '').lower()
s2=s2.replace(' ', '').lower()


check={}

for letter in s1:
if letter in check:
check[letter]+=1
else:
check[letter]=1

for letter in s2:
if letter in check:
check[letter]-=1
else:
check[letter]=1


for count in check:
if check[count]!=0:
return 'Not an Anagram'
else:
return 'It is an Anagram'

[govindsingh]

def check(str1, str2):
if(sorted(str1)== sorted(str2)):
print(" strings are anagram.")
else:
print(" strings aren't anagrams ")
str1 ="study"
str2 ="dusty"
check(str1, str2)

#sorted method returns the sorted string.

[fracturedillusions]

How about this ? I guess it solves the problem in O(n)

str1='Eleven7 plus two'
str2='Twelve77 plus one'
def anagram(str1, str2):
str1=str1.replace(' ', '').lower()
str2=str2.replace(' ', '').lower()
count=0
if len(str1)==len(str2):
for i in str1:
if i in str2:
count+=1
str2=str2.replace(i, '', 1)
else:
return False
else:
return False
if count==len(str1):
return True

[lpis]

Hi krish, i get on one Interview question, it was about decision trees :how tthe decision tree knows, What feature is in the first Node? This is Basic question the tree goes over All features, but I Red that decision tree set the most important features at the beginning of tree, Thanks for your answer

[nishu]

check weather Counter() (import it from collections) of both the words are same or not

[theoutlet]

def anagram(str1, str2):
a = sorted(str1.lower().replace(' ', ''))
b = sorted(str2.lower().replace(' ', ''))
if a == b:
return True
else:
return False

[akhilsoni]

def Anagram(str1, str2):
dict = {}
for i in range(len(str1)):
if ord(str1[i]) <= 122 and ord(str1[i])>=97:
if ord(str1[i])-32 not in dict:
dict[ord(str1[i])-32] = 1
else:
dict[ord(str1[i])-32] = dict[ord(str1[i])-32] + 1
else:
if ord(str1[i]) not in dict:
dict[ord(str1[i])] = 1
else:
dict[ord(str1[i])] = dict[ord(str1[i])] + 1

for i in range(len(str2)):
if ord(str2[i])<=122 and ord(str2[i])>=97:
if ord(str2[i])-32 in dict:
if dict[ord(str2[i])-32] != 0:
dict[ord(str2[i])-32] = dict[ord(str2[i])-32] - 1
else:
return 0
else:
return 0
else:
if ord(str2[i]) in dict:
if dict[ord(str2[i])] != 0:
dict[ord(str2[i])] = dict[ord(str2[i])] - 1
else:
return 0
else:
return 0

return 1



str1 = input()
str2 = input()
if Anagram(str1, str2):
print("Yes")
else:
print("No")

[sonalighadage]

str1 = "eleven plus two"
str2 = "twelve plus one"

def anagram(str1, str2):
str1 = str1.replace(" ", "").lower()
str2 = str2.replace(" ", "").lower()
if len(str1) == len(str2):
for i in str1:
if i in str2:
pass
return True
return False
print(anagram(str1, str2))

[jannatfardus]

str1 = input("Enter some text:")
str2 = input("Enter some text:")
str3=str1.lower()
str4=str2.lower()


remove = " "
def remo(c1):
result=""
for i in c1:
if i!=remove:
result+=i
return result
def char_count(value):
count={}
for char in value:
if char in count:
count[char]+=1
else:
count[char] = 1
return count
def aroma(value_1, value_2):
result1=""
result2=""
count1=char_count(value_1)
count2=char_count(value_2)
result1=remo(value_1)
result2=remo(value_2)
len1= len(result1)
len2= len(result2)

if len1 != len2:
print("Not anagram")

elif count1==count2:
print("Anagram")
else:
print("sorry not anagram")

aroma(str3, str4)

[dataanalyticsfun]

Str1 = input ('string1')
Str2= input ('string2')
If len(str1) == len(str2):
If sorted (str1)== sorted (str2):
Print('given string is anagrams')
Else:
Print ('given string are not anagrams')
Else:
Print('given string are not anagrams')
Sir this way is right to to do it

[KisaanTuber]

my solution. Although not optimized for competitive programming. Suggestions to improve further?

def counter_fn(str_char):
str_char = str_char.replace(" ", "").lower()
count_dict = {}
for entry in str_char:
if entry in count_dict.keys():
count_dict[entry] += 1
else:
count_dict[entry] = 1
return count_dict

if __name__ == '__main__':
str_1 = input()
str_2 = input()
count_dict1 = counter_fn(str_1)
count_dict2 = counter_fn(str_2)
if count_dict1 == count_dict2:
print('ANAGRAM')
else:
print('NOT ANAGRAM')

[Mujahidabbastarar]

I need someone to help me in my Python homeworks for free...😭😭