Problem 8 Solution-Anagram Check Analysis Using Python - Competitive Programming

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[krishnaik]

Revise The Competitive Programming Language playlist

[sonalighadage]

Thank you for the different test cases and types of errors you caught will be really helpful to write more efficient code.

[rankushvishwakarma]

Thankyou sir for such valuable knowledge

[ravitanwar]

basically same idea as yours but a bit more readable solution along with probably the best example to check the solution/

from collections import Counter
def anagram_ques(s1, s2):
s1=s1.replace(" ", '').lower()
s2=s2.replace(" ", '').lower()
return Counter(s1)==Counter(s2)
s1='tom marvolo riddle'
s2='i am lord voldemort'
anagram_ques(s1, s2)

answer to next question (group anagrams):-
def groupAnagrams(strs) :
h=defaultdict(list)
for i in strs:
", "")))].append(''.join(i))
return h.values()

[sumanthhabib]

a=input()
b=input()
def anagram(a, b):
for i in a:
if a.count(i)==b.count(i):
return print(f'{a} and {b} are anagram')
return print(f'{a} and {b} are not anagram')
anagram(a, b)

[Neon_-xogn]

Excellent learning platform for coder's

[nishu]

will the time complexity reduces if we use "group_letters.get(i)" instead of "i in group_letters" in line 4 ? (It return None if it is not present).

[navidbinahmed]

used sorted: def anagram(str1, str2):
return sorted(str1) == sorted(str2)

str1="elbow"
str2="below"

check = anagram(str1, str2)
print(check)

[VivekKumar-uphb]

Thanks for your solution, but I think its time complexity will be 3n. Correct me if I'm wrong.

[TopGHippie]

second one works if slightly modified:

str1 = "eleven0 plus two"
str2 = "twelve1 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
else:
return False
return True
return False
anagram(str1, str2)

[NikhilRaghav]

Answer for the last part of question:
List of anagrams:

h=['cat ', 'dog', 'go d', 'tac', 'act', 'OGd']
sortedwords=[]
for i in h:
i=i.replace(" ", "").lower()

print(sortedwords)

d={}
for i in unique_words:
c=[]
for j in range(len(sortedwords)):
if i==sortedwords[j]:
c.append(h[j])
d[i]=c
print(d)

[navidbinahmed]

This solution using ASCII conversion is not efficient too. Provides wrong output for an edge case like: "ab" and "Ab" strings.
On Leetcode, it passes 46 test cases out of 50. The below solution may be considered as an end to end edge-case solutions.

def isAnagram(self, str1: str, str2: str) -> bool:

str1 = str1.lower()
str2 = str2.lower()

if len(str1) != len(str2):
return False

char_count = {}

for char in str1:
if char in char_count:
char_count[char] += 1
else:
char_count[char] = 1

for char in str2:
if char in char_count:
char_count[char] -= 1
if char_count[char] < 0:
return False
else:
return False

return True

[akhilsoni]

My solution to today's Problem:
### Anagram Function:
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


### Function to find all the anagrams:
def Anagrams_In_List(List):
for i in range(len(List)):
if List[i] != "":
print(List[i])
for j in range(i+1, len(List)):
if List[j] != "":
if Anagram(List[i], List[j]):
print(List[j])
List[j] = ""
print('\n')

List = ['Cat', 'TAC', 'tac', 'god', 'DOg']
Anagrams_In_List(List)



You told us not to use any functions so instead of converting every character to the lower case by using the ".lower()" function, I used the ASCII values of all the characters.
I think we should use the ASCII value method because in some other common languages like C++ we don't usually use any functions except the sort function(probably)
let me know your views on it. :)
Also, The time complexity of your solution will be O(n) not O(n^2)

[deepum]

def anagram(string1, string2):
count = 0
if len(string1) == (string2):
for a in string1:
if a in string2:
pass
else:
return 'The entered strings are not anagram'
if string2[count] in string1:
pass
else:
return 'The entered strings are not anagram'
count += 1
return 'The entered strings are anagram'
else:
return 'The entered strings are not anagram'

string1 = input('Enter string one: ').replace(' ', '').lower()
string2 = input('Enter string two: ').replace(' ', '').lower()

anagram(string1, string2)

[hphp]

Abhi 2 minutes huye aur pura video dekhne ke pehle kisine Dislike kr diya!! LOL.. First watch and then try to give ur expression. video hi pura 19 minute ka hai 2 minute mai kaise dekh liya? bhai

[smilebig]

Hey krish, I like your videos however I have been observing for the past couple of days that you have left making videos on deep learning. You started paper to code series I guess but then you didn't carry on that series. I can understand that it might not be easy however I would love to see you continuing that series. Lets leave these code optimization tricks for other kids on youtube. xD

[Mhmd]

my answer is:

def anagram(str1, str2):
group_letters={}
for i in str1:
if i in group_letters:
group_letters[i]+=1
else:
group_letters[i]=1
for i in str2:
if i in group_letters:
group_letters[i]-=1
else:
group_letters[i]=1

for j in group_letters:
if group_letters[j]!=0:
return False
return True
Anagram_dictionay={}
list=["cat", "dog", "tac", "god", "act"]
count=1
for i in range(0, len(list)-1):

string1=list[i].replace(" ", "").lower()

if string1 in Anagram_dictionay:
continue
for k in range (i, len(list)-1):

string2=list[k+1]

string2=string2.replace(" ", "").lower()
var=anagram(string1, string2)
if(var):
list[k+1]=string1
print(string1)

if string1 in Anagram_dictionay:

else:
Anagram_dictionay[string1]=2

print(Anagram_dictionay)

[desinaveeravenkatasatyanar]

str1=input("Enter First String:").replace(" ", "").lower()
str2=input("Enter Second String:").replace(" ", "").lower()
for i in str1:
str2 = str2.replace(i, "", 1)
if str2 == "":
print("Anangram")
else:
print("Not Anangram")

[navidbinahmed]

words = ['car', 'bat', 'tac', 'tab', 'act', ]
anagrams = []
non_anagrams = []

for elements in range(0, len(words)):
found_anagram = False

for j in range (elements+1, len(words)):
if sorted(words[elements]) == sorted(words[j]):
print("The anagrams are: ", words[elements], "and", words[j])
found_anagram = True

if words[elements] not in anagrams:

if words[j] not in anagrams:
anagrams.append(words[j])

if not found_anagram and words[elements] not in anagrams:


print("The non-anagram words are: ", ', '.join(non_anagrams))

[Mujahidabbastarar]

can someone help me with my homeworks Python for free?? 🥺