Checking strings against each other (Anagrams)

Question

The assignment is to write a program that accepts two groups of words from the user and then prints a \"True\" statement if the two are anagrams (or at least if all the lett

Answer 1

Java Code for Anagram

static void anagram(String s1,String s2){
    if(s1.length()!=s2.length()){
        System.out.println("not anagram");
        return;
    }
    else{
        int []arr=new int[256];

        int size=s1.length();
        for(int i=0;i<size;i++){
            arr[s1.charAt(i)]++;
            arr[s2.charAt(i)]--;
        }
        for(int i=0;i<256;i++){
            if(arr[i]!=0){
                System.out.println("not anagram");
                return;
                }
        }
        System.out.println("anagram");
    }
    
}

Answer 2

Not sure if it was proposed up there, but I went with:

def is_anagram(a, b):
    return sorted(a.lower()) == sorted(b.lower())

Answer 3

To check if two strings are anagrams of each other using dictionaries: Note : Even Number, special characters can be used as an input

def anagram(s):
    string_list = []
    for ch in s.lower():
        string_list.append(ch)

    string_dict = {}
    for ch in string_list:
        if ch not in string_dict:
            string_dict[ch] = 1
        else:
            string_dict[ch] = string_dict[ch] + 1

    return string_dict



s1 = "master"
s2 = "stream"

a = anagram(s1)
b = anagram(s2)

if a == b:
    print "Anagram"
else:
    print "Not Anagram"

Answer 4

Here's one typical assignment solution:

def anagram(s1, s2):
""" (str, str) -> bool

Return True if s1 and s2 are anagrams

>>> anagram(s1, s2)
True
"""
    s1 = s1.replace(" ", "")
    s2 = s2.replace(" ", "")

    s1_new = list(s1)
    s2_new = list(s2)

    if len(s1_new) == len(s2_new):
        dupe_found = True
        while dupe_found:
            dupe_found = False
            for i in s1_new:
                for j in s2_new:
                    if i == j:
                        s1_new.remove(i)
                        s2_new.remove(j)
                        dupe_found = True
                        break
                break
    return s1_new == s2_new

Answer 5

This worked for me

str1="abcd"
str2="bcad"
word1=[]
word2=[]
for x in range(len(str1)):
    word1.append(str1[x])
for x in range(len(str2)):
    word2.append(str2[x])
if(len(word1)==len(word2)):
    for letter in word1:
        if letter in word2:
            word2.remove(letter)

if len(word2)==0:
    print "anagram"
else:
    print "not anagram"

Answer 6

def is_anagram(w1, w2):
    w1, w2 = list(w1.upper()), list(w2.upper())
    w2.sort()
    w1.sort()
    return w1 == w2