compare 2 strings while using dictionairies (anagrams)

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Check if two strings are anagrams. Write a function anagrams(s1,s2), which given two strings s1 and s2, returns True if they are anagrams, and False otherwise using dictionaries

  1. if len(s1) different than len(s2) then they are not anagrams
  2. i can't find a way to compare both strings using dictionaries.

Code:

D1={}
D2={}

def anagrams(s1,s2):
    if len(s1)!=len(s2):
        return False

    else:
         if D1==D2:
               return True
          else:
                return False

You can use a dict for each string to count the number of occurrences of each distinct character:

def anagrams(s1, s2):
    d = {}
    for s in s1, s2:
        d[s] = {}
        for c in s:
            d[s][c] = d[s].get(c, 0) + 1
    return d[s1] == d[s2]

Detecting if two strings are anagrams, To iterate over a string's chars it's recommended to use Python's The equivalence comparison will need to compare all keys, but it should fail as soon as it finds one key not in the other dictionary, therefore it is O(min(n,m)). Check whether two strings are anagram of each other; Given a sequence of words, print all anagrams together | Set 1; Given a sequence of words, print all anagrams together | Set 2; Split the given string into Primes : Digit DP; Print all possible combinations of the string by replacing '$' with any other digit from the string

You can load your words into a dictionary and compare the dictionaries' sorted values.

D1={}
D2={}

def anagrams(s1,s2):
    if len(s1)!=len(s2):
        return False

    else:
        elementNumber = 0
        for char in s1:                    #Load s1 into dictionary
            D1[elementNumber] = char
            elementNumber = elementNumber + 1

        elementNumber = 0
        for char in s2:                    #Load s2 into dictionary
            D2[elementNumber] = char
            elementNumber = elementNumber + 1

        print(sorted(D1.values()))         #Example output
        print(sorted(D2.values()))         #Example output

        if sorted(D1.values())==sorted(D2.values()): #Sort and compare
             return True
        else:
             return False

print("Anagrams: "+str(anagrams("Hello", "oHlel"))) #Returns True
print("Anagrams: "+str(anagrams("Hello", "xyzlo"))) #Returns False

Optimizing Java Anagram checker (compare 2 strings), When running this algorithm over generic sets of objects, one might store the counts in a dictionary keyed off the object's hashcode. However, because we're using� String s1 = “DELL” ; String s2 = “LLED”; These two strings are Anagram. Cause both have Same characters. String s3 = “RACE”; String s4 = “ARCER”; These strings are not Anagram. Cause both don’t have Same characters. String3 has 4 characters and String4 has 5 characters.

If you want to use dictionary, check the code below

def anagrams(s1,s2):
    s = s1+s2 #for example s = asd + dsa = asddsa
    l = list(s) #[a,s,d,d,s,a]
    dic = dict(enumerate(l)) # {0: 'a', 1: 's', 2: 'd', 3: 'd', 4: 's', 5: 'a'}
    if len(dic)%2 == 1: #if the two strings are anagrams, the length of the combined strings should be even number
        return False
    else: # now we just compare the two ends of all keys, in the above example, we compare 0 and 5 / 1 and 4 / 2 and 3
        # Notice: the sum of i and the its corresponding party is the len of dic
        i = 0
        while i < len(dic)/2:
            if dic[i] != dic[len(dic)-1-i]:
                return False
                break

            else:
                i += 1
        return True

Alternatively, you can use deque function, which satisfies the same purpose. The simple logic is to add the tow string together and compare the two ends

from collections import deque
def anagrams(s1,s2):
    s = s1+s2 # put them into one string and now we can simply compare if the far left and far right one is the same
    dq = deque(s) # apply deque function to it
    while len(dq) > 1: #if it equals to one
        if dq.popleft() != dq.pop():
            return False
    if len(dq) == 1:
        return False
    else:
        return True

3.4. An Anagram Detection Example — Problem Solving with , 3.7 Dictionaries � 3.8 Summary � 3.9 Key Terms � 3.10 Discussion Questions For the sake of simplicity, we will assume that the two strings in question are of equal Our first solution to the anagram problem will check the lengths of the strings is one simple iteration to compare the n characters after the sorting process. Thus our algorithm checks and compare frequency of each letter in both the strings. The strings to be anagram of each other, their length must be same. Let n1 be the length of first string & n2 be the length of second string. If (n1!=n2) Strings are not anagram. Return. Else Proceed to step2.

If you are just checking for an anagram try using python's Counter object. A single line is all you need.

From anagram with Counter

# Python code to check if two strings are 
# anagram 
from collections import Counter 

def anagram(input1, input2): 

    # Counter() returns a dictionary data 
    # structure which contains characters  
    # of input as key and their frequencies 
    # as it's corresponding value 
    return Counter(input1) == Counter(input2) 

An Anagram Detection Example, One string is an anagram of another if the second is a rearrangement of the letters in For the sake of simplicity, let's assume that the two strings in question use check that the strings are of the same length since our dictionary comparison� Anagram strings : An anagram string is formed by rearranging the characters of a string. For example, triangle and integral are anagram strings. Both strings should have the same set of characters. So, if we want to check if two strings are an anagram or not, we will have to check if both strings contain the same characters or not.

Given a sequence of words, print all anagrams together, With modulo sum, two non-anagram words may have same hash value. Use the index array to print the strings from the original array of strings. Compare individual words using strcmp() to sort In the following program, an array of structure “Word” is used to store both index together using dictionary. Check if two given strings are isomorphic to each other; Check if any permutation of string is a K times repeated string; Check whether two strings are anagrams of each other using unordered_map in C++; Check whether two strings contain same characters in same order; Check if two strings have a common substring

Find anagrams from a dictionar, Find anagrams from a dictionary. eg stop, pots are anagrams I had gone with creating a trie of dictionary and then searched the word in it. How about sorting the two words in lexicographic order and comparing them char-by-char? Given two strings S1 and S2 of size m and n respectively, you need to check whether the two strings are an anagram of each other or not. S1 is an anagram of S2 if the characters of S1 can be rearranged to form S2.

How we can check if an given two strings are anagrams?, The most efficient way is a combination of length comparison and using hash maps (dictionaries in Python). For each string, you create a hash map where the � Method 1: Check if Two Strings Are Anagram using Array This is the simplest of all methods. After getting the strings from the user and we need to first remove all the white space and convert them into the lower case for a non-case sensitive comparison. Now convert them into a character array and sort them alphabetically.

Comments
  • does your string contain spaces? what about special characters? Also I don't understand why you need a dictionaries.
  • Please help us and yourselve and formulate a good question. So far, there is none in you post.