Anagram

Intuition

gyan_pelo When overexcitedjyada_gyan_mat_pelo

Code

def is_anagram(s1: str, s2: str) -> bool:
    # Remove all whitespace and convert to lowercase
    s1 = s1.replace(" ", "").lower()
    s2 = s2.replace(" ", "").lower()
 
    # Check if the two strings have the same character frequency
    if len(s1) != len(s2):
        return False
    freq = {}
    for char in s1:
        if char not in freq:
            freq[char] = 1
        else:
            freq[char] += 1
    for char in s2:
        if char not in freq:
            return False
        freq[char] -= 1
        if freq[char] < 0:
            return False
    return True
 
# Example usage
s1 = "Tom Marvolo Riddle"
s2 = "I am Lord Voldemort"
print(is_anagram(s1, s2))  # True
 

Attempt ✅

/**
 
* @param {string} s
 
* @param {string} t
 
* @return {boolean}
 
*/
 
var isAnagram = function(s, t) {
 
/**
 
gyan pelo:
 
to check if an element is present dictionary can be used, to check the frequency as value of the key the frequency can be stored.
 
How to check if all letters have got used?
 
Have a count variable, maybe as edge case in the begining itself and return false if the count is not same, if the count is same and all elements of dictionary are present then it should be true only.
 
*/
 
  
 
let sLen = s.length;
 
let tLen = t.length;
 
if (sLen != tLen) {
 
return false;
 
}
 
var sDict = {};
 
for (let i = 0; i < sLen; i++) {
 
let cur = s[i];
 
if (sDict[cur]) {
 
sDict[cur] += 1;
 
} else {
 
sDict[cur] = 1;
 
}
 
}
 
  
 
for (let j = 0; j < sLen; j++) {
 
let cur = t[j];
 
if (sDict[cur]) {
 
sDict[cur] -= 1;
 
} else {
 
return false;
 
}
 
}
 
  
 
return true;
 
};