Given two strings s and t, return true if t is an anagram of s, and false otherwise.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
Approach 1
- Two unordered count map and a final loop to check if counts are same in both
Time: O(n) Space: O(n)
class Solution {
public:
bool isAnagram(string s, string t) {
if(s.size() != t.size()) {
return false;
}
unordered_map<char,int> countS;
unordered_map<char,int> countT;
for(int i = 0; i < s.size(); i++) {
countS[s[i]]++;
countT[t[i]]++;
}
for(const auto& pair: countS) {
if(countS[pair.first] != countT[pair.first]) {
return false;
}
}
return true;
}
};Approach 2
- Create an integer array of size 26
- Loop through string and add 1 to array if it is in first string and -1 for second
- Loop through the array and if it is not zero for any key then it is not anagram
class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length()) {
return false;
}
int[] help = new int[26];
for(int i = 0; i < s.length(); i++) {
help[s.charAt(i) - 'a']++;
help[t.charAt(i) - 'a']--;
}
for(int n: help) {
if(n != 0) {
return false;
}
}
return true;
}
}- The number 256 represents the total number of possible character codes in the **Extended ASCII character set**.
class Solution {
public boolean isAnagram(String s, String t) {
if(s.length() != t.length()) {
return false;
}
int[] count = new int[256];
for(int i = 0; i < s.length(); i++) {
count[s.charAt(i)]++;
count[t.charAt(i)]--;
}
for(int n: count) {
if(n != 0) {
return false;
}
}
return true;
}
}