| class betterSolution { | |
| public: | |
| bool isAnagram(string s, string t) { | |
| unordered_map <char,int> trackMap; | |
| if (t.size() != s.size()) { | |
| return false; // If lengths are not equal, impossible to be anagram | |
| } | |
| // for s, increment counts, for t decrement | |
| /* | |
| for (u_int i=0 ; i < s.size(); i++) causes a slower runtime, why? | |
| - Maybe conversion from u_int to int for operator[] location? | |
| */ | |
| for (int i = 0; i < s.size(); i++) { | |
| trackMap[s[i]]++; | |
| trackMap[t[i]]--; | |
| } | |
| for (auto kvPairs: trackMap) { | |
| if (kvPairs.second != 0) { | |
| return false; // characters did not match up exactly. | |
| } | |
| } | |
| return true; // All counts in hashmap are zero, meaning same letters and counts in both s and t | |
| } | |
| }; |
| class Solution { | |
| public: | |
| bool isAnagram(string s, string t) { | |
| unordered_map<char,int> s_map,t_map; | |
| for (int i = 0; i < s.size(); i++) { | |
| if (s_map.count(s[i])) { | |
| s_map.at(s[i])++; // increment number of times weve seen this character | |
| } | |
| else {// add first instance of char to map | |
| s_map.insert({s[i],1}); | |
| } | |
| } | |
| for (int i = 0; i < t.size(); i++) { | |
| if (t_map.count(t[i])) { | |
| t_map.at(t[i])++; // increment number of times weve seen this character | |
| } | |
| else {// add first instance of char to map | |
| t_map.insert({t[i],1}); | |
| } | |
| } | |
| return t_map == s_map; | |
| } | |
| }; |
Given two strings s and t, return true if the two strings are anagrams of each other, otherwise return false.
An anagram is a string that contains the exact same characters as another string, but the order of the characters can be different.
Example 1:
Input: s = "anagram", t = "nagaram"
Output: trueExample 2:
Input: s = "rat", t = "car"
Output: falseConstraints:
1 <= s.length, t.length <= 5 * 104
s and t consist of lowercase English letters.