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add solution: Valid Anagram
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valid-anagram/Gotprgmer.java

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// 각 s와 t의 알파벳 수를 카운팅 시켜서 마지막에 카운팅이 모두 같으면 true
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// 그렇지 않으면 false;
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// 시간복잡도 O(N) => 1중 for문
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// 공간복잡도 O(N) => N개 알파벳이 들어있는 Map 2개
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class SolutionGotprgmer {
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public boolean isAnagram(String s, String t) {
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Map<Character,Integer> sMap = new HashMap<>();
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Map<Character,Integer> tMap = new HashMap<>();
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if(s.length() != t.length()){
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return false;
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}
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for(int i=0;i<s.length();i++){
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char sChar = s.charAt(i);
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char tChar = t.charAt(i);
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sMap.put(sChar,sMap.getOrDefault(sChar,0)+1);
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tMap.put(tChar,tMap.getOrDefault(tChar,0)+1);
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}
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boolean flag = true;
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for(char c:sMap.keySet()){
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if(!sMap.getOrDefault(c,0).equals(tMap.getOrDefault(c,0))){
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flag = false;
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System.out.println(c);
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break;
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}
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}
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return flag;
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}
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}

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