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+ class Solution {
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+ public boolean isAnagram (String s , String t ) {
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+ /**
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+ 1. λ¬Έμ μ΄ν΄.
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+ - μλκ·Έλ¨: λ¬Έμμ μμλ§ λ°λμμ λ λμΌν κΈμ
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+ - κ΅¬μ± λ¬Έμμ μ«μλ§ λμΌνλ©΄ μλκ·Έλ¨
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+ 2. νμ΄ λ°©μ
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+ - comapre s.length() with t.length()
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+ - loop over s and t, count each word's character and it's count, and then save it as map(key-value pair structure)
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+ - compare map of s and t
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+ 3. Complexity
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+ - time complexity: O(N)
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+ - space complexity: O(N)
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+ */
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+
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+ if (s .length () != s .length ()) return false ;
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+
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+ Map <Character , Integer > sMap = new HashMap <>();
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+ Map <Character , Integer > tMap = new HashMap <>();
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+
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+ for (char c : s .toCharArray ()) {
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+ sMap .put (c , sMap .getOrDefault (c , 0 ) + 1 );
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+ }
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+ for (char c : t .toCharArray ()) {
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+ tMap .put (c , tMap .getOrDefault (c , 0 ) + 1 );
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+ }
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+
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+ return Objects .equals (sMap , tMap );
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+ }
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+ }
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+
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