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Anagram.java
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102 lines (86 loc) · 2.04 KB
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import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
/*
* 1. Write a function boolean isAnagram(String lhs, String rhs).
* An anagram is direct word switch or word play, the result of rearranging the letters of a word
* or phrase to produce a new word or phrase, using all the original letters exactly once;
* for example, the word ‘anagram' can be rearranged into ‘nagaram’.
*/
/**
* @author Manjeer
*
*/
public class Anagram {
public static void main(String[] args) {
System.out.println(new Anagram().isAnagramBySort("abcdefg", "fgcdeab"));
System.out.println(new Anagram().isAnagramByHash("abcdefgz", "abcdefgz"));
}
/**
* @param s1
* @param s2
* @return
*
* Using Sort O(NlogN)
*/
public boolean isAnagramBySort(String s1, String s2) {
if (s1 == null && s2 == null) {
return true;
} else if (s1 == null || s2 == null) {
return false;
} else if (s1.equals(s2)) {
return true;
}
char[] s1a = s1.toCharArray();
char[] s2a = s2.toCharArray();
Arrays.sort(s1a);
Arrays.sort(s2a);
return Arrays.equals(s1a, s2a);
}
/**
* @param s1
* @param s2
* @return
*
* Using HashMap O(n)
*/
public boolean isAnagramByHash(String s1, String s2) {
if (s1 == null && s2 == null) {
return true;
} else if (s1 == null || s2 == null) {
return false;
} else if (s1.equals(s2)) {
return true;
}
Map<Character, Integer> m1 = new HashMap<>();
Map<Character, Integer> m2 = new HashMap<>();
for (int i = 0; i < s1.length(); i++) {
char charAt = s1.charAt(i);
Integer count = m1.get(charAt);
if (count == null) {
m1.put(charAt, 1);
} else {
m1.put(charAt, count + 1);
}
}
for (int i = 0; i < s2.length(); i++) {
char charAt = s2.charAt(i);
Integer count = m2.get(charAt);
if (count == null) {
m2.put(charAt, 1);
} else {
m2.put(charAt, count + 1);
}
}
for (char c : m1.keySet()) {
if (m1.get(c) != m2.get(c)) {
return false;
}
m2.remove(c);
}
if (m2.isEmpty()) {
return true;
}
return false;
}
}