How to find if two Strings are anagrams in Java?
The answer is that you can determine whether two strings are anagrams through sorting or character frequency statistics. First, spaces are removed and converted to lower case. If the lengths are different, false is returned directly; the sorting method sorts the character arrays and compares them, with a time complexity of O(n log n); the frequency statistics method uses an array to record the number of occurrences of each character, traverses the two strings to increase or decrease the frequency, and finally checks whether the frequency arrays are all zero, with a time complexity of O(n).
To check if two strings are anagrams in Java, you need to determine if they contain the same characters in the same quantity, regardless of order. Here's how you can do it effectively.
Understanding Anagrams
Two strings are anagrams if:
- They have the same length
- They contain identical characters with the same frequency
- Case and spaces may or may not be ignored, depending on requirements
Method 1: Using Sorting
A simple approach is to sort the characters of both strings and compare them.
public static boolean areAnagrams(String str1, String str2) {
// Remove spaces and convert to lowercase
str1 = str1.replaceAll("\\s", "").toLowerCase();
str2 = str2.replaceAll("\\s", "").toLowerCase();
// Convert to char arrays and sort
char[] chars1 = str1.toCharArray();
char[] chars2 = str2.toCharArray();
java.util.Arrays.sort(chars1);
java.util.Arrays.sort(chars2);
//Compare sorted arrays
return java.util.Arrays.equals(chars1, chars2);
}
This method works well and is easy to understand. Time complexity is O(n log n) due to sorting.
Method 2: Using Character Frequency Count
You can count character frequencies using a HashMap or an array (for ASCII characters).
public static boolean areAnagrams(String str1, String str2) {
str1 = str1.replaceAll("\\s", "").toLowerCase();
str2 = str2.replaceAll("\\s", "").toLowerCase();
if (str1.length() != str2.length()) {
return false;
}
int[] freq = new int[256]; // For all ASCII characters
for (int i = 0; i This approach runs in O(n) time and is more efficient than sorting for large inputs.
Testing the Methods
Here's a quick example to test:
System.out.println(areAnagrams("listen", "silent")); // true
System.out.println(areAnagrams("hello", "bello")); // false
Make sure to handle edge cases like null inputs or empty strings if needed.
Basically just pick the sorting method for simplicity or frequency counting for better performance. Both work reliably for most use cases.
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