Write a function to check whether two given strings are Permutation of each other or not. A Permutation of a string is another string that contains same characters, only the order of characters can be different. For example, “abcd” and “dabc” are Permutation of each other.

Method 1 (Use Sorting) 
1) Sort both strings 
2) Compare the sorted strings 

// Java program to check whether two strings are
// Permutations of each other
import java.util.*;
class GfG {

/* function to check whether two strings are
Permutation of each other */
static boolean arePermutation(String str1, String str2)
{
	// Get lengths of both strings
	int n1 = str1.length();
	int n2 = str2.length();

	// If length of both strings is not same,
	// then they cannot be Permutation
	if (n1 != n2)
	return false;
	char ch1[] = str1.toCharArray();
	char ch2[] = str2.toCharArray();

	// Sort both strings
	Arrays.sort(ch1);
	Arrays.sort(ch2);

	// Compare sorted strings
	for (int i = 0; i < n1; i++)
	if (ch1[i] != ch2[i])
		return false;

	return true;
}

/* Driver program to test to print printDups*/
public static void main(String[] args)
{
	String str1 = "test";
	String str2 = "ttew";
	if (arePermutation(str1, str2))
	System.out.println("Yes");
	else
	System.out.println("No");

}
}

Output: 

No

Time Complexity: Time complexity of this method depends upon the sorting technique used. In the above implementation, quickSort is used which may be O(n^2) in worst case. If we use a O(nLogn) sorting algorithm like merge sort, then the complexity becomes O(nLogn)

Method 2 (Count characters) 
This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters. 

1) Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0. 
2) Iterate through every character of both strings and increment the count of character in the corresponding count arrays. 
3) Compare count arrays. If both count arrays are same, then return true.

// JAVA program to check if two strings
// are Permutations of each other
import java.io.*;
import java.util.*;

class GFG{
	
	static int NO_OF_CHARS = 256;
	
	/* function to check whether two strings
	are Permutation of each other */
	static boolean arePermutation(char str1[], char str2[])
	{
		// Create 2 count arrays and initialize
		// all values as 0
		int count1[] = new int [NO_OF_CHARS];
		Arrays.fill(count1, 0);
		int count2[] = new int [NO_OF_CHARS];
		Arrays.fill(count2, 0);
		int i;

		// For each character in input strings,
		// increment count in the corresponding
		// count array
		for (i = 0; i

Output: 

Yes

The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are Permutation of each other. Thanks to Ace for suggesting this optimization.

// Java function to check whether two strings are
// Permutations of each other
static boolean arePermutation(char str1[], char str2[])
{
	
	// Create a count array and initialize all
	// values as 0
	int count[] = new int[NO_OF_CHARS];
	int i;
	
	// For each character in input strings,
	// increment count in the corresponding
	// count array
	for (i = 0; str1[i] && str2[i]; i++)
	{
		count[str1[i]]++;
		count[str2[i]]--;
	}
	
	// If both strings are of different length.
	// Removing this condition will make the
	// program fail for strings like "aaca" and
	// "aca"
	if (str1[i] || str2[i])
	return false;
	
	// See if there is any non-zero value in
	// count array
	for (i = 0; i < NO_OF_CHARS; i++)
		if (count[i] != 0)
			return false;
	return true;
}

// This code is contributed by divyesh072019.

If the possible set of characters contains only English alphabets, then we can reduce the size of arrays to 58 and use str[i] – ‘A’ as an index for count arrays because ASCII value of ‘A’ is 65 , ‘B’ is 66, ….. , Z is 90 and ‘a’ is 97 , ‘b’ is 98 , …… , ‘z’ is 122. This will further optimize this method.

Time Complexity: O(n)