Given a string, determine if a permutation of the string could form a palindrome

I am solving this problem on Leetcode and I have seen a solution, but I am having trouble understanding some syntax used in the solution. If anyone can explain me, I would really appreciate it.

Problem:
<https://leetcode.com/problems/palindrome-permutation/>

Solution:

class Solution {
    public boolean canPermutePalindrome(String s) {
        
        int count=0;
        int[]arr=new int[128];
        for(int i=0;i<s.length();i++){
            arr[s.charAt(i)]++;
       
        if(arr[s.charAt(i)]%2==0)
            count--;
        
        else 
            count++;
    
        }
        return count <=1;
    }
}

My questions:

1. I have seen other similar problems initialize an array of size 128 or 256. Why do we use 128 or 256 for array size in problems like this? is there any particular reason?

2. arr[s.charAt(i)]++; << I don't understand this syntax, can anyone explain me what exactly is going in here? If anyone can refer me to a link for understanding syntax like this, it would be nice.

1. Why do we use 128 or 256 for array size in problems like this? This is used for character instance counting. Characters are expected to be in the range of 0-127 or 0-255
2. arr[s.charAt(i)]++; << i don't understand this syntax, can anyone explain me what exactly is going in here - this is incrementing the count for character each character in the string (s.charAt(i))

For a properly formed palindrome permutation, there can be no more than one character that has an odd number of instances, as the only place to put a character with an odd number of instances is at the midpoint of the palindrome.

1. The array length is taken as 128 or 256 on basis of whether the character set is ascii or extended ascii. please refer http://www.asciitable.com/
2. arr[s.charAt(i)]++ means a character increment. In Java, char is a numeric type. When you add 1 to a char, you get to the next unicode code point.
   arr[s.charAt(i)]++ translates to :
   arr[s.charAt(i)] = arr[s.charAt(i)] +1;

Try this.

• it does a frequency count of the characters.
• Then it counts the number of characters that have an odd frequency.
• if you have more than one set of characters that occur an odd number of times it
  can't be rearranged into a palindrome.

public static boolean isConvertable(String a) {
	long count = Arrays.stream(a.split(""))
			.collect(Collectors.groupingBy(str -> str,
					Collectors.counting()))
			.values().stream().filter(val -> (val % 2) == 1)
			.count();
	
    return count <= 1;
}

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