code logic not working to find non repetitive char and its count

Hi Manisha,

You can write something like this. This will give you the first non-repetitive character inside the arrray. In your case, it is 'h'

public class Countchar {

	public static void main(String[] args) {
		countcharacter("mmaasshaa");

	}

	public static void countcharacter(String word1) {
		//char ch[] = word1.toCharArray();
		//System.out.println("The number of characters in array are " + ch.length);
		// Assuming total number of characters inside Array
	    int NO_OF_CHARS = 256; 

		// Initialize all characters as absent. 
		int arr[] = new int[NO_OF_CHARS]; 
		for (int i = 0;i < NO_OF_CHARS; i++) 
	        arr[i] = -1; 
		
		// After below loop, the value of arr[x] is going to be index x if x appears only once 
		// else the value is going to be either -1 or -2. 
		// This is a way to find out elements inside Arrays using charAt() function
		for (int i = 0; i < word1.length(); i++) {
			if (arr[word1.charAt(i)] == -1) 
	            arr[word1.charAt(i)] = i; 
	        else
	            arr[word1.charAt(i)] = -2; 
	    } 
		
		 int res = Integer.MAX_VALUE; 
		 for (int i = 0; i < NO_OF_CHARS; i++) 
		        // If the character occurs only once and appears before the current result, then update the result 
		        if (arr[i] >= 0) 
		            res = Math.min(res, arr[i]); 
	
System.out.println("The position of non-repetive character inside array is " + res);
	
	 int index = res; 
	    if (index == Integer.MAX_VALUE) 
	        System.out.print("Either all characters are " +  
	                       "repeating or string is empty"); 
	    else
	        System.out.print("First non-repeating character"+  
	                             " is " + word1.charAt(index)); 

	/*for (int j = 0; j < word1.length(); j++) {
				if (ch[i] == ch[j]) {
					count++;
				}
			}
			System.out.println(ch[i] + count);*/
		}
}

If you want something more from the code, Just let me know.