public class Library {
    
    public static int[] histogram(int a[],int M) {
        int[] newArr = new int[M];
        
        //Fill the new array
        try {
        if(a.length<M)
            System.out.println("Array lenght should be "
                      "bigger than the number");
        else
            
            for(int i = 0; i < a.length; i  ){
                int count = a[i];
                
                newArr[count]   ;
            }
        
        }
        catch (Exception e) {
            // TODO: handle exception
            System.out.println(e.getMessage());
        }
        
        //return the array
        return newArr;
    }
    
    public void printArray(int s[]) {
        int position = 0;
        for (int i = 0; i < s.length; i  ) {
            System.out.print(position " ");
            position  ;
        }
        System.out.println();
        
        for (int i = 0; i < s.length; i  ) {
            System.out.print(s[i] " ");
        }
    }
    
    
    
    public static void main(String[] args) {
        
        Library l1 = new Library();
        
        int J = 5;
        int[] w = {1,2,0,1,2,3};
        
        l1.printArray(histogram(w,J));
        
    }
}

I wrote this and some of the parts I looked at from google and other sources but I couldn't understand the else part in public static int[] histogram

        else
            
            for(int i = 0; i < a.length; i  ){
                int count = a[i];
                
                newArr[count]   ;
            }

How does this newArr[count] ; works can someone explain to me, please

CodePudding user response：

` How does this newArr[count] ; works can someone explain to me, please

Two thins are happening on this one line.

1. We get the reference to value in position count from newArr array.
2. We increment it by 1.

Therefore if you have array newArr = [1,2,3] calling newArr[0] would result in you having the following state of newArr as [2,2,3].

If something is still unclear, respond with a comment to this answer.