Checking if array is symmetric

Question

public class symm
{


/* 
 * Returns true if array A is symmetric.
 * Returns false otherwise.
 * n is the number of elements A contains.
 *
 * The running time of your algorithm is O(  ).
 * You may add a brief explanation here if you wish.
 */

 public static boolean symmetric( int[] A, int n )
 {
 return symmHelper(A, n, 0);

 }

private static boolean symmHelper(int[] A, int n, int i) {
if(n==1)
    return true;
if((n==2) && (A[i] == A[n-1-i]))
    return true;
if((i == n-1-i) && (A[i] == A[n-1-i] ))
    return true;    

if(A[i] == A[n-1-i] && i < n/2 )
    return symmHelper(A, n, i+1);

return false;
}  


}

Test cases: I passed all the tests ecxept the fitst on I get no whenever I run it, I think the problem is that there are two 2s in the middle. And I'm not really sure about the code, I think it can be simplified. Is the running time o(log n)?

5 8 2 2 8 5 YES

10 7 50 16 20 16 50 7 10 YES

5 8 5 YES

1000 1000 YES

6000 YES

10 7 50 16 20 16 50 7 1000 NO

10 7 50 16 20 16 50 700 10 NO

10 7 50 16 20 16 5000 7 10 NO

10 7 50 16 20 1600 50 7 10 NO

10 7 50 16 1600 50 7 10 NO

Add a print command to your function so you can see what i and n are each time through the recursion. Then you can use that to help walk through your if() statements and see when each one is firing. — DiMono, Mar 08 '13 at 17:06
BTw, This check `if((i == n-1-i) && (A[i] == A[n-1-i] )) return true;` is redudant and can be reduced to `if(i == n-1-i) return true;` You could even combine this case with the `n == 1` case using the `||` operator. — Tuxdude, Mar 08 '13 at 17:10

score 2 · Accepted Answer · answered Mar 08 '13 at 20:30

Complex code makes for more mistakes. Thus, simplify it. Also, look for inequalities rather than equalities; it's easier to check for one mistake than for everything to be correct.

// A = array, n = size of array, i = looking at now
private static boolean symmHelper(int[] A, int n, int i) {

    if (i > n/2)     // If we're more than halfway without returning false yet, we win
        return true;

    else if (A[i] != A[n-1-i])    // If these two don't match, we lose
        return false;

    else    // If neither of those are the case, try again
        return symmHelper(A, n, i+1);
}

If I remember my O() notation right, I think this should be O(n+1). There are other tweaks you can make to this to remove the +1, but it'll make the code run slower overall.

@DiMono Isn't O(n+1)=O(n). Only the highest power of the poly in the O notation matters. — Dheeraj Bhaskar, Apr 20 '13 at 15:02

score 1 · Answer 2 · answered Mar 08 '13 at 17:12

1

if(A[i] == A[n-1-i] && i < n/2 )

That line right there is the problem. Because you're using an even number > 2 of values, when it gets to this line it skips over it because at that point i = n/2, rather than being less than it. So the function skips that and continues on to return false. Change it to this and you should be fine:

if(A[i] == A[n-1-i] && i <= n/2 )

answered Mar 08 '13 at 17:12

DiMono

3,308
2
19
40

Not really. Truth is you should return true if `i >= n/2` – Ivaylo Strandjev Mar 08 '13 at 17:14
@DiMono both are not working! The problem is as you said, because I'm using an even number > 2, I'm not really sure how to deal with that! – Moe Mar 08 '13 at 18:26

score 1 · Answer 3 · answered Dec 20 '16 at 07:37

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner input = new Scanner(System.in);

    int N;
    int i;
    boolean sym = true;

    N=input.nextInt();
    int [] numbers = new int [N];

    for (i=0; i<N; i++){
        numbers[i]= input.nextInt();
    }
    for(i=0;i<N;i++){
        if(numbers[i]!= numbers[N-1-i]){
        sym=false;}
    }

   if(sym==true){
       System.out.println("The array is a symetrical array");
   }
   else{
       System.out.println("The array is NOT a symetrical array");
   }
}

}

score 0 · Answer 4 · answered Mar 08 '13 at 17:13

0

This check is useless:

if((i == n-1-i) && (A[i] == A[n-1-i] ))
    return true;

Of course if the two indices are the same the values there will match.

Also you need to split this if in two:

if(A[i] == A[n-1-i] && i < n/2 )
    return symmHelper(A, n, i+1);

And return true if i >= n/2.

Otherwise what happens is that after i > n/2 (which means you already know your array is symmetrical), you do not go into that if and thus return false, which is wrong.

answered Mar 08 '13 at 17:13

Ivaylo Strandjev

69,226
18
123
176

if I split them, it will return true if i >= n/2, even if the array in symmetric, isn't? – Moe Mar 08 '13 at 18:22

score 0 · Answer 5 · answered Dec 20 '16 at 08:06

public static void main(String[] args) {
    // TODO Auto-generated method stub
    Scanner input = new Scanner(System.in);

    int N;
    int i;


    N=input.nextInt();
    int [] numbers = new int [N];

    for (i=0; i<N; i++){
        numbers[i]= input.nextInt();
    }
    i=0;
    while (i<N/2&& numbers[i] == numbers [N-1-i]){i++;
    }

   if(i==N/2){
       System.out.println("The array is a symetrical array");
   }
   else{
       System.out.println("The array is NOT a symetrical array");
   }
        }

Checking if array is symmetric

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