Java: Is there a way to count occurrence of int in 2d array?

Question

I want to know if there is a faster way than mine, to count how many zeros are inside a 2d array. Here is how I am doing it now.

static int zeroInside = 0;
static int[][] board = new int[][]{
          {5,0,0,0,1,0,0,7,0},
          {1,0,0,0,8,0,3,0,0},
          {0,6,0,2,3,0,0,5,0},
          {2,5,0,7,0,0,6,3,0},
          {9,0,0,3,0,1,0,0,5},
          {0,3,1,0,0,8,0,2,7},
          {0,2,0,0,9,5,0,1,0},
          {0,0,9,0,6,0,0,0,2},
          {0,8,0,0,7,0,0,0,6}}; 
static void initialZeroCounting(){
     for (int outer = 0; outer < board.length;outer++){
         for(int inner = 0; inner < board[outer].length; inner++){
             if (board[outer][inner] ==0){
                 zeroInside++;
             }
         }
     }

I tried many variation of this snippet but without results.

     List<int[]> list = Arrays.asList(board);
     int count = Collections.frequency(list, 0);

Your solution is O(n^2). You can't possibly do better in computational complexity level because you have to inspect n^2 elements. Best you can do is perhaps reduce the coefficient, but you're not going to notice a real performance change by doing so unless the array is relatively small, and even then the change will be minimal. — Kon
– Kon, Commented Nov 6, 2013 at 13:55
Rather than making your zeroInside static and your function void, make your function return an int, and use a local variable to do the counting. — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Nov 6, 2013 at 13:56
If you stick with your current approach, you may see a performance improvement by assigning board.length and board[outer].length to variables. At present, the length will be evaluated for each iteration of the loop, unless the compiler optimises it out. — pinoyyid
– pinoyyid, Commented Nov 6, 2013 at 14:15
@Antoniossss I get 0 with the .frequency method for the reasons stated by Marko Topolnik — pinolo
– pinolo, Commented Nov 7, 2013 at 6:13

Marko Topolnik · Accepted Answer · 2013-11-06 14:26:47Z

7

Your approach with

List<int[]> list = Arrays.asList(board);
int count = Collections.frequency(list, 0);

fails because you are searching for 0 in a list of int[]. Naturally, the frequency will always be zero. This would seem as a possible avenue:

for (int[] row : board) count += frequency(asList(row), 0);

but unfortunately it fails because Arrays.asList with a primitive-typed argument results in just a single-element List<int[]>. This could be fixed by introducing your own asList to autobox the ints, but that would not be worth it since then performance would suffer quite a bit.

You could also exchange your int[][] for an Integer[][] (the initializer stays the same). This would result in less performance penalty, but now the memory consumption for the array would increase by a factor of 6—again not worth it in my book.

Conclusion: just live with the nested for-loop.

edited Nov 6, 2013 at 14:26

answered Nov 6, 2013 at 14:00

Marko Topolnik

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2 Comments

Sage Over a year ago

No, it didn't work for me!! i have tried it first time but it seems no frequency is counted!

Marko Topolnik Over a year ago

True, asList with a primitive-typed array will not work: it will result in a single-member List<int[]>.

Adam Arold · Accepted Answer · 2013-11-06 13:54:47Z

5

Since you have to inspect all elements to tell whether it is 0 or not there is no other way then iterating through all of them.

answered Nov 6, 2013 at 13:54

Adam Arold

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Lukas Warsitz · Accepted Answer · 2013-11-06 13:56:29Z

1

IMO you did it in the best way because you cant avoid to iterate through all items of your array

answered Nov 6, 2013 at 13:56

Lukas Warsitz

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pinoyyid · Accepted Answer · 2013-11-06 14:04:58Z

0

You could convert your array to a comma separated String using Arrays.deepToString() and then use commons StringUtils.countMatches to count the occurrences of the letter 'zero'.

I'm not suggesting this is any way more performant that iterating the array, just fewer lines of code.

answered Nov 6, 2013 at 14:04

pinoyyid

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