CODEWITHSUNDEEP
javaarrays
Arpit Patel
Arpit Patel

Reputation: 8047

Find the most common value from a 2D integer array

I am looking for the most common value from the 2D array and how many times it occurs. I tried this solution but it's not working. I tried searching but not able to find a proper example. Please help me solve this problem.

Here is my code:

private void commonElevation(int[][] data) {
    for (int i = 0; i < data.length; i++) {
        for (int j = 0; j < data[i].length; j++) {
            if (i + 1 < data.length) {
                if (data[i][j] == data[i + 1][j]) {
                    System.out.println(data[i][j] + " = " + data[i + 1][j]);
                }
            }
        }
    }
}

Upvotes: 1

Views: 534

Answers (2)

Dani Mesejo
Dani Mesejo

Reputation: 61910

You could use the Stream API:

int[][] data = {{1, 2, 3}, {2, 2, 2}, {4, 5, 6}};

Map<Integer, Long> counts = Arrays.stream(data).flatMapToInt(Arrays::stream).boxed()
        .collect(groupingBy(Function.identity(), counting()));

Optional<Map.Entry<Integer, Long>> max = counts.entrySet().stream().max(Comparator.comparing(Map.Entry::getValue));

max.ifPresent(System.out::println);

Output

2=4

Given the new constraints a brute-force approach will work:

// find the maximum
int value = 0, max = Integer.MIN_VALUE;
for (int i = 0; i < data.length; i++) {
    for (int j = 0; j < data[i].length; j++) {

        // search for counts
        int currentCount = 0;
        for (int k = 0; k < data.length; k++) {
            for (int l = 0; l < data[k].length; l++) {
                if(data[k][l] == data[i][j]) {
                    currentCount++;
                }
            }
        }

        if (currentCount > max) {
            value = data[i][j];
            max = currentCount;
        }
    }
}

System.out.println(value + "=" + max);

Output

2=4

Basically iter over all values and count the appearances of each of those values. This approach (brute-force) is very inefficient.

Upvotes: 2

Tim Biegeleisen
Tim Biegeleisen

Reputation: 521249

One possibility would be to use a hash map to keep track of the values along with the number of times each occurs:

private void commonElevation(int[][] data) {
    Map<Integer, Integer> counts = new HashMap<>();

    for (int i=0; i < data.length; i++) {
        for (int j=0; j < data[i].length; j++) {
            int count = counts.get(data[i][j]) == null ? 0 : counts.get(data[i][j]);
            counts.put(data[i][j], ++count);
        }
    }

    int frequent = Integer.MIN_VALUE;
    for (Integer value : counts.values()) {
        if (value > frequent) frequent = value;
    }

    System.out.println("most frequent value is: " + frequent);
}

Upvotes: 0

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