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HomeSoftware EngineeringDiscover Depend of Most Frequent Merchandise in an Array in Java

# Discover Depend of Most Frequent Merchandise in an Array in Java

## The problem#

Full the operate to search out the rely of probably the most frequent merchandise of an array. You possibly can assume that enter is an array of integers. For an empty array return “

Instance

``````enter array: [3, -1, -1, -1, 2, 3, -1, 3, -1, 2, 4, 9, 3]
ouptut: 5
``````

## The answer in Java code#

Possibility 1:

``````import java.util.*;

public static int mostFrequentItemCount(int[] assortment) {
if (assortment.size==0) return 0;
Map<Integer, Integer> m = new HashMap<>();
for (Integer i : assortment)
m.merge(i, 1, Integer::sum);
return Collections.max(m.values());
}
}
``````

Possibility 2:

``````import java.util.stream.Stream;
import java.util.Arrays;
import java.util.stream.Collectors;
import static java.util.operate.Operate.id;
import static java.util.stream.Collectors.counting;

public static int mostFrequentItemCount(int[] assortment) {
return (int)Arrays.stream(assortment).mapToObj(i -> i)
.gather(Collectors.groupingBy(id(), counting()))
.values().stream().mapToLong(c -> (lengthy)c).max().orElse(0);
}
}
``````

Possibility 3:

``````import java.util.Collections;
import java.util.ArrayList;

public static int mostFrequentItemCount(int[] assortment) {
ArrayList<Integer> arr = new ArrayList<>();
for (Integer i : assortment)
int most = 0;
for (Integer i: arr) {
int b = Collections.frequency(arr,i);
if (b>most)
most=b;
}
return most;
}
}
``````

## Take a look at instances to validate our answer#

``````import org.junit.Take a look at;
import static org.junit.Assert.assertEquals;

public class FrequentExampleTests {
@Take a look at
public void assessments() {
assertEquals(2, Answer.mostFrequentItemCount(new int[] {3, -1, -1}));
assertEquals(5, Answer.mostFrequentItemCount(new int[] {3, -1, -1, -1, 2, 3, -1, 3, -1, 2, 4, 9, 3}));
}
}
``````
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