Java Program to Count Number of Occurrences of a Given Number using Binary Search

This is a Java Program to find number of occurences of a given number using binary search approach. The time complexity of the following program is O (log n).

Here is the source code of the Java program to find number of occurences of a given number using binary search approach. The Java program is successfully compiled and run on a Windows system. The program output is also shown below.

```
/*
```

 *    Java Program to Find the Number of occurrences of a given Number using Binary Search approach

```
 */
```
```
 
```
```
import java.util.Scanner;
```
```
 
```
```
public class NumberOfOccurences
```
```
{
```

    public static void main(String[] args)

```
    {
```

        Scanner scan = new Scanner(System.in);

        System.out.println("Enter number of elements in sorted array");

```
        int N = scan.nextInt();
```
```
        int[] arr = new int[ N ];
```
```
        /* Accept N elements */
```

        System.out.println("Enter "+ N +" sorted elements");

```
        for (int i = 0; i < N; i++)
```
```
            arr[i] = scan.nextInt();
```

        System.out.println("Enter number to find occurences");

```
        int num = scan.nextInt();
```
```
 
```
```
        int f = occur(arr, num);
```
```
        if (f == -1)
```

            System.out.println("No occurence");

```
        else 
```

            System.out.println("Occurences = "+ f);

```
    }    
```

    public static int occur(int[] arr, int num)

```
    {
```
```
        /* find first index */
```
```
        int l1 = first(arr, num);
```
```
        /* find last index */
```
```
        int l2 = last(arr, num);
```
```
        if (l1 == -1 || l2 == -1)
```
```
            return -1;
```
```
        return l2 - l1 + 1;
```
```
    }
```

    public static int first(int[] arr, int num)

```
    {
```
```
        if (arr[0] == num)
```
```
            return 0;
```

        int start = 0, end = arr.length - 1;

```
        int mid = (start + end) / 2;
```
```
        int flag = 0;
```

        while (!(arr[mid] == num && arr[mid - 1] < arr[mid]))

```
        {
```
```
            if (start == end)
```
```
            {
```
```
                flag = 1;
```
```
                break;
```
```
            }
```
```
            if (arr[mid] >= num)
```
```
                end = mid - 1;
```
```
            if (arr[mid] < num)
```
```
                start = mid + 1;
```
```
            mid = (start + end) / 2;
```
```
        }
```
```
        if (flag == 0)
```
```
            return mid;
```
```
        return -1;        
```
```
    }
```

    public static int last(int[] arr, int num)

```
    {
```
```
        if (arr[arr.length - 1] == num)
```
```
            return arr.length - 1;
```

        int start = 0, end = arr.length - 1;

```
        int mid = (start + end) / 2;
```
```
        int flag = 0;
```

        while (!(arr[mid] == num && arr[mid + 1] > arr[mid]))

```
        {
```
```
            if (start == end)
```
```
            {
```
```
                flag = 1;
```
```
                break;
```
```
            }
```
```
            if (arr[mid] > num)
```
```
                end = mid - 1;
```
```
            if (arr[mid] <= num)
```
```
                start = mid + 1;
```
```
            mid = (start + end) / 2;
```
```
        }
```
```
        if (flag == 0)
```
```
            return mid;
```
```
        return -1;        
```
```
    }
```
```
}
```

Enter number of elements in sorted array
10
Enter 10 sorted elements
1 1 3 3 3 3 4 4 4 5
Enter number to find occurences
3
Occurences = 4
 
 
Enter number of elements in sorted array
10
Enter 10 sorted elements
1 1 3 3 3 3 4 4 4 5
Enter number to find occurences
5
Occurences = 1

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