## First Element to Occur K Times in java

Given an array of ‘n’ integers. Print the first element that occurs `k` number of times. If there is no element that occurs for at least `k` number of times print -1.

### Input Format

Line 1: contains two integers n and k.

Line 2: contains n-spaced integers denoting elements of the array.

### Output Format

Print a single integer denoting the first element in the array which occurs at least `k` times. If no such element exists, print `-1`.

### Example 1

Input

```7 2
1 7 4 3 4 8 7
```

Output

```4
```

Explanation

As we traverse the array the first number whose frequency becomes greater than or equal to k(2) is 4. Hence, the answer is 4.

### Example 2

Input

```9 4
2 4 1 2 2 19 3
```

Output

```-1
```

Explanation

As no element in array has a frequency greater than or equal to k(4), the output will be -1.

### Constraints

1 <= n <= 10^6

1 <= arr[i] <= 10^6

## Solution of First Element to Occur K Times in java:–

```import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int A[] = new int[n];
int freq[] = new int[1000001];

for(int i=0;i<n;i++){
A[i] = sc.nextInt();
}
for(int i=0;i<n;i++){
freq[A[i]]++;
if(freq[A[i]]==k){
System.out.println(A[i]);
return;
}
}
System.out.println(-1);
}
}```