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 { //your code here 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); } }

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