Given an array of ints, compute recursively the number of times that the value 11 appears in the array. We’ll use the convention of considering only the part of the array that begins at the given index. In this way, a recursive call can pass index+1 to move down the array. The initial call will pass in index as 0.

array11([1, 2, 11], 0) → 1 array11([11, 11], 0) → 2 array11([1, 2, 3, 4], 0) → 0

**Input Format** The first line contains the number n the size of array The second line contains N integers

**Output Format** Print the number of 11 in the array.

** Example ** 5 1 3 11 11 2 ** Output ** 2 **Constraints ** 2 <= N <= 3000 0 <= A[i] <= 5000

## Solution of array11:–

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[] arr=new int[n]; for(int i=0;i<n;i++) { arr[i]=sc.nextInt(); } System.out.print(array11(arr,n)); } public static int array11(int[] arr,int n) { if(n<1) return 0; if(arr[n-1]==11) return 1+array11(arr,n-1); return array11(arr,n-1); } }

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