# Check if the given number K is enough to reach the end of an array

Given an array arr[] of n elements and a number K. The task is to determine if it is possible to reach the end of the array by doing the below operations:

Traverse the given array and,

• If any element is found to be non-prime then decrement the value of K by 1.
• If any element is prime then refill the value of K to its initial value.

If it is possible to reach the end of array with (K > 0), then print YES otherwise print NO.

Examples:

```Input : K = 2   ,  arr[]={ 6, 3, 4, 5, 6} ;
Output : Yes
Explanation :
1- arr[0] is not prime, so K = K-1 = 1
2- arr[1] is prime so K will be refilled to its
initial value. Therefore,  K = 2.
3- arr[2] is not prime.
Therefore,  K = 2-1 = 1
4- arr[3] is prime so K will be refilled to its
initial value. Therefore,  K = 2.
5- arr[4] is not prime.
Therefore,  K = 2-1 = 1
6- Since the end of the array is reached with K>=0
So output is YES

Input :  n=6 , k=3;
arr[]={ 1, 2, 10, 4, 6, 8};
Output : No
```

# Recommended: Please solve it on “PRACTICE “first , before moving on to the solution.

Simple Approach:

• Traverse each element of the array and Check if the value of current element is prime or not.
• If it is Prime then refill the power of K else decrement by 1.
• If it is possible to reach the end of the array with (K > 0) then print “YES” otherwise “NO”.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if it is possible ` `// to reach the end of the array ` `#include ` `using` `namespace` `std; ` ` `  `// Function To check number is prime or not ` `bool` `is_Prime( ``int` `num )  ` `{ ` `    ``// because 1 is not prime  ` `    ``if``(num == 1) ` `    ``return` `false``; ` `     `  `    ``for``(``int` `i=2 ; i*i <= num ; i++ ) ` `    ``{ ` `        ``if``( num % i == 0 ) ` `        ``return` `false``; ` `    ``} ` `     `  `return` `true``;  ` `} ` ` `  `// Function to check whether it is possible ` `// to reach the end of the array or not      ` `bool` `isReachable( ``int` `arr[] , ``int` `n , ``int` `k) ` `{    ` `    ``// store initial value of K ` `    ``int` `x = k ; ` `             `  `    ``for``(``int` `i=0 ; i < n ; i++ ) ` `    ``{ ` `        ``// Call is_prime function to ` `        ``// check if a number is prime.  ` `        ``if``( is_Prime(arr[i]) ) ` `        ``{ ` `            ``// Refill K to initial value ` `            ``k = x; ` `        ``}                  ` `        ``else` `        ``{ ` `            ``// Decrement k by 1 ` `            ``k-- ;                  ` `        ``} ` ` `  `     `  `        ``if``( k <= 0 && i < (n-1) && (!is_Prime(arr[i+1])) ) ` `            ``return` `false` `; ` `    ``} ` `             `  `    ``return` `true` `; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 3, 4, 5, 6}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]) ; ` `    ``int` `k = 2 ; ` ` `  `         `  `    ``isReachable( arr , n , k ) ? cout << ``"Yes"` `<< endl :  ` `                                    ``cout << ``"No"` `<< endl ;  ` `         `  `    ``return` `0 ; ` `}  `

## Java

 `// Java program to check if  ` `// it is possible to reach ` `// the end of the array ` `import` `java.io.*; ` `import` `java.util.*; ` `import` `java.lang.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function To check  ` `// number is prime or not ` `static` `boolean` `is_Prime(``int` `num)  ` `{ ` `    ``// because 1 is not prime  ` `    ``if``(num == ``1``) ` `    ``return` `false``; ` `     `  `    ``for``(``int` `i = ``2` `; ` `            ``i * i <= num ; i++ ) ` `    ``{ ` `        ``if``(num % i == ``0``) ` `        ``return` `false``; ` `    ``} ` `     `  `return` `true``;  ` `} ` ` `  `// Function to check whether  ` `// it is possible to reach ` `// the end of the array or not      ` `static` `boolean` `isReachable(``int` `arr[] ,  ` `                           ``int` `n , ``int` `k) ` `{  ` `    ``// store initial value of K ` `    ``int` `x = k ; ` `             `  `    ``for``(``int` `i = ``0` `; i < n ; i++ ) ` `    ``{ ` `        ``// Call is_prime function to ` `        ``// check if a number is prime.  ` `        ``if``(is_Prime(arr[i])) ` `        ``{ ` `            ``// Refill K to  ` `            ``// initial value ` `            ``k = x; ` `        ``}                  ` `        ``else` `        ``{ ` `            ``// Decrement k by 1 ` `            ``k-- ;                  ` `        ``} ` ` `  `     `  `        ``if``(k <= ``0` `&& i < (n - ``1``) &&  ` `          ``(is_Prime(arr[i + ``1``]) != ``true``)) ` `            ``return` `false` `; ` `    ``} ` `             `  `    ``return` `true` `; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = ``new` `int``[]{ ``6``, ``3``, ``4``, ``5``, ``6``}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``2` `; ` ` `  `    ``if``(isReachable(arr, n, k) == ``true``) ` `        ``System.out.print(``"Yes"` `+ ``"\n"``); ` `     ``else` `        ``System.out.print(``"No"` `+ ``"\n"``);  ` `}  ` `} `

## Python3

 `# Python 3 program to check if it is  ` `# possible to reach the end of the array ` `from` `math ``import` `sqrt ` ` `  `# Function To check number is prime or not ` `def` `is_Prime(num): ` `     `  `    ``# because 1 is not prime  ` `    ``if``(num ``=``=` `1``): ` `        ``return` `False` `    ``k ``=` `int``(sqrt(num)) ``+` `1` ` `  `    ``for` `i ``in` `range``(``2``, k, ``1``): ` `        ``if``(num ``%` `i ``=``=` `0``): ` `            ``return` `False` `             `  `    ``return` `True` ` `  `# Function to check whether it is possible ` `# to reach the end of the array or not  ` `def` `isReachable(arr, n , k): ` `     `  `    ``# store initial value of K ` `    ``x ``=` `k ` `             `  `    ``for` `i ``in` `range``(``0``, n, ``1``): ` `         `  `        ``# Call is_prime function to ` `        ``# check if a number is prime.  ` `        ``if``( is_Prime(arr[i])): ` `             `  `            ``# Refill K to initial value ` `            ``k ``=` `x      ` `        ``else``: ` `             `  `            ``# Decrement k by 1 ` `            ``k ``-``=` `1`         `     `  `        ``if``(k <``=` `0` `and` `i < (n ``-` `1``) ``and`  `          ``(is_Prime(arr[i ``+` `1``])) ``=``=` `False``): ` `            ``return` `False` `             `  `    ``return` `True` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[``6``, ``3``, ``4``, ``5``, ``6``] ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `2` ` `  `    ``if` `(isReachable( arr , n , k )): ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) ` ` `  `# This code is contributed by ` `# Sahil_Shelangia `

## C#

 `// C# program to check if  ` `// it is possible to reach ` `// the end of the array ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function To check  ` `// number is prime or not ` `static` `bool` `is_Prime(``int` `num)  ` `{ ` `    ``// because 1 is not prime  ` `    ``if``(num == 1) ` `    ``return` `false``; ` `     `  `    ``for``(``int` `i = 2 ; ` `            ``i * i <= num ; i++ ) ` `    ``{ ` `        ``if``(num % i == 0) ` `        ``return` `false``; ` `    ``} ` `     `  `return` `true``;  ` `} ` ` `  `// Function to check whether  ` `// it is possible to reach ` `// the end of the array or not  ` `static` `bool` `isReachable(``int` `[]arr ,  ` `                        ``int` `n , ``int` `k) ` `{  ` `    ``// store initial  ` `    ``// value of K ` `    ``int` `x = k ; ` `             `  `    ``for``(``int` `i = 0 ; i < n ; i++ ) ` `    ``{ ` `        ``// Call is_prime function  ` `        ``// to check if a number  ` `        ``// is prime.  ` `        ``if``(is_Prime(arr[i])) ` `        ``{ ` `            ``// Refill K to  ` `            ``// initial value ` `            ``k = x; ` `        ``}              ` `        ``else` `        ``{ ` `            ``// Decrement k by 1 ` `            ``k-- ;              ` `        ``} ` ` `  `     `  `        ``if``(k <= 0 && i < (n - 1) &&  ` `        ``(is_Prime(arr[i + 1]) != ``true``)) ` `            ``return` `false` `; ` `    ``} ` `             `  `    ``return` `true` `; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = ``new` `int``[]{ 6, 3, 4, 5, 6}; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 2 ; ` ` `  `    ``if``(isReachable(arr, n, k) == ``true``) ` `        ``Console.WriteLine(``"Yes"` `+ ``"\n"``); ` `    ``else` `        ``Console.WriteLine(``"No"` `+ ``"\n"``);  ` `}  ` `} ` ` `  `// This code is contributed by vt_m `

## PHP

 ` `

Output:

```Yes
```

Time complexity: O(N(sqrt N))

Efficient Approach: The above approach can be optimized by using Sieve of Eratosthenes to check if a number is prime or not.

Below is the implementation of the efficient approach:

## C++

 `// CPP program to check if it is possible ` `// to reach the end of the array ` `#include ` `#define MAX 1000000 ` `using` `namespace` `std; ` ` `  `// Function for Sieve of Eratosthenes ` `void` `SieveOfEratosthenes( ``int` `sieve[], ``int` `max )  ` `{    ` `    ``for``(``int` `i=0; i

## Java

 `// Java program to check if it is possible ` `// to reach the end of the array ` `class` `GFG  ` `{ ` ` `  `    ``static` `int` `MAX = ``1000000``; ` ` `  `    ``// Function for Sieve of Eratosthenes ` `    ``static` `void` `SieveOfEratosthenes(``int` `sieve[], ``int` `max)  ` `    ``{ ` `        ``for` `(``int` `i = ``0``; i < max; i++)  ` `        ``{ ` `            ``sieve[i] = ``1``; ` `        ``} ` ` `  `        ``for` `(``int` `i = ``2``; i * i < max; i++)  ` `        ``{ ` `            ``if` `(sieve[i] == ``1``) ` `            ``{ ` `                ``for` `(``int` `j = i * ``2``; j < max; j += i)  ` `                ``{ ` `                    ``sieve[j] = ``0``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Function to check if it is possible to ` `    ``// reach end of the array      ` `    ``static` `boolean` `isReachable(``int` `arr[], ``int` `n,  ` `                                ``int` `sieve[], ``int` `k)  ` `    ``{ ` `        ``// store initial value of K ` `        ``int` `x = k; ` ` `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``if` `(sieve[arr[i]] == ``1``)  ` `            ``{ ` `                ``// Refill K to initial value ` `                ``k = x; ` `            ``}  ` `            ``else` `            ``{ ` `                ``// Decrement k by 1 ` `                ``k -= ``1``; ` `            ``} ` ` `  `            ``if` `((k <= ``0``) && (i < (n - ``1``)) ` `                    ``&& (sieve[arr[i + ``1``]] == ``0``))  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `arr[] = {``6``, ``3``, ``4``, ``5``, ``6``}; ` `        ``int``[] sieve = ``new` `int``[MAX]; ` `        ``int` `n = arr.length; ` `        ``int` `k = ``2``; ` ` `  `        ``SieveOfEratosthenes(sieve, MAX); ` ` `  `        ``if` `(isReachable(arr, n, sieve, k))  ` `        ``{ ` `            ``System.out.println(``"Yes"``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``System.out.println(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## C#

 `// C# program to check if it is possible ` `// to reach the end of the array ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `MAX = 1000000; ` ` `  `    ``// Function for Sieve of Eratosthenes ` `    ``static` `void` `SieveOfEratosthenes(``int` `[]sieve, ``int` `max)  ` `    ``{ ` `        ``for` `(``int` `i = 0; i < max; i++)  ` `        ``{ ` `            ``sieve[i] = 1; ` `        ``} ` ` `  `        ``for` `(``int` `i = 2; i * i < max; i++)  ` `        ``{ ` `            ``if` `(sieve[i] == 1) ` `            ``{ ` `                ``for` `(``int` `j = i * 2; j < max; j += i)  ` `                ``{ ` `                    ``sieve[j] = 0; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Function to check if it is possible to ` `    ``// reach end of the array  ` `    ``static` `bool` `isReachable(``int` `[]arr, ``int` `n,  ` `                                ``int` `[]sieve, ``int` `k)  ` `    ``{ ` `        ``// store initial value of K ` `        ``int` `x = k; ` ` `  `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``if` `(sieve[arr[i]] == 1)  ` `            ``{ ` `                ``// Refill K to initial value ` `                ``k = x; ` `            ``}  ` `            ``else` `            ``{ ` `                ``// Decrement k by 1 ` `                ``k -= 1; ` `            ``} ` ` `  `            ``if` `((k <= 0) && (i < (n - 1)) ` `                    ``&& (sieve[arr[i + 1]] == 0))  ` `            ``{ ` `                ``return` `false``; ` `            ``} ` `        ``} ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `[]arr = {6, 3, 4, 5, 6}; ` `        ``int``[] sieve = ``new` `int``[MAX]; ` `        ``int` `n = arr.Length; ` `        ``int` `k = 2; ` ` `  `        ``SieveOfEratosthenes(sieve, MAX); ` ` `  `        ``if` `(isReachable(arr, n, sieve, k))  ` `        ``{ ` `            ``Console.WriteLine(``"Yes"``); ` `        ``}  ` `        ``else` `        ``{ ` `            ``Console.WriteLine(``"No"``); ` `        ``} ` `    ``} ` `} ` ` `  `/* This code contributed by ajit. */`

Output:

```Yes
```

Time complexity: O(?Max * loglog(Max)) + O(n)

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