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

• Difficulty Level : Easy
• Last Updated : 05 May, 2021

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.

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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 :  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 the current element is prime or not.
• If it is Prime then refill the power of K else decrements 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

 ``

## Javascript

 ``
Output:
`Yes`

Time complexity: O(N(sqrt N))

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

Below is the implementation of the efficient approach:

## C++

 `// C++ 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 */`

## Python3

 `# Python3 program to check if it is``# possible to reach the end of the``# array``import` `math` `# Function for Sieve of Eratosthenes``def` `SieveOfEratosthenes(sieve, ``max``):``    ` `    ``for` `i ``in` `range``(``0``, ``max``):``        ``sieve[i] ``=` `1``        ` `    ``sqt ``=` `int``(math.sqrt(``max``))``    ` `    ``for` `i ``in` `range``(``2``, sqt):``        ``if` `(sieve[i] ``=``=` `1``):``            ``for` `j ``in` `range``(i ``*` `2``, ``max``, i):``                ``sieve[j] ``=` `0` `# Function to check if it is possible to``# reach end of the array    ``def` `isReachable(arr, n, sieve, k):``    ` `    ``# store initial value of K``    ``x ``=` `k``    ``for` `i ``in` `range``(``0``, n):``        ``if` `(sieve[arr[i]] !``=` `0``):``            ` `            ``# Refill K to initial value``            ``k ``=` `x       ``        ``else``:``            ` `            ``# Decrement k by 1``            ``k ``-``=` `1``        ``if` `((k <``=` `0``) ``and` `(i < (n ``-` `1``)) ``and``           ``(sieve[arr[i ``+` `1``]] ``=``=` `0``)):``            ``return` `0``            ` `    ``return` `1` `# Driver Code``arr ``=` `[ ``6``, ``3``, ``4``, ``5``, ``6` `]``sieve ``=` `[``0` `for` `x ``in` `range``(``1000000``)]` `n ``=` `len``(arr)``k ``=` `2` `SieveOfEratosthenes(sieve, ``1000000``)` `ch ``=` `isReachable(arr, n, sieve, k)` `if` `(ch):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)``    ` `# This code is contributed by Stream_Cipher`

## 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. */`

## Javascript

 ``
Output:
`Yes`

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

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