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Number of 0s and 1s at prime positions in the given array

  • Last Updated : 31 May, 2021
Geek Week

Given an array arr[] of size N where each element is either 0 or 1. The task is to find the count of 0s and 1s which are at prime indices.

Examples:  

Input: arr[] = {1, 0, 1, 0, 1} 
Output: 
Number of 0s = 1 
Number of 1s = 1

Input: arr[] = {1, 0, 1, 1} 
Output: 
Number of 0s = 0 
Number of 1s = 2  

Approach: Traverse the array and for every 0 encountered update the count of 0s if the current index is prime and update the count of 1s for all the 1s which are at prime indices.



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
 
// Function that returns true
// if n is prime
bool isPrime(int n)
{
    if (n <= 1)
        return false;
         
    // Check from 2 to n
    for(int i = 2; i < n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count
// of 0s and 1s at prime indices
void countPrimePosition(int arr[], int n)
{
     
    // To store the count of 0s and 1s
    int c0 = 0, c1 = 0;
    for(int i = 0; i < n; i++)
    {
         
        // If current 0 is at
        // prime position
        if (arr[i] == 0 && isPrime(i))
            c0++;
             
        // If current 1 is at
        // prime position
        if (arr[i] == 1 && isPrime(i))
            c1++;
    }
    cout << "Number of 0s = " << c0 << endl;
    cout << "Number of 1s = " << c1;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 0, 1, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    countPrimePosition(arr, n);
    return 0;
}
 
// This code is contributed by noob2000

Java




// Java implementation of the approach
class GFG {
 
    // Function that returns true
    // if n is prime
    static boolean isPrime(int n)
    {
        if (n <= 1)
            return false;
 
        // Check from 2 to n
        for (int i = 2; i < n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    // Function to find the count
    // of 0s and 1s at prime indices
    static void countPrimePosition(int arr[])
    {
 
        // To store the count of 0s and 1s
        int c0 = 0, c1 = 0;
        int n = arr.length;
        for (int i = 0; i < n; i++) {
 
            // If current 0 is at
            // prime position
            if (arr[i] == 0 && isPrime(i))
                c0++;
 
            // If current 1 is at
            // prime position
            if (arr[i] == 1 && isPrime(i))
                c1++;
        }
        System.out.println("Number of 0s = " + c0);
        System.out.println("Number of 1s = " + c1);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 0, 1, 0, 1 };
        countPrimePosition(arr);
    }
}

Python3




# Python3 implementation of the approach
 
# Function that returns true
# if n is prime
def isPrime(n) :
     
    if (n <= 1) :
        return False;
 
    # Check from 2 to n
    for i in range(2, n) :
        if (n % i == 0) :
            return False;
         
    return True;
 
# Function to find the count
# of 0s and 1s at prime indices
def countPrimePosition(arr) :
 
    # To store the count of 0s and 1s
    c0 = 0; c1 = 0;
    n = len(arr);
     
    for i in range(n) :
     
        # If current 0 is at
        # prime position
        if (arr[i] == 0 and isPrime(i)) :
            c0 += 1;
     
        # If current 1 is at
        # prime position
        if (arr[i] == 1 and isPrime(i)) :
            c1 += 1;
             
    print("Number of 0s =", c0);
    print("Number of 1s =", c1);
         
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 0, 1, 0, 1 ];
    countPrimePosition(arr);
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
    // Function that returns true
    // if n is prime
    static bool isPrime(int n)
    {
        if (n <= 1)
            return false;
 
        // Check from 2 to n
        for (int i = 2; i < n; i++)
        {
            if ((n % i) == 0)
                return false;
        }
        return true;
    }
 
    // Function to find the count
    // of 0s and 1s at prime indices
    static void countPrimePosition(int []arr)
    {
 
        // To store the count of 0s and 1s
        int c0 = 0, c1 = 0;
        int n = arr.Length;
        for (int i = 0; i < n; i++)
        {
 
            // If current 0 is at
            // prime position
            if ((arr[i] == 0) && (isPrime(i)))
                c0++;
 
            // If current 1 is at
            // prime position
            if ((arr[i] == 1) && (isPrime(i)))
                c1++;
        }
        Console.WriteLine("Number of 0s = " + c0);
        Console.WriteLine("Number of 1s = " + c1);
    }
 
    // Driver code
    static public void Main ()
    {
         
        int[] arr = { 1, 0, 1, 0, 1 };
        countPrimePosition(arr);
    }
}
 
// This code is contributed by ajit.

Javascript




<script>
// Javascript implementation of the approach
 
// Function that returns true
// if n is prime
function isPrime(n) {
    if (n <= 1)
        return false;
 
    // Check from 2 to n
    for (let i = 2; i < n; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count
// of 0s and 1s at prime indices
function countPrimePosition(arr, n) {
 
    // To store the count of 0s and 1s
    let c0 = 0, c1 = 0;
 
    for (let i = 0; i < n; i++) {
 
        // If current 0 is at
        // prime position
        if (arr[i] == 0 && isPrime(i))
            c0++;
 
        // If current 1 is at
        // prime position
        if (arr[i] == 1 && isPrime(i))
            c1++;
    }
    document.write("Number of 0s = " + c0 + "<br>");
    document.write("Number of 1s = " + c1);
}
 
// Driver code
let arr = [1, 0, 1, 0, 1];
 
let n = arr.length;
countPrimePosition(arr, n);
 
// This code is contributed by _saurabh_jaiswal
</script>
Output: 
Number of 0s = 1
Number of 1s = 1

 

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