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 = 1Input: 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 primebool 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 indicesvoid 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 codeint 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 approachclass 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 approachusing 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 primefunction 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 indicesfunction 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 codelet 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
Time Complexity: O(n2)
Auxiliary Space: O(1)
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