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