Remove all 1 s from the adjacent left of 0 s in a Binary Array | Java implementation of the above approach ; Function to find the maximum number of 1 's before 0 ; Traverse the array ; If value is 1 ; If consecutive 1 followed by 0 , then update the maxCnt ; Print the maximum consecutive 1 's  followed by 0 ; Driver Code ; Function Call ; Function Call
class GFG { static void noOfMoves ( int arr [ ] , int n ) { int cnt = 0 ; int maxCnt = 0 ; for ( int i = 0 ; i < n ; i ++ ) { if ( arr [ i ] == 1 ) { cnt ++ ; } else { if ( cnt != 0 ) { maxCnt = Math . max ( maxCnt , cnt ) ; cnt = 0 ; } } } System . out . print ( maxCnt + "NEW_LINE"); } public static void main ( String [ ] args ) { int arr [ ] = { 0 , 1 , 1 , 1 , 1 , 0 , 0 , 1 , 1 , 0 , 0 , 1 } ; int N = arr . length ; noOfMoves ( arr , N ) ; int arr1 [ ] = { 1 , 0 , 1 , 0 , 1 , 0 , 1 , 0 } ; N = arr1 . length ; noOfMoves ( arr1 , N ) ; } }