public class CountNumBinaryStr {

public static long startTime;

public static long endTime;

public static void startAlgo() {

startTime=System.currentTimeMillis();

}

public static long endAlgo() {

endTime=System.currentTimeMillis();

return endTime-startTime;

}

public static int numStrIS(int n) {

int[] zeros=new int[n];

int []ones=new int[n];

//seed

zeros[0]=1;

ones[0]=1;

for(int i=1;i<n;i++) {

zeros[i]=zeros[i-1]+ones[i-1];

ones[i]=zeros[i-1];

}

int ans=zeros[n-1]+ones[n-1];

return ans;

}

private class Binary{

int ones;

int zeros;

int ans;

Binary(int ones,int zeros){

this.ones=ones;

this.zeros=zeros;

this.ans=0;

}

Binary(){}

}

public Binary numStrR(int n) {

if(n==1) {

Binary br=new Binary(1,1);

return br;

}

Binary mr=new Binary();

Binary rr=numStrR(n-1);

mr.zeros=rr.zeros+rr.ones;

mr.ones=rr.zeros;

mr.ans=mr.zeros+mr.ones;

return mr;

}

public static int countStrings(int n, int lastDigit)

{

if (n == 0) {

return 0;

}

// if only one digit is left

if (n == 1) {

return (lastDigit == 1) ? 1: 2;

}

// if last digit is 0, we can have both 0 and 1 at current pos

if (lastDigit == 0) {

return countStrings(n - 1, 0) + countStrings(n - 1, 1);

}

// if last digit is 1, we can have only 0 at current position

else {

return countStrings(n - 1, 0);

}

}

}