JavaAlgorithms/DynamicProgramming/CountNumBinaryStrings

package DynamicProgramming;
/*
* here is a important algo in this we have to count  
* maximum no. of different binary strings which doesnot have 
* consectuive 1s 



Test Case:

int n=30;
		
		startAlgo();
		System.out.println(numStrIS(n));
		System.out.println(endAlgo()+"ms");

		startAlgo();
		CountNumBinaryStr out=new CountNumBinaryStr();
		System.out.println(out.numStrR(n).ans);
		System.out.println(endAlgo()+"ms");
		
		startAlgo();
		System.out.println(countStrings(n,0));
		System.out.println(endAlgo()+"ms");
		
		
		
*/
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);
		}
	}


}