Create CountNumBinaryStrings

Author: Ritik2604

DynamicProgramming/CountNumBinaryStrings

@@ -0,0 +1,69 @@
+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);
+		}
+	}
+
+
+}