diff --git a/DynamicProgramming/CountNumBinaryStrings b/DynamicProgramming/CountNumBinaryStrings
deleted file mode 100644
index 3dd0bf7de..000000000
--- a/DynamicProgramming/CountNumBinaryStrings
+++ /dev/null
@@ -1,97 +0,0 @@
-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);
-		}
-	}
-
-
-}