added dp solution of no. of binary string without consecutive 1

CountNumberOfBinaryWithoutConsecutive1s.java

@@ -0,0 +1,46 @@
+/**
+ * Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1’s.
+ *
+ * Input: N = 3
+ * Output: 5
+ * The 5 strings are 000, 001, 010, 100, 101
+ *
+ * It is really a straight up fibonacci series with values
+ * 1,2,3,5,8,13....
+ * Look how we assign a[i] value of a[i-1] + b[i-1] and then b[i] becomes a[i]
+ */
+public class CountNumberOfBinaryWithoutConsecutive1s {
+
+    public int count(int n){
+        int a[] = new int[n];
+        int b[] = new int[n];
+
+        a[0] = 1;
+        b[0] = 1;
+
+        for(int i=1; i < n; i++){
+            a[i] = a[i-1] + b[i-1];
+            b[i] = a[i-1];
+        }
+
+        return a[n-1] + b[n-1];
+    }
+
+    public int countSimple(int n){
+        int a = 1;
+        int b = 1;
+
+        for(int i=1; i < n; i++){
+            int tmp = a;
+            a = a + b;
+            b = tmp;
+        }
+
+        return a + b;
+    }
+
+    public static void main(String args[]){
+        CountNumberOfBinaryWithoutConsecutive1s cnb = new CountNumberOfBinaryWithoutConsecutive1s();
+        System.out.println(cnb.count(3));
+    }
+}