diff --git a/special binary string.cpp b/special binary string.cpp
new file mode 100644
index 0000000..1a9e640
--- /dev/null
+++ b/special binary string.cpp	
@@ -0,0 +1,41 @@
+/*
+
+CITRIX IIT Guwahti 2019
+
+Special binary strings are binary strings with the following two properties:
+
+The number of 0's is equal to the number of 1's.
+Every prefix of the binary string has at least as many 1's as 0's.
+Given a special string S, a move consists of choosing two consecutive, non-empty, 
+special substrings of S, and swapping them. (Two strings are consecutive if the last character 
+of the first string is exactly one index before the first character of the second string.)
+
+At the end of any number of moves, what is the lexicographically largest resulting string possible?
+
+Solution:
+Recursively try to find the lexicographically largest strings and sort them using a comparator function.
+
+*/
+
+    string makeLargestSpecial(string S) {
+        int i=0, count=0;
+        vector<string> res;
+        for(int j=0;j<S.length();j++){
+            if(S[j]=='1')
+                count++;
+            else
+                count--;
+            if(count==0){
+                res.push_back('1' + makeLargestSpecial(S.substr(i+1,j-i-1)) +'0');
+                i = j+1;
+            }
+            
+        }
+        sort(res.begin(),res.end(),greater<string>());
+        
+        string r = "";
+        for(auto s:res){
+            r+=s;
+        }
+        return r;
+    }
diff --git a/wormholes.cpp b/wormholes.cpp
new file mode 100644
index 0000000..23bc4b3
--- /dev/null
+++ b/wormholes.cpp
@@ -0,0 +1,61 @@
+/*
+Samsung Bangalore NIT Agartala 2019
+
+There is one spaceship. X and Y co-odinate of source of spaceship and destination spaceship is given.
+There are N number of warmholes; each warmhole has 5 values.
+First 2 values are starting co-ordinate of warmhole and after that value no. 3 and 4 represents ending
+co-ordinate of warmhole and last 5th value is represents cost to pass through this warmhole. 
+Now these warmholes are bi-directional. Now the to go from (x1,y1) to (x2,y2) is abs(x1-x2)+abs(y1-y2).
+The main problem here is to find minimum distance to reach spaceship from source to destination
+co-ordinate using any number of warm-hole. 
+It is ok if you wont use any warmhole.
+*/
+
+#include<iostream>
+using namespace std;
+
+int ans,mask[10],w[10][5],n;
+
+int distance(int sx,int sy,int dx,int dy) {
+	int xd=(sx>dx)?(sx-dx):(dx-sx);
+	int yd=(sy>dy)?(sy-dy):(dy-sy);
+	return (xd+yd);
+}
+
+void cal(int sx,int sy,int dx,int dy,int dis) {
+
+	ans=min(ans,distance(sx,sy,dx,dy)+dis);
+	
+	for(int i=0;i<n;i++) {
+		if(mask[i]==0) {
+			mask[i]=1;
+			int temp=distance(sx,sy,w[i][0],w[i][1])+dis+w[i][4];
+			cal(w[i][2],w[i][3],dx,dy,temp);
+			temp=distance(sx,sy,w[i][2],w[i][3])+dis+w[i][4];
+			cal(w[i][0],w[i][1],dx,dy,temp);
+			mask[i]=0;
+		}
+	}
+}
+int main() {
+	int t,index=1;
+	cin>>t;
+	while(t--) {
+
+		cin>>n;
+		int sx,sy,dx,dy;
+		cin>>sx>>sy>>dx>>dy;
+		
+		for(int i=0;i<n;i++) {
+			mask[i]=0;
+			for(int j=0;j<5;j++) {
+				cin>>w[i][j];
+			}
+		}
+		ans=999999;
+		cal(sx,sy,dx,dy,0);
+		cout<<"#"<<index<<" "<<ans<<endl;
+		index++;
+	}
+	return 0;
+}