diff --git a/Problem-1.py b/Problem-1.py
new file mode 100644
index 0000000..b3dc39f
--- /dev/null
+++ b/Problem-1.py
@@ -0,0 +1,62 @@
+#Approach
+# This uses to find the minimum spanning tree . Have the zero node where well at itself is connected
+# find the union algorithm and if it ithe parents are diffrent parent then add to the nodes
+
+
+#Complexities
+#Time: O(m*n)log(M+n)
+#Space: O(n)
+
+
+from typing import List
+
+
+class Solution:
+    def minCostToSupplyWater(self, n: int, wells: List[int], pipes: List[List[int]]) -> int:
+        self.parent = list(range(n+1))
+        self.rank = [1]*(n+1)
+        for i  in range(len(wells)):
+            pipes.append([0,i+1,wells[i]])
+
+        pipes = sorted(pipes, key=lambda x: x[2])
+
+        cost =0
+        for houseA,houseB,pipelength in pipes:
+            if self.union(houseA,houseB):
+                cost+=pipelength
+                n-=1
+                if n==0:
+                    return cost
+
+
+
+
+    def find(self,u):
+        parent = self.parent[u]
+        if parent != u:
+            self.parent[u]=self.find(parent)
+        return self.parent[u]
+
+    def union(self,u,v):
+        pu,pv = self.find(u),self.find(v)
+
+        if pu==pv:
+            return False
+
+        if self.rank[pu]>=self.rank[pv]:
+            self.rank[pu]+=1
+            self.parent[pv]=pu
+        elif self.rank[pv]>self.rank[pu]:
+            self.rank[pv]+=1
+            self.parent[pu]=pv
+        return True
+
+
+
+
+s = Solution()
+print(s.minCostToSupplyWater(3,[1,2,2],[[1,2,1],[2,3,1]]))
+print(s.minCostToSupplyWater(2,[1,1],[[1,2,1],[1,2,2]]))
+
+
+
diff --git a/Problem-2.py b/Problem-2.py
new file mode 100644
index 0000000..bb2935d
--- /dev/null
+++ b/Problem-2.py
@@ -0,0 +1,19 @@
+#Apporach
+#check the pair of persons and return the knows and store the person who is popular left behind
+
+
+#Complexities
+#Time: O(N)
+#Space: O(1)
+
+class Solution:
+    def findCelebrity(self, n: int) -> int:
+        ans = 0
+        for i in range(1, n):
+            if knows(ans, i):
+                ans = i
+        for i in range(n):
+            if ans != i:
+                if knows(ans, i) or not knows(i, ans):
+                    return -1
+        return ans
\ No newline at end of file