diff --git a/celebrity.py b/celebrity.py
new file mode 100644
index 0000000..3a279ef
--- /dev/null
+++ b/celebrity.py
@@ -0,0 +1,22 @@
+# findCelebrity:
+# - Check each person to see if they could be a celebrity.
+# - A celebrity knows no one and is known by everyone else.
+# - Use nested loops to verify the conditions for each person.
+
+# TC: O(n^2) due to nested checks with knows(i, j).
+# SC: O(n) for storing possible celebrities.
+
+
+class Solution:
+    def findCelebrity(self, n: int) -> int:
+        
+        isCelebrity = [True] * n
+        for i in range(n):
+            if isCelebrity[i]:
+                for j in range(n):
+                    if j != i and (knows(i, j) or not knows(j, i)):
+                        isCelebrity[i] = False
+                        break
+                if isCelebrity[i]:
+                    return i
+        return -1
\ No newline at end of file
diff --git a/optimize.py b/optimize.py
new file mode 100644
index 0000000..2274da5
--- /dev/null
+++ b/optimize.py
@@ -0,0 +1,43 @@
+# minCostToSupplyWater:
+# - Treat wells as edges from a virtual node 0 to each house.
+# - Use Kruskal’s algorithm with Union-Find to build a minimum spanning tree.
+# - Sort all edges (wells + pipes) and greedily connect components.
+
+# TC: O(E log E), E = n + len(pipes) (for sorting edges).
+# SC: O(n) for Union-Find structure.
+
+
+class Solution:
+    def minCostToSupplyWater(self, n: int, wells: List[int], pipes: List[List[int]]) -> int:
+        n += 1
+        parents = [-1] * n
+
+        def find(a):
+            if parents[a] == -1:
+                return a
+            else:
+                parents[a] = find(parents[a])
+                return parents[a]
+
+        def union(a, b, cost):
+            parentA = find(a)
+            parentB = find(b)
+
+            if parentA != parentB:
+                parents[parentB] = parentA
+                return cost
+
+            return 0
+
+        edgesList = [(wells[i - 1], 0, i) for i in range(1, n)]
+
+        for cityOne, cityTwo, cost in pipes:
+            edgesList.append((cost, cityOne, cityTwo))
+
+        edgesList.sort()
+
+        totalCost = 0
+        for cost, cityOne, cityTwo in edgesList:
+            totalCost += union(cityOne, cityTwo, cost)
+
+        return totalCost
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