diff --git a/find-celebrity.java b/find-celebrity.java
new file mode 100644
index 0000000..e50491c
--- /dev/null
+++ b/find-celebrity.java
@@ -0,0 +1,27 @@
+// TC: O(n)
+// SC: O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+public int findCelebrity(int n) {
+    int start = 0;
+    int end = n - 1;
+    
+    // Narrow down the candidate
+    while (start < end) {
+        if (knows(start, end)) {
+            start++;
+        } else {
+            end--;
+        }
+    }
+    
+    // Verify the candidate
+    for (int i = 0; i < n; i++) {
+        if (start != i && (knows(start, i) || !knows(i, start))) {
+            return -1;
+        }
+    }
+    
+    return start;
+}
diff --git a/water-in-village.java b/water-in-village.java
new file mode 100644
index 0000000..b8da948
--- /dev/null
+++ b/water-in-village.java
@@ -0,0 +1,78 @@
+// TC: O(n log n)
+// SC: O(n)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+class Solution {
+    public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
+        // List to hold all edges (both well edges and pipe edges)
+        List<Edge> edges = new ArrayList<>();
+        
+        // Virtual water source is node 0.
+        // For each house (1-indexed in our union-find), add an edge from node 0 with cost wells[i].
+        for (int i = 0; i < n; i++) {
+            edges.add(new Edge(0, i + 1, wells[i]));
+        }
+        
+        // Add all the pipe edges.
+        for (int[] pipe : pipes) {
+            // Each pipe is represented by [house1, house2, cost].
+            // Houses are numbered from 1 to n.
+            int u = pipe[0], v = pipe[1], cost = pipe[2];
+            edges.add(new Edge(u, v, cost));
+        }
+        
+        // Sort edges by cost (ascending order).
+        Collections.sort(edges, (a, b) -> a.cost - b.cost);
+        
+        // Initialize union-find for n + 1 nodes (including virtual node 0).
+        UnionFind uf = new UnionFind(n + 1);
+        int totalCost = 0;
+        
+        // Process each edge in order; if it connects two components, add its cost.
+        for (Edge edge : edges) {
+            if (uf.union(edge.u, edge.v)) {
+                totalCost += edge.cost;
+            }
+        }
+        
+        return totalCost;
+    }
+    
+    // Edge class to store an edge in the graph.
+    class Edge {
+        int u, v, cost;
+        public Edge(int u, int v, int cost) {
+            this.u = u;
+            this.v = v;
+            this.cost = cost;
+        }
+    }
+    
+    // Union-Find (Disjoint Set) data structure.
+    class UnionFind {
+        int[] parent;
+        
+        public UnionFind(int n) {
+            parent = new int[n];
+            for (int i = 0; i < n; i++) {
+                parent[i] = i;
+            }
+        }
+        
+        public int find(int x) {
+            if (parent[x] != x) {
+                parent[x] = find(parent[x]);
+            }
+            return parent[x];
+        }
+        
+        public boolean union(int x, int y) {
+            int rootX = find(x);
+            int rootY = find(y);
+            if (rootX == rootY) return false;
+            parent[rootX] = rootY;
+            return true;
+        }
+    }
+}