diff --git a/05- May/21- Shortest Bridge/21- Shortest Bridge (Ahmed Gamal).cpp b/05- May/21- Shortest Bridge/21- Shortest Bridge (Ahmed Gamal).cpp
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+// Author: Ahmed Gamal
+
+// for this problem, we just need to find the shortest path between the two islands
+// we can use bfs to find the shortest path, with starting from all the cells in the first island
+// and stop when we reach the second island
+// to find the first island, we can use dfs (flood fill) and mark all the cells in the first island, and add them to the queue
+
+class Solution {
+    // dx, dy: arrays to represent the four directions
+    using pii = pair<int, int>;
+    vector<int> dx {0, 0, 1, -1}, dy {1, -1, 0, 0};
+public:
+    int shortestBridge(vector<vector<int>>& grid) {
+        // n: size of the grid
+        // q: queue to store the cells for bfs
+        // vis: array to keep track of the visited cells
+        const int n = int(grid.size());
+        queue<pii> q;
+        vector vis(n, vector(n, false));
+
+        // valid: lambda function to check if the cell is valid or not
+        auto valid = [&](int i, int j) -> bool {
+            return i >= 0 and i < n and j >= 0 and j < n and not vis[i][j];
+        };
+
+        // dfs: lambda function to mark all the cells in the first island (flood fill)
+        auto dfs = [&](auto dfs, int i, int j) -> void {
+            vis[i][j] = true;
+            q.emplace(i, j);
+
+            for(int d = 0; d < 4; d++) {
+                int new_i = i + dx[d], new_j = j + dy[d];
+                if(valid(new_i, new_j) and grid[new_i][new_j]) {
+                    dfs(dfs, new_i, new_j);
+                }
+            }
+        };
+
+        // find the first island
+        for(int i = 0; i < n; i++) {
+            bool found = false;
+            for(int j = 0; j < n; j++) {
+                if(grid[i][j]) {
+                    // mark all the cells in the first island, add them to the queue, and break
+                    dfs(dfs, i, j);
+                    found = true;
+                    break;
+                }
+            }
+            if(found) {
+                break;
+            }
+        }
+
+        // bfs to find the shortest path
+        // dist: distance from the first island to the second island
+        int dist = 0;
+        while(not q.empty()) {
+            // size: number of cells in the current level
+            int size = int(q.size());
+            while(size--) {
+                // i, j: current cell
+                auto [i, j] = q.front();
+                q.pop();
+
+                // if we reached the second island, return the distance - 1 (we don't count the last cell)
+                if(grid[i][j] and dist) {
+                    return --dist;
+                }
+
+                // add all the valid cells to the queue
+                for(int d = 0; d < 4; d++) {
+                    int new_i = i + dx[d], new_j = j + dy[d];
+                    if(valid(new_i, new_j)) {
+                        vis[new_i][new_j] = true;
+                        q.emplace(new_i, new_j);
+                    }
+                }
+            }
+            // increase the distance
+            dist++;
+        }
+
+        // return any value (we will never reach this line)
+        return n;
+    }
+};
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