Python efficeient O((n+m)*k) solution Real Solution , Fast , Python3 , dfs solution
Intuition\nThe problem involves two graphs represented by edges1 and edges2. We need to determine the maximum number of nodes reachable within k steps from each
# Intuition\nThe problem involves two graphs represented by `edges1` and `edges2`. We need to determine the maximum number of nodes reachable within `k` steps from each node in both graphs and return a list of these values.\n\nTo achieve this, a depth-first search (DFS) is used to explore nodes up to a depth of `k` for each starting node.\n\n# Approach\n1. **Data Structures**:\n   - Use two `defaultdict` objects (`dp1` and `dp2`) to represent adjacency lists for `edges1` and `edges2`.\n\n2. **DFS Functions**:\n   - `dfs_spl`: Computes the number of nodes reachable within `k` steps from a given node in the second graph (`dp2`).\n   - `dfs_spl2`: Computes the number of nodes reachable within `k` steps from a given node in the first graph (`dp1`).\n\n3. **Main Logic**:\n   - For each node in `edges2`, calculate the reachable nodes within `k` steps and store them in the `ans` list.\n   - For each node in `edges1`, calculate the reachable nodes within `k` steps and store them in the `ans2` list.\n   - Find the maximum value from `ans` and update `ans2` with this value when `k > 0`.\n\n4. **Return**:\n   - Return the final list `ans2`.\n\n# Complexity\n- **Time Complexity**:  \n  $$O((n + m) \\times k)$$  \n  - Constructing the adjacency lists takes \\(O(n + m)\\), where \\(n\\) is the number of nodes and \\(m\\) is the number of edges.  \n  - Each DFS explores nodes up to a depth of `k`, and we perform DFS for each node in both graphs. Hence, the total time complexity is \\(O((n + m) \\times k)\\).\n\n- **Space Complexity**:  \n  $$O(n + m)$$  \n  Space is used for adjacency lists (`dp1` and `dp2`), recursion stack (which can go up to depth `k`), and the result lists (`ans` and `ans2`). Hence, the space complexity is linear with respect to the number of nodes and edges.\n\n# Code\n```python\nfrom typing import List\nfrom collections import defaultdict\n\nclass Solution:\n    def maxTargetNodes(self, edges1: List[List[int]], edges2: List[List[int]], k: int) -> List[int]:\n        dp1 = defaultdict(list)\n        dp2 = defaultdict(list)\n\n        # Build adjacency lists for both graphs\n        for i, j in edges1:\n            dp1[i].append(j)\n            dp1[j].append(i)\n\n        for i, j in edges2:\n            dp2[i].append(j)\n            dp2[j].append(i)\n\n        ans = [1] * (len(edges2) + 1)\n\n        # DFS for the second graph\n        def dfs_spl(i, par, k):\n            if k <= 0:\n                return 0\n            ct = 0\n            for j in dp2[i]:\n                if j != par:\n                    ct += (1 + dfs_spl(j, i, k - 1))\n            return ct\n\n        # DFS for the first graph\n        def dfs_spl2(i, par, k):\n            if k <= 0:\n                return 0\n            ct = 0\n            for j in dp1[i]:\n                if j != par:\n                    ct += (1 + dfs_spl2(j, i, k - 1))\n            return ct\n\n        # Calculate reachable nodes for graph 2\n        for i in range(len(edges2) + 1):\n            ans[i] += dfs_spl(i, -1, k - 1)\n\n        ans2 = [1] * (len(edges1) + 1)\n\n        # Calculate reachable nodes for graph 1\n        for i in range(len(edges1) + 1):\n            ans2[i] += dfs_spl2(i, -1, k)\n\n        # Find max reachable nodes in graph 2 and add to graph 1\'s result\n        q = max(ans)\n        for i in range(len(ans2)):\n            if k > 0:\n                ans2[i] += q\n\n        return ans2\n
['Depth-First Search', 'Python3']