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| 1 | +class Solution { |
| 2 | +public: |
| 3 | + int getNodesAtDistance(int curr,int k,unordered_map<int,vector<int>>&tree,vector<int>&dist){ |
| 4 | + int cnt = 0; |
| 5 | + queue<int> q; |
| 6 | + unordered_map<int, bool> visited; |
| 7 | + visited[curr] = true; |
| 8 | + q.push(curr); |
| 9 | + dist[curr] = 0; |
| 10 | + while (!q.empty()) { |
| 11 | + int node = q.front(); |
| 12 | + q.pop(); |
| 13 | + for (auto it : tree[node]) { |
| 14 | + if (!visited[it]) { |
| 15 | + visited[it] = true; |
| 16 | + dist[it] = dist[node] + 1; |
| 17 | + q.push(it); |
| 18 | + if (dist[it] % 2 == k) { |
| 19 | + cnt++; |
| 20 | + } |
| 21 | + } |
| 22 | + } |
| 23 | + } |
| 24 | + return cnt; |
| 25 | + } |
| 26 | + vector<int> maxTargetNodes(vector<vector<int>>& edges1, vector<vector<int>>& edges2) { |
| 27 | + unordered_map<int,vector<int>>tree1,tree2; |
| 28 | + int n1 = INT_MIN, n2 = INT_MIN; |
| 29 | + for (auto& edge : edges1) { |
| 30 | + tree1[edge[0]].push_back(edge[1]); |
| 31 | + tree1[edge[1]].push_back(edge[0]); |
| 32 | + n1 = max(n1, max(edge[0], edge[1])); |
| 33 | + } |
| 34 | + for (auto& edge : edges2) { |
| 35 | + tree2[edge[0]].push_back(edge[1]); |
| 36 | + tree2[edge[1]].push_back(edge[0]); |
| 37 | + n2 = max(n2, max(edge[0], edge[1])); |
| 38 | + } |
| 39 | + vector<int>ans(n1+1,0); |
| 40 | + vector<int>distInTree1(n1+1,-1),distInTree2(n2+1,-1); |
| 41 | + int oddInTree2 = getNodesAtDistance(0,1,tree2,distInTree2); |
| 42 | + int maxInTree2 = 0; |
| 43 | + for(int i=0;i<=n2;i++){ |
| 44 | + int currDist = distInTree2[i]; |
| 45 | + int canVisit = 0; |
| 46 | + if(currDist%2==0){ |
| 47 | + canVisit = oddInTree2; |
| 48 | + } else { |
| 49 | + canVisit = n2 + 1 - oddInTree2; |
| 50 | + } |
| 51 | + maxInTree2 = max(maxInTree2,canVisit); |
| 52 | + } |
| 53 | + int evenInTree1 = 1 + getNodesAtDistance(0,0,tree1,distInTree1); |
| 54 | + for(int i=0;i<=n1;i++){ |
| 55 | + int currDist = distInTree1[i]; |
| 56 | + int canVisit = 0; |
| 57 | + if(currDist % 2== 0){ |
| 58 | + canVisit = evenInTree1; |
| 59 | + } else { |
| 60 | + canVisit = n1 + 1 - evenInTree1; |
| 61 | + } |
| 62 | + ans[i] = canVisit + maxInTree2; |
| 63 | + } |
| 64 | + return ans; |
| 65 | + } |
| 66 | +}; |
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