<p>There is an undirected connected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> edges.</p>

<p>You are given the integer <code>n</code> and the array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>Return an array <code>answer</code> of length <code>n</code> where <code>answer[i]</code> is the sum of the distances between the <code>i<sup>th</sup></code> node in the tree and all other nodes.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist1.jpg" style="width: 304px; height: 224px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
<strong>Output:</strong> [8,12,6,10,10,10]
<strong>Explanation:</strong> The tree is shown above.
We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
equals 1 + 1 + 2 + 2 + 2 = 8.
Hence, answer[0] = 8, and so on.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist2.jpg" style="width: 64px; height: 65px;" />
<pre>
<strong>Input:</strong> n = 1, edges = []
<strong>Output:</strong> [0]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist3.jpg" style="width: 144px; height: 145px;" />
<pre>
<strong>Input:</strong> n = 2, edges = [[1,0]]
<strong>Output:</strong> [1,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>The given input represents a valid tree.</li>
</ul>

// https://www.youtube.com/watch?v=kM3A6AjHvjM
class Solution {
public:
    int N;
    void dfs(int node, vector<int>& ans,vector<int>& count, vector<vector<int>>& adj,vector<bool>& visited,int parent){
        // if(visited[node])
        //     return;
        
        // visited[node] = true;

        for(auto x:adj[node]){
            if(x==parent)
                continue;
            dfs(x,ans,count,adj,visited,node);
            ans[node] = ans[node] + ans[x] + count[x];
            count[node] = count[node] + count[x];
        }
    }

    void dfs2(int node, vector<int>& ans,vector<int>& count, vector<vector<int>>& adj,vector<bool>& visited,int parent){
        // if(visited[node])
        //     return;
        
        // visited[node] = true;

        for(auto x:adj[node]){
            if(x==parent)
                continue;
            
            ans[x] = (ans[node]-count[x]) + (N-count[x]); // Note it should be N-count[x] to account for all the remaining nodes. count[node]-count[x] would not work because what if the parent had another parent, then we would be missing those nodes. so it should be (total nodes - nodes in this subtree)

            dfs2(x,ans,count,adj,visited,node);  
        }
    }

    
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        N=n;
        vector<int> count(n,1);
        vector<int> ans(n,0);
        vector<bool> visited(n,false);
        vector<vector<int>> adj(n);

        for(int i=0;i<edges.size();i++){
            adj[edges[i][0]].push_back(edges[i][1]);
            adj[edges[i][1]].push_back(edges[i][0]);
        }


        // first dfs to get the ans of the root
        dfs(0,ans,count,adj,visited,-1);

        visited = vector<bool>(n,false);
        // second dfs to get the ans of the other nodes
        dfs2(0,ans,count,adj,visited,-1);

        return ans;
    }
    
    // FLoyd warshall would not work - constraint does not allow
    vector<int> sumOfDistancesInTree2(int n, vector<vector<int>>& edges) {
        vector<vector<int>> dist(n,vector<int>(n,1e6));

        for(int i=0;i<n;i++){
            dist[i][i] = 0;
        }

        for(int i=0;i<edges.size();i++){
            dist[edges[i][0]][edges[i][1]] = 1;
            dist[edges[i][1]][edges[i][0]] = 1;
        }

        vector<int> result;
        for(int k=0;k<n;k++){
            int curr =0;

            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    if(i!=j && i!=k && j!=k){
                        if(dist[i][k] + dist[k][j] < dist[i][j]){
                            dist[i][j] = dist[i][k]+dist[k][j];
                        }
                    }
                }
            }
        }

        for(int i=0;i<n;i++){
            long long curr = 0;
            for(int j=0;j<n;j++){
                if(dist[i][j] == 1e6){
                   // cout<<"-1 ";;
                } else{
                    curr+= dist[i][j];
                  //  cout<<dist[i][j]<<" ";
                }
            }
           // cout<<endl;
            result.push_back(curr);
        }

        return result;
    }
};
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