# Lowest Common Ancestor for a Set of Nodes in a Rooted Tree

Given a rooted tree with N nodes, the task is to find the Lowest Common Ancestor for a given set of nodes V of that tree.

Examples:

```Input:
1
/  |  \
2   3   4
/  \  |   |
5   6  7   10
/ \
8   9
V[] = {7, 3, 8, 9}
Output: 3

Input:
1
/  |  \
2   3   4
/  \  |   |
5   6  7   10
/ \
8   9
V[] = {4, 6, 7}
Output: 1
```

Approach: We can observe that

1. the Lowest Common Ancestors for any set of nodes will have in-time less than or equal to that of all nodes in the set and out-time greater than or equal(equal if LCA is present in the set) to that of all nodes in the set.
2. Thus, in order to solve the problem, we need to traverse the entire tree starting from the root node using Depth First Search Traversal and store the in-time, out-time, and level for every node.
3. The node with in-time less than or equal to all the nodes in the set and out-time greater than or equal to all the nodes in the set and with maximum level possible is the answer.

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the LCA ` `// in a rooted tree for a given ` `// set of nodes ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Set time 1 initially ` `int` `T = 1; ` `void` `dfs(``int` `node, ``int` `parent, ` `         ``vector<``int``> g[], ` `         ``int` `level[], ``int` `t_in[], ` `         ``int` `t_out[]) ` `{ ` ` `  `    ``// Case for root node ` `    ``if` `(parent == -1) { ` `        ``level[node] = 1; ` `    ``} ` `    ``else` `{ ` `        ``level[node] = level[parent] + 1; ` `    ``} ` ` `  `    ``// In-time for node ` `    ``t_in[node] = T; ` `    ``for` `(``auto` `i : g[node]) { ` `        ``if` `(i != parent) { ` `            ``T++; ` `            ``dfs(i, node, g, ` `                ``level, t_in, t_out); ` `        ``} ` `    ``} ` `    ``T++; ` ` `  `    ``// Out-time for the node ` `    ``t_out[node] = T; ` `} ` ` `  `int` `findLCA(``int` `n, vector<``int``> g[], ` `            ``vector<``int``> v) ` `{ ` ` `  `    ``// level[i]--> Level of node i ` `    ``int` `level[n + 1]; ` ` `  `    ``// t_in[i]--> In-time of node i ` `    ``int` `t_in[n + 1]; ` ` `  `    ``// t_out[i]--> Out-time of node i ` `    ``int` `t_out[n + 1]; ` ` `  `    ``// Fill the level, in-time ` `    ``// and out-time of all nodes ` `    ``dfs(1, -1, g, ` `        ``level, t_in, t_out); ` ` `  `    ``int` `mint = INT_MAX, maxt = INT_MIN; ` `    ``int` `minv = -1, maxv = -1; ` `    ``for` `(``auto` `i = v.begin(); i != v.end(); i++) { ` `        ``// To find minimum in-time among ` `        ``// all nodes in LCA set ` `        ``if` `(t_in[*i] < mint) { ` `            ``mint = t_in[*i]; ` `            ``minv = *i; ` `        ``} ` `        ``// To find maximum in-time among ` `        ``// all nodes in LCA set ` `        ``if` `(t_out[*i] > maxt) { ` `            ``maxt = t_out[*i]; ` `            ``maxv = *i; ` `        ``} ` `    ``} ` ` `  `    ``// Node with same minimum ` `    ``// and maximum out time ` `    ``// is LCA for the set ` `    ``if` `(minv == maxv) { ` `        ``return` `minv; ` `    ``} ` ` `  `    ``// Take the minimum level as level of LCA ` `    ``int` `lev = min(level[minv], level[maxv]); ` `    ``int` `node, l = INT_MIN; ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` `        ``// If i-th node is at a higher level ` `        ``// than that of the minimum among ` `        ``// the nodes of the given set ` `        ``if` `(level[i] > lev) ` `            ``continue``; ` ` `  `        ``// Compare in-time, out-time ` `        ``// and level of i-th node ` `        ``// to the respective extremes ` `        ``// among all nodes of the given set ` `        ``if` `(t_in[i] <= mint ` `            ``&& t_out[i] >= maxt ` `            ``&& level[i] > l) { ` `            ``node = i; ` `            ``l = level[i]; ` `        ``} ` `    ``} ` ` `  `    ``return` `node; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 10; ` `    ``vector<``int``> g[n + 1]; ` `    ``g[1].push_back(2); ` `    ``g[2].push_back(1); ` `    ``g[1].push_back(3); ` `    ``g[3].push_back(1); ` `    ``g[1].push_back(4); ` `    ``g[4].push_back(1); ` `    ``g[2].push_back(5); ` `    ``g[5].push_back(2); ` `    ``g[2].push_back(6); ` `    ``g[6].push_back(2); ` `    ``g[3].push_back(7); ` `    ``g[7].push_back(3); ` `    ``g[4].push_back(10); ` `    ``g[10].push_back(4); ` `    ``g[8].push_back(7); ` `    ``g[7].push_back(8); ` `    ``g[9].push_back(7); ` `    ``g[7].push_back(9); ` ` `  `    ``vector<``int``> v = { 7, 3, 8 }; ` ` `  `    ``cout << findLCA(n, g, v) << endl; ` `} `

Output:

```3
```

Time Complexity: O(N)
Space Complexity: O(N)

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