# Number of distinct pair of edges such that it partitions both trees into same subsets of nodes

Given two trees each of N nodes. Removing an edge of the tree partitions the tree in two subsets.

Find the total maximum number of distinct edges (e1, e2): e1 from the first tree and e2 from the second tree such that it partitions both the trees into subsets with same nodes.

Examples: Input : Same as the above figure N = 6
Tree 1 : {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Tree 2 :{(1, 2), (2, 3), (1, 6), (6, 5), (5, 4)}
Output : 1
We can remove edge 3-4 in the first graph and edge 1-6 in the second graph.
The subsets will be {1, 2, 3} and {4, 5, 6}.

Input : N = 4
Tree 1 : {(1, 2), (1, 3), (1, 4)}
Tree 2 : {(1, 2), (2, 4), (1, 3)}
Output : 2
We can select an edge 1-3 in the first graph and 1-3 in the second graph.
The subsets will be {3} and {1, 2, 4} for both graphs.
Also we can select an edge 1-4 in the first graph and 2-4 in the second graph.
The subsets will be {4} and {1, 2, 3} for both the graphs

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

• The idea is to use hashing on trees, We will root both of trees at node 1, then We will assign random values to each node of the tree.
• We will do a dfs on the tree and, Suppose we are at node x, then we will keep a variable
subtree[x] that will store the hash value of all the nodes in its subtree.
• One we did the above two steps, we are just left with storing the value of each subtree of nodes for both the trees the we get.
• We can use unordered map for it. The last step is to find how many common values of subtree[x] are there is both trees.
• Increase the count of distinct edges by +1 for every common values of subtree[x] for both trees.

Below is the implementation of above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `long` `long` `p = 97, MAX = 300005; ` ` `  `// This function checks whether ` `// a node is leaf node or not. ` `bool` `leaf1(``long` `long` `NODE, ``long` `long` `int` `deg1[]) ` `{ ` `    ``if` `(deg1[NODE] == 1 and NODE != 1) ` `        ``return` `true``; ` `    ``return` `false``; ` `} ` ` `  `// This function calculates the Hash sum ` `// of all the children of a ` `// particular node for subtree 1 ` `void` `dfs3(``long` `long` `curr, ``long` `long` `par, ` `          ``vector<``long` `long` `int``> tree1[], ` `          ``long` `long` `int` `subtree1[], ``long` `long` `int` `deg1[], ` `          ``long` `long` `int` `node[]) ` `{ ` `    ``for` `(``auto``& child : tree1[curr]) { ` `        ``if` `(child == par) ` `            ``continue``; ` `        ``dfs3(child, curr, tree1, subtree1, deg1, node); ` `    ``} ` ` `  `    ``// If the node is leaf node then ` `    ``// there is no child, so hash sum ` `    ``// will be same as the ` `    ``// hash value for the node. ` `    ``if` `(leaf1(curr, deg1) == ``true``) { ` `        ``subtree1[curr] = node[curr]; ` `        ``return``; ` `    ``} ` `    ``long` `long` `sum = 0; ` ` `  `    ``// Else calculate hash sum of all the ` `    ``// children of a particular node, this is done ` `    ``// by iterating on all the children of a node. ` `    ``for` `(``auto``& child : tree1[curr]) { ` `        ``sum = sum + subtree1[child]; ` `    ``} ` ` `  `    ``// store the hash value for ` `    ``// all the subtree of current node ` `    ``subtree1[curr] = node[curr] + sum; ` `    ``return``; ` `} ` ` `  `// This function checks whether ` `// a node is leaf node or not. ` `bool` `leaf2(``long` `long` `NODE, ``long` `long` `int` `deg2[]) ` `{ ` `    ``if` `(deg2[NODE] == 1 and NODE != 1) ` `        ``return` `true``; ` `    ``return` `false``; ` `} ` ` `  `// This function calculates the Hash ` `// sum of all the children of ` `// a particular node for subtree 2. ` `void` `dfs4(``long` `long` `curr, ``long` `long` `par, ` `          ``vector<``long` `long` `int``> tree2[], ` `          ``long` `long` `int` `subtree2[], ``long` `long` `int` `deg2[], ` `          ``long` `long` `int` `node[]) ` `{ ` `    ``for` `(``auto``& child : tree2[curr]) { ` `        ``if` `(child == par) ` `            ``continue``; ` `        ``dfs4(child, curr, tree2, subtree2, deg2, node); ` `    ``} ` ` `  `    ``// If the node is leaf node then ` `    ``// there is no child, so hash sum will be ` `    ``// same as the hash value for the node. ` `    ``if` `(leaf2(curr, deg2) == ``true``) { ` `        ``subtree2[curr] = node[curr]; ` `        ``return``; ` `    ``} ` `    ``long` `long` `sum = 0; ` ` `  `    ``// Else calculate hash sum of all ` `    ``// the children of a particular node, this is ` `    ``// done by iterating on all the children of a node. ` `    ``for` `(``auto``& child : tree2[curr]) { ` `        ``sum = sum + subtree2[child]; ` `    ``} ` ` `  `    ``// store the hash value for ` `    ``// all the subtree of current node ` `    ``subtree2[curr] = node[curr] + sum; ` `} ` ` `  `// Calculates x^y in logN time. ` `long` `long` `exp``(``long` `long` `x, ``long` `long` `y) ` `{ ` `    ``if` `(y == 0) ` `        ``return` `1; ` `    ``else` `if` `(y & 1) ` `        ``return` `x * ``exp``(x, y / 2) * ``exp``(x, y / 2); ` `    ``else` `        ``return` `exp``(x, y / 2) * ``exp``(x, y / 2); ` `} ` ` `  `// This function helps in building the tree ` `void` `Insertt(vector<``long` `long` `int``> tree1[], ` `             ``vector<``long` `long` `int``> tree2[], ` `             ``long` `long` `int` `deg1[], ``long` `long` `int` `deg2[]) ` `{ ` `    ``// Building Tree 1 ` `    ``tree1.push_back(2); ` `    ``tree1.push_back(1); ` `    ``tree1.push_back(3); ` `    ``tree1.push_back(2); ` `    ``tree1.push_back(4); ` `    ``tree1.push_back(3); ` `    ``tree1.push_back(5); ` `    ``tree1.push_back(4); ` `    ``tree1.push_back(6); ` `    ``tree1.push_back(5); ` ` `  `    ``// Since 6 is a leaf node for tree 1 ` `    ``deg1 = 1; ` ` `  `    ``// Building Tree 2 ` `    ``tree2.push_back(2); ` `    ``tree2.push_back(1); ` `    ``tree2.push_back(3); ` `    ``tree2.push_back(2); ` `    ``tree2.push_back(6); ` `    ``tree2.push_back(1); ` `    ``tree2.push_back(5); ` `    ``tree2.push_back(6); ` `    ``tree2.push_back(4); ` `    ``tree2.push_back(5); ` ` `  `    ``// since both 3 and 4 are leaf nodes of tree 2 . ` `    ``deg2 = 1; ` `    ``deg2 = 1; ` `} ` ` `  `// Function to make the hash values ` `void` `TakeHash(``long` `long` `n, ``long` `long` `int` `node[]) ` `{ ` `    ``// Take a very high prime ` `    ``long` `long` `p = 97 * 13 * 19; ` ` `  `    ``// Initialize random values to each node . ` `    ``for` `(``long` `long` `i = 1; i <= n; ++i) { ` ` `  `        ``// A good random function is ` `        ``// choosen for each node . ` `        ``long` `long` `val = (``rand``() * ``rand``() * ``rand``()) ` `                        ``+ ``rand``() * ``rand``() + ``rand``(); ` `        ``node[i] = val * p * ``rand``() + p * 13 * 19 * ``rand``() * ``rand``() * 101 * p; ` `        ``p *= p; ` `        ``p *= p; ` `    ``} ` `} ` ` `  `// Function that returns the required answer ` `void` `solve(``long` `long` `n, vector<``long` `long` `int``> tree1[], ` `           ``vector<``long` `long` `int``> tree2[], ``long` `long` `int` `subtree1[], ` `           ``long` `long` `int` `subtree2[], ``long` `long` `int` `deg1[], ` `           ``long` `long` `int` `deg2[], ``long` `long` `int` `node[]) ` `{ ` `    ``// Do dfs on both trees to ` `    ``// get subtree[x] for each node. ` `    ``dfs3(1, 0, tree1, subtree1, deg1, node); ` `    ``dfs4(1, 0, tree2, subtree2, deg2, node); ` ` `  `    ``// cnt_tree1 and cnt_tree2 is used ` `    ``// to store the count of all ` `    ``// the hashes of every node . ` `    ``unordered_map<``long` `long``, ``long` `long``> ` `        ``cnt_tree1, cnt_tree2; ` `    ``vector<``long` `long``> values; ` `    ``for` `(``long` `long` `i = 1; i <= n; ++i) { ` `        ``long` `long` `value1 = subtree1[i]; ` `        ``long` `long` `value2 = subtree2[i]; ` ` `  `        ``// Store the subtree value of tree 1 ` `        ``// in a vector to compare it later ` `        ``// with subtree value of tree 2. ` `        ``values.push_back(value1); ` ` `  `        ``// increment the count of hash ` `        ``// value for a subtree of a node. ` `        ``cnt_tree1[value1]++; ` `        ``cnt_tree2[value2]++; ` `    ``} ` ` `  `    ``// Stores the sum of all the hash values ` `    ``// of children for root node of subtree 1. ` `    ``long` `long` `root_tree1 = subtree1; ` `    ``long` `long` `root_tree2 = subtree2; ` ` `  `    ``// Stores the sum of all the hash values ` `    ``// of children for root node of subtree 1. ` `    ``cnt_tree1[root_tree1] = 0; ` `    ``cnt_tree2[root_tree2] = 0; ` `    ``long` `long` `answer = 0; ` `    ``for` `(``auto``& x : values) { ` ` `  `        ``// Check if for a given hash value for ` `        ``// tree 1 is there any hash value which ` `        ``// matches to hash value of tree 2 ` `        ``// If yes, then its possible to divide ` `        ``// the tree for this hash value ` `        ``// into two equal subsets. ` `        ``if` `(cnt_tree1[x] != 0 and cnt_tree2[x] != 0) ` `            ``++answer; ` `    ``} ` `    ``cout << answer << endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``vector<``long` `long` `int``> tree1[MAX], tree2[MAX]; ` `    ``long` `long` `int` `node[MAX], deg1[MAX], deg2[MAX]; ` `    ``long` `long` `int` `subtree1[MAX], subtree2[MAX]; ` `    ``long` `long` `n = 6; ` ` `  `    ``// To generate a good random function ` `    ``srand``(``time``(NULL)); ` `    ``Insertt(tree1, tree2, deg1, deg2); ` `    ``TakeHash(n, node); ` `    ``solve(n, tree1, tree2, subtree1, subtree2, deg1, deg2, node); ` ` `  `    ``return` `0; ` `} `

Output:

```1
```

Time Complexity : O(N)

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