# 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

** 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 <bits/stdc++.h> ` `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[1].push_back(2); ` ` ` `tree1[2].push_back(1); ` ` ` `tree1[2].push_back(3); ` ` ` `tree1[3].push_back(2); ` ` ` `tree1[3].push_back(4); ` ` ` `tree1[4].push_back(3); ` ` ` `tree1[4].push_back(5); ` ` ` `tree1[5].push_back(4); ` ` ` `tree1[5].push_back(6); ` ` ` `tree1[6].push_back(5); ` ` ` ` ` `// Since 6 is a leaf node for tree 1 ` ` ` `deg1[6] = 1; ` ` ` ` ` `// Building Tree 2 ` ` ` `tree2[1].push_back(2); ` ` ` `tree2[2].push_back(1); ` ` ` `tree2[2].push_back(3); ` ` ` `tree2[3].push_back(2); ` ` ` `tree2[1].push_back(6); ` ` ` `tree2[6].push_back(1); ` ` ` `tree2[6].push_back(5); ` ` ` `tree2[5].push_back(6); ` ` ` `tree2[5].push_back(4); ` ` ` `tree2[4].push_back(5); ` ` ` ` ` `// since both 3 and 4 are leaf nodes of tree 2 . ` ` ` `deg2[3] = 1; ` ` ` `deg2[4] = 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[1]; ` ` ` `long` `long` `root_tree2 = subtree2[1]; ` ` ` ` ` `// 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

1

** Time Complexity **: O(N)

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