# Maximum weighted edge in path between two nodes in an N-ary tree using binary lifting

Given an N-ary tree with weighted edge and Q queries where each query contains two nodes of the tree. The task is to find the maximum weighted edge in the simple path between these two nodes.
Examples: Naive Approach: A simple solution is to traverse the whole tree for each query and find the path between the two nodes.
Efficient Approach: The idea is to use binary lifting to pre-compute the maximum weighted edge from every node to every other node at distance of some . We will store the maximum weighted edge till level. and where

• j is the node and
• i is the distance of • dp[i][j] stores the parent of j at • distance if present, else it will store 0
• mx[i][j] stores the maximum edge from node j to the parent of this node at • distance.

We’ll do a depth-first search to find all the parents at distance and their weight and then precompute parents and maximum edges at every distance.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the` `// maximum weighted edge in the simple` `// path between two nodes in N-ary Tree`   `#include `   `using` `namespace` `std;`   `const` `int` `N = 100005;`   `// Depths of Nodes` `vector<``int``> level(N);` `const` `int` `LG = 20;`   `// Parent at every 2^i level` `vector > dp(LG, vector<``int``>(N));`   `// Maximum node at every 2^i level` `vector > mx(LG, vector<``int``>(N));`   `// Graph that stores destinations` `// and its weight` `vector > > v(N);` `int` `n;`   `// Function to traverse the nodes` `// using the Depth-First Search Traversal` `void` `dfs_lca(``int` `a, ``int` `par, ``int` `lev)` `{` `    ``dp[a] = par;` `    ``level[a] = lev;` `    ``for` `(``auto` `i : v[a]) {`   `        ``// Condition to check if its` `        ``// equal to its parent then skip` `        ``if` `(i.first == par)` `            ``continue``;` `        ``mx[i.first] = i.second;`   `        ``// DFS Recursive Call` `        ``dfs_lca(i.first, a, lev + 1);` `    ``}` `}`   `// Function to find the ansector` `void` `find_ancestor()` `{`   `    ``// Loop to set every 2^i distance` `    ``for` `(``int` `i = 1; i < LG; i++) {` `        ``// Loop to calculate for` `        ``// each node in the N-ary tree` `        ``for` `(``int` `j = 1; j <= n; j++) {` `            ``dp[i][j]` `                ``= dp[i - 1][dp[i - 1][j]];`   `            ``// Storing maximum edge` `            ``mx[i][j]` `                ``= max(mx[i - 1][j],` `                      ``mx[i - 1][dp[i - 1][j]]);` `        ``}` `    ``}` `}`   `int` `getMax(``int` `a, ``int` `b)` `{` `    ``// Swaping if node a is at more depth` `    ``// than node b because we will` `    ``// always take at more depth` `    ``if` `(level[b] < level[a])` `        ``swap(a, b);`   `    ``int` `ans = 0;`   `    ``// Diffeence between the depth of` `    ``// the two given nodes` `    ``int` `diff = level[b] - level[a];` `    ``while` `(diff > 0) {` `        ``int` `log` `= log2(diff);` `        ``ans = max(ans, mx[``log``][b]);`   `        ``// Changing Node B to its` `        ``// parent at 2 ^ i distance` `        ``b = dp[``log``][b];`   `        ``// Subtracting distance by 2^i` `        ``diff -= (1 << ``log``);` `    ``}`   `    ``// Take both a, b to its` `    ``// lca and find maximum` `    ``while` `(a != b) {` `        ``int` `i = log2(level[a]);`   `        ``// Loop to find the maximum 2^ith` `        ``// parent the is differnet` `        ``// for both a and b` `        ``while` `(i > 0` `               ``&& dp[i][a] == dp[i][b])` `            ``i--;`   `        ``// Updating ans` `        ``ans = max(ans, mx[i][a]);` `        ``ans = max(ans, mx[i][b]);`   `        ``// Changing value to its parent` `        ``a = dp[i][a];` `        ``b = dp[i][b];` `    ``}` `    ``return` `ans;` `}`   `// Function to compute the Least` `// common Ansector` `void` `compute_lca()` `{` `    ``dfs_lca(1, 0, 0);` `    ``find_ancestor();` `}`   `// Driver Code` `int` `main()` `{` `    ``// Undirected tree` `    ``n = 5;` `    ``v.push_back(make_pair(2, 2));` `    ``v.push_back(make_pair(1, 2));` `    ``v.push_back(make_pair(3, 5));` `    ``v.push_back(make_pair(1, 5));` `    ``v.push_back(make_pair(4, 3));` `    ``v.push_back(make_pair(3, 4));` `    ``v.push_back(make_pair(5, 1));` `    ``v.push_back(make_pair(3, 1));`   `    ``// Computing LCA` `    ``compute_lca();`   `    ``int` `queries[]` `        ``= { { 3, 5 },` `            ``{ 2, 3 },` `            ``{ 2, 4 } };` `    ``int` `q = 3;`   `    ``for` `(``int` `i = 0; i < q; i++) {` `        ``int` `max_edge = getMax(queries[i],` `                              ``queries[i]);` `        ``cout << max_edge << endl;` `    ``}` `    ``return` `0;` `}`

## Python3

 `# Python3 implementation to ` `# find the maximum weighted ` `# edge in the simple path ` `# between two nodes in N-ary Tree` `import` `math` `N ``=` `100005``;` ` `  `# Depths of Nodes` `level ``=` `[``0` `for` `i ``in` `range``(N)]` `LG ``=` `20``;` ` `  `# Parent at every 2^i level` `dp ``=` `[[``0` `for` `j ``in` `range``(N)] ` `         ``for` `i ``in` `range``(LG)]` ` `  `# Maximum node at every 2^i level` `mx ``=` `[[``0` `for` `j ``in` `range``(N)] ` `         ``for` `i ``in` `range``(LG)]` ` `  `# Graph that stores destinations` `# and its weight` `v ``=` `[[] ``for` `i ``in` `range``(N)]` `n ``=` `0` ` `  `# Function to traverse the ` `# nodes using the Depth-First ` `# Search Traversal` `def` `dfs_lca(a, par, lev):`   `    ``dp[``0``][a] ``=` `par;` `    ``level[a] ``=` `lev;` `    `  `    ``for` `i ``in` `v[a]:` ` `  `        ``# Condition to check ` `        ``# if its equal to its ` `        ``# parent then skip` `        ``if` `(i[``0``] ``=``=` `par):` `            ``continue``;` `        ``mx[``0``][i[``0``]] ``=` `i[``1``];` ` `  `        ``# DFS Recursive Call` `        ``dfs_lca(i[``0``], a, lev ``+` `1``);`   `# Function to find the ansector` `def` `find_ancestor():` ` `  `    ``# Loop to set every 2^i distance` `    ``for` `i ``in` `range``(``1``, ``16``):` `    `  `        ``# Loop to calculate for` `        ``# each node in the N-ary tree` `        ``for` `j ``in` `range``(``1``, n ``+` `1``):` `        `  `            ``dp[i][j] ``=` `dp[i ``-` `1``][dp[i ``-` `1``][j]];` ` `  `            ``# Storing maximum edge` `            ``mx[i][j] ``=` `max``(mx[i ``-` `1``][j],` `                           ``mx[i ``-` `1``][dp[i ``-` `1``][j]]);`   `def` `getMax(a, b):`   `    ``# Swaping if node a is at more depth` `    ``# than node b because we will` `    ``# always take at more depth` `    ``if` `(level[b] < level[a]):` `        ``a, b ``=` `b, a` ` `  `    ``ans ``=` `0``;` ` `  `    ``# Diffeence between the ` `    ``# depth of the two given ` `    ``# nodes` `    ``diff ``=` `level[b] ``-` `level[a];` `    `  `    ``while` `(diff > ``0``):` `        ``log ``=` `int``(math.log2(diff));` `        ``ans ``=` `max``(ans, mx[log][b]);` ` `  `        ``# Changing Node B to its` `        ``# parent at 2 ^ i distance` `        ``b ``=` `dp[log][b];` ` `  `        ``# Subtracting distance by 2^i` `        ``diff ``-``=` `(``1` `<< log);` `     `  `    ``# Take both a, b to its` `    ``# lca and find maximum` `    ``while` `(a !``=` `b):` `        ``i ``=` `int``(math.log2(level[a]));` ` `  `        ``# Loop to find the maximum 2^ith` `        ``# parent the is differnet` `        ``# for both a and b` `        ``while` `(i > ``0` `and` `               ``dp[i][a] ``=``=` `dp[i][b]):` `            ``i``-``=``1` ` `  `        ``# Updating ans` `        ``ans ``=` `max``(ans, mx[i][a]);` `        ``ans ``=` `max``(ans, mx[i][b]);` ` `  `        ``# Changing value to ` `        ``# its parent` `        ``a ``=` `dp[i][a];` `        ``b ``=` `dp[i][b];` `    `  `    ``return` `ans;` ` `  `# Function to compute the Least` `# common Ansector` `def` `compute_lca():` `    `  `    ``dfs_lca(``1``, ``0``, ``0``);` `    ``find_ancestor();`   `# Driver code` `if` `__name__``=``=``"__main__"``:` `    `  `    ``# Undirected tree` `    ``n ``=` `5``;` `    ``v[``1``].append([``2``, ``2``]);` `    ``v[``2``].append([``1``, ``2``]);` `    ``v[``1``].append([``3``, ``5``]);` `    ``v[``3``].append([``1``, ``5``]);` `    ``v[``3``].append([``4``, ``3``]);` `    ``v[``4``].append([``3``, ``4``]);` `    ``v[``3``].append([``5``, ``1``]);` `    ``v[``5``].append([``3``, ``1``]);` ` `  `    ``# Computing LCA` `    ``compute_lca();` ` `  `    ``queries``=` `[[``3``, ``5``], [``2``, ``3``], [``2``,``4``]]` `    ``q ``=` `3``;` `    `  `    ``for` `i ``in` `range``(q):` `        ``max_edge ``=` `getMax(queries[i][``0``],` `                          ``queries[i][``1``]);` `        ``print``(max_edge)` `        `  `# This code is contributed by Rutvik_56`

Output:

```1
5
5

```

Time Complexity:  My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : rutvik_56