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Maximum weighted edge in path between two nodes in an N-ary tree using binary lifting
  • Difficulty Level : Hard
  • Last Updated : 09 Nov, 2020

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 

2^{i}

. We will store the maximum weighted edge till 



2^{i}

level.

dp[i][j] = dp[i - 1][dp[i - 1][j]]

and 
 

mx[i][j] = max(mx[i - 1][j], mx[i - 1][dp[i - 1][j]])

where

  • j is the node and
  • i is the distance of 

2^{i}

  • dp[i][j] stores the parent of j at 

2^{i}



  • 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 

2^{i}

  • distance.

We’ll do a depth-first search to find all the parents at 

2^{0}

distance and their weight and then precompute parents and maximum edges at every 

2^{i}

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 <bits/stdc++.h>
 
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<vector<int> > dp(LG, vector<int>(N));
 
// Maximum node at every 2^i level
vector<vector<int> > mx(LG, vector<int>(N));
 
// Graph that stores destinations
// and its weight
vector<vector<pair<int, int> > > 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[0][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[0][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[1].push_back(make_pair(2, 2));
    v[2].push_back(make_pair(1, 2));
    v[1].push_back(make_pair(3, 5));
    v[3].push_back(make_pair(1, 5));
    v[3].push_back(make_pair(4, 3));
    v[4].push_back(make_pair(3, 4));
    v[3].push_back(make_pair(5, 1));
    v[5].push_back(make_pair(3, 1));
 
    // Computing LCA
    compute_lca();
 
    int queries[][2]
        = { { 3, 5 },
            { 2, 3 },
            { 2, 4 } };
    int q = 3;
 
    for (int i = 0; i < q; i++) {
        int max_edge = getMax(queries[i][0],
                              queries[i][1]);
        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: 

O(N*logN)

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