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Maximum absolute difference between any two level sum in a N-ary Tree

  • Last Updated : 21 Jun, 2021

Given an N-ary Tree having N nodes with positive and negative values and (N – 1) edges, the task is to find the maximum absolute difference of level sum in it.

Examples:

Input: N = 8, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}}, Value[] = {4,2, 3, -5,-1, 3, -2, 6}, Below is the graph: 
 

Output: 6
Explanation:
Sum of all nodes of 0th level is 4.
Sum of all nodes of 1st level is 0.
Sum of all nodes of 2nd level is 6.
Hence, maximum absolute difference of level sum = (6 – 0) = 6.



Input: N = 10, Edges[][2] = {{0, 1}, {0, 2}, {0, 3}, {1, 4}, {1, 5}, {3, 6}, {6, 7}, {6, 8}, {6, 9}}, Value[] = {1, 2, -1, 3, 4, 5, 8, 6, 12, 7}, Below is the graph:

Output: 24

Approach: To find the maximum absolute difference of level sum, first find the maximum level sum and minimum level sum because the absolute difference of maximum level sum and minimum level sum always gives us maximum absolute difference i.e.,

Maximum absolute difference = abs(Maximum level sum – Minimum level sum)

Below are the steps:

  1. Perform the BFS Traversal on the give N-ary tree.
  2. While doing the BFS traversal, process nodes of different levels separately.
  3. For every level being processed, compute the sum of nodes in the level and keep track of maximum and minimum level sum.
  4. After the above traversal, find the absolute difference of maximum and minimum level sum.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// absolute difference of level sum
void maxAbsDiffLevelSum(int N, int M,
                        vector<int> cost,
                        int Edges[][2])
{
    // Create the adjacency list
    vector<int> adj[N];
 
    for (int i = 0; i < M; i++) {
        int u = Edges[i][0];
        int v = Edges[i][1];
        adj[u].push_back(v);
    }
 
    // Initialize value of maximum
    // and minimum level sum
    int maxSum = cost[0], minSum = cost[0];
 
    // Do Level order traversal keeping
    // track of nodes at every level
    queue<int> q;
    q.push(0);
 
    while (!q.empty()) {
 
        // Get the size of queue when
        // the level order traversal
        // for one level finishes
        int count = q.size();
 
        int sum = 0;
 
        // Iterate for all the nodes
        // in the queue currently
        while (count--) {
 
            // Dequeue an node from queue
            int temp = q.front();
            q.pop();
 
            sum = sum + cost[temp];
 
            // Enqueue the children of
            // dequeued node
            for (int i = 0;
                 i < adj[temp].size(); i++) {
 
                q.push(adj[temp][i]);
            }
        }
 
        // Update the maximum level
        // sum value
        maxSum = max(sum, maxSum);
 
        // Update the minimum level
        // sum value
        minSum = min(sum, minSum);
    }
 
    // Return the result
    cout << abs(maxSum - minSum);
}
 
// Driver Code
int main()
{
    // Number of nodes and edges
    int N = 10, M = 9;
 
    // Edges of the N-ary tree
    int Edges[][2] = { { 0, 1 }, { 0, 2 },
                       { 0, 3 }, { 1, 4 },
                       { 1, 5 }, { 3, 6 },
                       { 6, 7 }, { 6, 8 },
                       { 6, 9 } };
 
    // Given cost
    vector<int> cost = { 1, 2, -1, 3, 4,
                         5, 8, 6, 12, 7 };
 
    // Function Call
    maxAbsDiffLevelSum(N, M, cost, Edges);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the maximum
// absolute difference of level sum
static void maxAbsDiffLevelSum(int N, int M,
                               int []cost,
                               int Edges[][])
{
     
    // Create the adjacency list
    @SuppressWarnings("unchecked")
    Vector<Integer> []adj = new Vector[N];
    for(int i = 0; i < adj.length; i++)
        adj[i] = new Vector<Integer>();
         
    for(int i = 0; i < M; i++)
    {
        int u = Edges[i][0];
        int v = Edges[i][1];
        adj[u].add(v);
    }
 
    // Initialize value of maximum
    // and minimum level sum
    int maxSum = cost[0], minSum = cost[0];
 
    // Do Level order traversal keeping
    // track of nodes at every level
    Queue<Integer> q = new LinkedList<Integer>();
    q.add(0);
 
    while (!q.isEmpty())
    {
         
        // Get the size of queue when
        // the level order traversal
        // for one level finishes
        int count = q.size();
 
        int sum = 0;
 
        // Iterate for all the nodes
        // in the queue currently
        while (count-- > 0)
        {
 
            // Dequeue an node from queue
            int temp = q.peek();
            q.remove();
 
            sum = sum + cost[temp];
 
            // Enqueue the children of
            // dequeued node
            for(int i = 0;
                    i < adj[temp].size();
                    i++)
            {
                q.add(adj[temp].get(i));
            }
        }
 
        // Update the maximum level
        // sum value
        maxSum = Math.max(sum, maxSum);
 
        // Update the minimum level
        // sum value
        minSum = Math.min(sum, minSum);
    }
 
    // Return the result
    System.out.print(Math.abs(maxSum - minSum));
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Number of nodes and edges
    int N = 10, M = 9;
 
    // Edges of the N-ary tree
    int Edges[][] = { { 0, 1 }, { 0, 2 },
                      { 0, 3 }, { 1, 4 },
                      { 1, 5 }, { 3, 6 },
                      { 6, 7 }, { 6, 8 },
                      { 6, 9 } };
 
    // Given cost
    int []cost = { 1, 2, -1, 3, 4,
                   5, 8, 6, 12, 7 };
 
    // Function call
    maxAbsDiffLevelSum(N, M, cost, Edges);
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program for the above approach
from collections import deque
 
# Function to find the maximum
# absolute difference of level sum
def maxAbsDiffLevelSum(N, M, cost, Edges):
 
    # Create the adjacency list
    adj = [[] for i in range(N)]
 
    for i in range(M):
        u = Edges[i][0]
        v = Edges[i][1]
        adj[u].append(v)
 
    # Initialize value of maximum
    # and minimum level sum
    maxSum = cost[0]
    minSum = cost[0]
 
    # Do Level order traversal keeping
    # track of nodes at every level
    q = deque()
    q.append(0)
 
    while len(q) > 0:
 
        # Get the size of queue when
        # the level order traversal
        # for one level finishes
        count = len(q)
 
        sum = 0
 
        # Iterate for all the nodes
        # in the queue currently
        while (count):
 
            # Dequeue an node from queue
            temp = q.popleft()
            # q.pop()
 
            sum = sum + cost[temp]
 
            # Enqueue the children of
            # dequeued node
            for i in adj[temp]:
                q.append(i)
                 
            count -= 1
 
        # Update the maximum level
        # sum value
        maxSum = max(sum, maxSum)
 
        # Update the minimum level
        # sum value
        minSum = min(sum, minSum)
 
    # Return the result
    print(abs(maxSum - minSum))
 
# Driver Code
if __name__ == '__main__':
     
    # Number of nodes and edges
    N = 10
    M = 9
 
    # Edges of the N-ary tree
    Edges = [ [ 0, 1 ], [ 0, 2 ],
              [ 0, 3 ], [ 1, 4 ],
              [ 1, 5 ], [ 3, 6 ],
              [ 6, 7 ], [ 6, 8 ],
              [ 6, 9 ] ]
 
    # Given cost
    cost = [ 1, 2, -1, 3, 4,
             5, 8, 6, 12, 7 ]
 
    # Function call
    maxAbsDiffLevelSum(N, M, cost, Edges)
 
# This code is contributed by mohit kumar 29

C#




// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to find the maximum
// absolute difference of level sum
static void maxAbsDiffLevelSum(int N, int M,
                               int []cost,
                               int [,]Edges)
{
  // Create the adjacency list
  List<int> []adj = new List<int>[N];
  for(int i = 0; i < adj.Length; i++)
    adj[i] = new List<int>();
 
  for(int i = 0; i < M; i++)
  {
    int u = Edges[i, 0];
    int v = Edges[i, 1];
    adj[u].Add(v);
  }
 
  // Initialize value of maximum
  // and minimum level sum
  int maxSum = cost[0], minSum = cost[0];
 
  // Do Level order traversal keeping
  // track of nodes at every level
  Queue<int> q = new Queue<int>();
  q.Enqueue(0);
 
  while (q.Count!=0)
  {
    // Get the size of queue when
    // the level order traversal
    // for one level finishes
    int count = q.Count;
 
    int sum = 0;
 
    // Iterate for all the nodes
    // in the queue currently
    while (count-- > 0)
    {
      // Dequeue an node from queue
      int temp = q.Peek();
      q.Dequeue();
 
      sum = sum + cost[temp];
 
      // Enqueue the children of
      // dequeued node
      for(int i = 0; i < adj[temp].Count; i++)
      {
        q.Enqueue(adj[temp][i]);
      }
    }
 
    // Update the maximum level
    // sum value
    maxSum = Math.Max(sum, maxSum);
 
    // Update the minimum level
    // sum value
    minSum = Math.Min(sum, minSum);
  }
 
  // Return the result
  Console.Write(Math.Abs(maxSum - minSum));
}
 
// Driver Code
public static void Main(String[] args)
{   
  // Number of nodes and edges
  int N = 10, M = 9;
 
  // Edges of the N-ary tree
  int [,]Edges = {{0, 1}, {0, 2},
                  {0, 3}, {1, 4},
                  {1, 5}, {3, 6},
                  {6, 7}, {6, 8},
                  {6, 9}};
 
  // Given cost
  int []cost = {1, 2, -1, 3, 4,
                5, 8, 6, 12, 7};
 
  // Function call
  maxAbsDiffLevelSum(N, M, cost, Edges);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
    // JavaScript implementation of the above approach
     
    // Function to find the maximum
    // absolute difference of level sum
    function maxAbsDiffLevelSum(N, M, cost, Edges)
    {
      // Create the adjacency list
      let adj = new Array(N);
      for(let i = 0; i < adj.length; i++)
        adj[i] = [];
 
      for(let i = 0; i < M; i++)
      {
        let u = Edges[i][0];
        let v = Edges[i][1];
        adj[u].push(v);
      }
 
      // Initialize value of maximum
      // and minimum level sum
      let maxSum = cost[0], minSum = cost[0];
 
      // Do Level order traversal keeping
      // track of nodes at every level
      let q = [];
      q.push(0);
 
      while (q.length!=0)
      {
        // Get the size of queue when
        // the level order traversal
        // for one level finishes
        let count = q.length;
 
        let sum = 0;
 
        // Iterate for all the nodes
        // in the queue currently
        while (count-- > 0)
        {
          // Dequeue an node from queue
          let temp = q[0];
          q.shift();
 
          sum = sum + cost[temp];
 
          // Enqueue the children of
          // dequeued node
          for(let i = 0; i < adj[temp].length; i++)
          {
            q.push(adj[temp][i]);
          }
        }
 
        // Update the maximum level
        // sum value
        maxSum = Math.max(sum, maxSum);
 
        // Update the minimum level
        // sum value
        minSum = Math.min(sum, minSum);
      }
 
      // Return the result
      document.write(Math.abs(maxSum - minSum));
    }
     
    // Number of nodes and edges
    let N = 10, M = 9;
 
    // Edges of the N-ary tree
    let Edges = [[0, 1], [0, 2],
                    [0, 3], [1, 4],
                    [1, 5], [3, 6],
                    [6, 7], [6, 8],
                    [6, 9]];
 
    // Given cost
    let cost = [1, 2, -1, 3, 4, 5, 8, 6, 12, 7];
 
    // Function call
    maxAbsDiffLevelSum(N, M, cost, Edges);
     
</script>
Output: 
24

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

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