Count the nodes of the tree which make a pangram when concatenated with the sub-tree nodes

Given a tree, and the weights (in the form of strings) of all the nodes, the task is to count the nodes whose weighted string when concatenated with the strings of the sub-tree nodes becomes a pangram.
Pangram: A pangram is a sentence containing every letter of the English Alphabet.

Examples:

Input:

Output: 1
Only the weighted string of sub-tree of node 1 makes the pangram.

Approach: Perform dfs on the tree and update the weight of every node such that it stores its weight concatenated with the weights of the sub-tree nodes. Then, count the nodes whose updated weighted string forms a pangram.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
vector<int> graph[100];
vector<string> weight(100);
  
// Function that returns if the
// string x is a pangram
bool Pangram(string x)
{
    map<char, int> mp;
    int n = x.size();
  
    for (int i = 0; i < n; i++)
        mp[x[i]]++;
    if (mp.size() == 26)
        return true;
    else
        return false;
}
  
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
int countTotalPangram(int n)
{
    int cnt = 0;
    for (int i = 1; i <= n; i++)
        if (Pangram(weight[i]))
            cnt++;
    return cnt;
}
  
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
void dfs(int node, int parent)
{
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
        weight[node] += weight[to];
    }
}
  
// Driver code
int main()
{
    int n = 6;
  
    // Weights of the nodes
    weight[1] = "abcde";
    weight[2] = "fghijkl";
    weight[3] = "abcdefg";
    weight[4] = "mnopqr";
    weight[5] = "stuvwxy";
    weight[6] = "zabcdef";
  
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
    graph[5].push_back(6);
  
    dfs(1, 1);
  
    cout << countTotalPangram(n);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach
graph = [[] for i in range(100)]
weight = [0] * 100
  
# Function that returns if the
# string x is a pangram
def Pangram(x):
    mp = {}
    n = len(x)
    for i in range(n):
        if x[i] not in mp:
            mp[x[i]] = 0
        mp[x[i]] += 1
    if (len(mp)== 26):
        return True
    else:
        return False
  
# Function to return the count of nodes
# which make pangram with the
# sub-tree nodes
def countTotalPangram(n):
    cnt = 0
    for i in range(1, n + 1):
        if (Pangram(weight[i])):
            cnt += 1
    return cnt
  
# Function to perform dfs and update the nodes
# such that weight[i] will store the weight[i]
# concatenated with the weights of
# all the nodes in the sub-tree
def dfs(node, parent):
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
        weight[node] += weight[to]
  
# Driver code
n = 6
  
# Weights of the nodes
weight[1] = "abcde"
weight[2] = "fghijkl"
weight[3] = "abcdefg"
weight[4] = "mnopqr"
weight[5] = "stuvwxy"
weight[6] = "zabcdef"
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
  
dfs(1, 1)
print(countTotalPangram(n))
  
# This code is contributed by SHUBHAMSINGH10

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Output:

1

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