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++
// 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; } |
Java
// Java implementation of the approach import java.util.*; class GFG{ @SuppressWarnings ( "unchecked" ) static Vector<Integer> []graph = new Vector[ 100 ]; static String []weight = new String[ 100 ]; // Function that returns if the // String x is a pangram static boolean Pangram(String x) { HashMap<Character, Integer> mp = new HashMap<>(); int n = x.length(); for ( int i = 0 ; i < n; i++) { if (mp.containsKey(x.charAt(i))) { mp.put(x.charAt(i), mp.get(x.charAt(i)) + 1 ); } else { mp.put(x.charAt(i), 1 ); } } if (mp.size() == 26 ) return true ; else return false ; } // Function to return the count of nodes // which make pangram with the // sub-tree nodes static 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 static void dfs( int node, int parent) { for ( int to : graph[node]) { if (to == parent) continue ; dfs(to, node); weight[node] += weight[to]; } } // Driver code public static void main(String[] args) { 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" ; for ( int i = 0 ; i < graph.length; i++) graph[i] = new Vector<Integer>(); // Edges of the tree graph[ 1 ].add( 2 ); graph[ 2 ].add( 3 ); graph[ 2 ].add( 4 ); graph[ 1 ].add( 5 ); graph[ 5 ].add( 6 ); dfs( 1 , 1 ); System.out.print(countTotalPangram(n)); } } // This code is contributed by Amit Katiyar |
Python3
# 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 |
C#
// C# implementation of // the above approach using System; using System.Collections.Generic; class GFG{ static List< int > []graph = new List< int >[100]; static String []weight = new String[100]; // Function that returns if the // String x is a pangram static bool Pangram(String x) { Dictionary< char , int > mp = new Dictionary< char , int >(); int n = x.Length; for ( int i = 0 ; i < n; i++) { if (mp.ContainsKey(x[i])) { mp[x[i]] = mp[x[i]] + 1; } else { mp.Add(x[i], 1); } } if (mp.Count == 26) return true ; else return false ; } // Function to return the // count of nodes which // make pangram with the // sub-tree nodes static 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 static void dfs( int node, int parent) { foreach ( int to in graph[node]) { if (to == parent) continue ; dfs(to, node); weight[node] += weight[to]; } } // Driver code public static void Main(String[] args) { 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" ; for ( int i = 0; i < graph.Length; i++) graph[i] = new List< int >(); // Edges of the tree graph[1].Add(2); graph[2].Add(3); graph[2].Add(4); graph[1].Add(5); graph[5].Add(6); dfs(1, 1); Console.Write(countTotalPangram(n)); } } // This code is contributed by shikhasingrajput |
1
Complexity Analysis:
- Time Complexity: O(N*S).
In dfs, every node of the tree is processed once, and hence the complexity due to the dfs is O(N) if there are total N nodes in the tree. Also, for processing each node the Pangram() function is used for every node which has a complexity of O(S) where S is the sum of the length of all weight strings in a subtree and since this is done for every node, the overall time complexity for this part becomes O(N*S). Therefore, the final time complexity is O(N*S). - Auxiliary Space: O(1).
Any extra space is not required, so the space complexity is constant.
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