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 is an anagram with the given string str.
Examples:
Input:
str = “geek”
Output: 2
Only the weighted strings of the nodes 2 and 6
are anagrams of the given string “geek”.
Approach: Perform dfs on the tree and for every node, check if it’s weighted string is anagram with the given string or not, If not then increment the count.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
string s; int cnt = 0;
vector< int > graph[100];
vector<string> weight(100); // Function that return true if both // the strings are anagram of each other bool anagram(string x, string s)
{ sort(x.begin(), x.end());
sort(s.begin(), s.end());
if (x == s)
return true ;
else
return false ;
} // Function to perform dfs void dfs( int node, int parent)
{ // If current node's weighted
// string is an anagram of
// the given string s
if (anagram(weight[node], s))
cnt += 1;
for ( int to : graph[node]) {
if (to == parent)
continue ;
dfs(to, node);
}
} // Driver code int main()
{ s = "geek" ;
// Weights of the nodes
weight[1] = "eeggk" ;
weight[2] = "geek" ;
weight[3] = "gekrt" ;
weight[4] = "tree" ;
weight[5] = "eetr" ;
weight[6] = "egek" ;
// 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 << cnt;
return 0;
} |
Java
// Java implementation of the approach import java.util.*;
class GFG
{ static String s;
static int cnt = 0 ;
static Vector<Integer>[] graph = new Vector[ 100 ];
static String[] weight = new String[ 100 ];
// Function that return true if both
// the Strings are anagram of each other
static boolean anagram(String x, String s)
{
x = sort(x);
s = sort(s);
if (x.equals(s))
return true ;
else
return false ;
}
static String sort(String inputString)
{
// convert input string to char array
char tempArray[] = inputString.toCharArray();
// sort tempArray
Arrays.sort(tempArray);
// return new sorted string
return new String(tempArray);
}
// Function to perform dfs
static void dfs( int node, int parent)
{
// If current node's weighted
// String is an anagram of
// the given String s
if (anagram(weight[node], s))
cnt += 1 ;
for ( int to : graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
// Driver code
public static void main(String[] args)
{
s = "geek" ;
for ( int i = 0 ; i < 100 ; i++)
graph[i] = new Vector<Integer>();
// Weights of the nodes
weight[ 1 ] = "eeggk" ;
weight[ 2 ] = "geek" ;
weight[ 3 ] = "gekrt" ;
weight[ 4 ] = "tree" ;
weight[ 5 ] = "eetr" ;
weight[ 6 ] = "egek" ;
// 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(cnt);
}
} // This code is contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach cnt = 0
graph = [[] for i in range ( 100 )]
weight = [ 0 ] * 100
# Function that return true if both # the strings are anagram of each other def anagram(x, s):
x = sorted ( list (x))
s = sorted ( list (s))
if (x = = s):
return True
else :
return False
# Function to perform dfs def dfs(node, parent):
global cnt, s
# If weight of the current node
# string is an anagram of
# the given string s
if (anagram(weight[node], s)):
cnt + = 1
for to in graph[node]:
if (to = = parent):
continue
dfs(to, node)
# Driver code s = "geek"
# Weights of the nodes weight[ 1 ] = "eeggk"
weight[ 2 ] = "geek"
weight[ 3 ] = "gekrt"
weight[ 4 ] = "tree"
weight[ 5 ] = "eetr"
weight[ 6 ] = "egek"
# 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 (cnt)
# This code is contributed by SHUBHAMSINGH10 |
C#
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ static String s;
static int cnt = 0;
static List< int >[] graph = new List< int >[100];
static String[] weight = new String[100];
// Function that return true if both
// the Strings are anagram of each other
static bool anagram(String x, String s)
{
x = sort(x);
s = sort(s);
if (x.Equals(s))
return true ;
else
return false ;
}
static String sort(String inputString)
{
// convert input string to char array
char []tempArray = inputString.ToCharArray();
// sort tempArray
Array.Sort(tempArray);
// return new sorted string
return new String(tempArray);
}
// Function to perform dfs
static void dfs( int node, int parent)
{
// If current node's weighted
// String is an anagram of
// the given String s
if (anagram(weight[node], s))
cnt += 1;
foreach ( int to in graph[node])
{
if (to == parent)
continue ;
dfs(to, node);
}
}
// Driver code
public static void Main(String[] args)
{
s = "geek" ;
for ( int i = 0; i < 100; i++)
graph[i] = new List< int >();
// Weights of the nodes
weight[1] = "eeggk" ;
weight[2] = "geek" ;
weight[3] = "gekrt" ;
weight[4] = "tree" ;
weight[5] = "eetr" ;
weight[6] = "egek" ;
// 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(cnt);
}
} // This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the approach let s; let cnt = 0; let graph = new Array(100);
let weight = new Array(100);
for (let i = 0; i < 100; i++)
{ graph[i] = [];
weight[i] = 0;
} const sort1 = str => str.split( '' ).sort((a, b) => a.localeCompare(b)).join( '' );
// Function that return true if both // the strings are anagram of each other function anagram(x, s)
{ x = sort1(x);
s = sort1(s);
if (x == s)
return true ;
else
return false ;
} // Function to perform dfs function dfs(node, parent)
{ // If current node's weighted
// string is an anagram of
// the given string s
if (anagram(weight[node], s))
cnt += 1;
for (let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue
dfs(graph[node][to], node);
}
} // Driver code s = "geek" ;
// Weights of the nodes
weight[1] = "eeggk" ;
weight[2] = "geek" ;
weight[3] = "gekrt" ;
weight[4] = "tree" ;
weight[5] = "eetr" ;
weight[6] = "egek" ;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
graph[5].push(6);
dfs(1, 1);
document.write(cnt);
// This code is contributed by Dharanendra L V.
</script> |
Output:
2
Complexity Analysis:
-
Time Complexity : O(N*(S*log(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 sort() function is used which has a complexity of O(S*log(S)) where S is the length of the weighted string. Therefore, the time complexity is O(N*(S*log(S))) where S is the maximum length of the weight string in the tree. -
Auxiliary Space : O(1).
Any extra space is not required, so the space complexity is constant.