Given an N-ary tree, and the weights which are in the form of strings of all the nodes, the task is to count the number of leaf nodes whose weights are palindrome.
Examples:
Input: 1(ab) / \ (abca)2 5 (aba) / \ (axxa)3 4 (geeks) Output: 2 Explanation: Only the weights of the leaf nodes "axxa" and "aba" are palindromes. Input: 1(abx) / 2(abaa) / 3(amma) Output: 1 Explanation: Only the weight of the leaf node "amma" is palindrome.
Approach: To solve the problem mentioned above follow the steps given below:
- Depth First Search can be used to traverse the complete tree.
- We will keep track of parent while traversing to avoid the visited node array.
- Initially for every node we can set a flag and if the node have at least one child (i.e. non-leaf node) then we will reset the flag.
- The nodes with no children are the leaf nodes. For every leaf node, we will check if it’s string is palindrome or not. If yes 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;
int cnt = 0;
vector< int > graph[100];
vector<string> weight(100); // Function that returns true // if x is a palindrome bool isPalindrome(string x)
{ int n = x.size();
for ( int i = 0; i < n / 2; i++) {
if (x[i] != x[n - 1 - i])
return false ;
}
return true ;
} // Function to perform DFS on the tree void dfs( int node, int parent)
{ int flag = 1;
// Iterating the children of current node
for ( int to : graph[node]) {
// There is at least a child
// of the current node
if (to == parent)
continue ;
flag = 0;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1) {
// Weight of the current node
string x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
} // Driver code int main()
{ // Weights of the node
weight[1] = "ab" ;
weight[2] = "abca" ;
weight[3] = "axxa" ;
weight[4] = "geeks" ;
weight[5] = "aba" ;
// Edges of the tree
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
dfs(1, 1);
cout << cnt;
return 0;
} |
Java
// Java implementation of the approach import java.util.*;
class GFG{
static int cnt = 0 ;
static Vector<Integer> []graph = new Vector[ 100 ];
static String []weight = new String[ 100 ];
// Function that returns true // if x is a palindrome static boolean isPalindrome(String x)
{ int n = x.length();
for ( int i = 0 ; i < n / 2 ; i++)
{
if (x.charAt(i) != x.charAt(n - 1 - i))
return false ;
}
return true ;
} // Function to perform DFS on the tree static void dfs( int node, int parent)
{ int flag = 1 ;
// Iterating the children of current node
for ( int to : graph[node])
{
// There is at least a child
// of the current node
if (to == parent)
continue ;
flag = 0 ;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1 )
{
// Weight of the current node
String x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1 ;
}
} // Driver code public static void main(String[] args)
{ for ( int i = 0 ; i < graph.length;i++)
graph[i] = new Vector<Integer>();
// Weights of the node
weight[ 1 ] = "ab" ;
weight[ 2 ] = "abca" ;
weight[ 3 ] = "axxa" ;
weight[ 4 ] = "geeks" ;
weight[ 5 ] = "aba" ;
// Edges of the tree
graph[ 1 ].add( 2 );
graph[ 2 ].add( 3 );
graph[ 2 ].add( 4 );
graph[ 1 ].add( 5 );
dfs( 1 , 1 );
System.out.print(cnt);
} } // This code is contributed by amal kumar choubey |
Python3
# Python3 implementation of the approach cnt = 0
graph = [ 0 ] * 100
for i in range ( 100 ):
graph[i] = []
weight = [ 0 ] * 100
# Function that returns true # if x is a palindrome def isPalindrome(x: str ) - > bool :
n = len (x)
for i in range (n / / 2 ):
if (x[i] ! = x[n - 1 - i]):
return False
return True
# Function to perform DFS on the tree def dfs(node: int , parent: int ) - > None :
global cnt, graph, weight
flag = 1
# Iterating the children of current node
for to in graph[node]:
# There is at least a child
# of the current node
if (to = = parent):
continue
flag = 0
dfs(to, node)
# Current node is connected to only
# its parent i.e. it is a leaf node
if (flag = = 1 ):
# Weight of the current node
x = weight[node]
# If the weight is a palindrome
if (isPalindrome(x)):
cnt + = 1
# Driver code if __name__ = = "__main__" :
# Weights of the node
weight[ 1 ] = "ab"
weight[ 2 ] = "abca"
weight[ 3 ] = "axxa"
weight[ 4 ] = "geeks"
weight[ 5 ] = "aba"
# Edges of the tree
graph[ 1 ].append( 2 )
graph[ 2 ].append( 3 )
graph[ 2 ].append( 4 )
graph[ 1 ].append( 5 )
dfs( 1 , 1 )
print (cnt)
# This code is contributed by sanjeev2552 |
C#
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG{
static int cnt = 0;
static List< int > []graph = new List< int >[100];
static String []weight = new String[100];
// Function that returns true // if x is a palindrome static bool isPalindrome(String x)
{ int n = x.Length;
for ( int i = 0; i < n / 2; i++)
{
if (x[i] != x[n - 1 - i])
return false ;
}
return true ;
} // Function to perform DFS on the tree static void dfs( int node, int parent)
{ int flag = 1;
// Iterating the children of
// current node
foreach ( int to in graph[node])
{
// There is at least a child
// of the current node
if (to == parent)
continue ;
flag = 0;
dfs(to, node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1)
{
// Weight of the current node
String x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
} // Driver code public static void Main(String[] args)
{ for ( int i = 0; i < graph.Length; i++)
graph[i] = new List< int >();
// Weights of the node
weight[1] = "ab" ;
weight[2] = "abca" ;
weight[3] = "axxa" ;
weight[4] = "geeks" ;
weight[5] = "aba" ;
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
Console.Write(cnt);
} } // This code is contributed by amal kumar choubey |
Javascript
<script> // JavaScript implementation of the approach
let cnt = 0;
let graph = new Array(100);
let weight = new Array(100);
// Function that returns true
// if x is a palindrome
function isPalindrome(x)
{
let n = x.length;
for (let i = 0; i < parseInt(n / 2, 10); i++)
{
if (x[i] != x[n - 1 - i])
return false ;
}
return true ;
}
// Function to perform DFS on the tree
function dfs(node, parent)
{
let flag = 1;
// Iterating the children of current node
for (let to = 0; to < graph[node].length; to++)
{
// There is at least a child
// of the current node
if (graph[node][to] == parent)
continue ;
flag = 0;
dfs(graph[node][to], node);
}
// Current node is connected to only
// its parent i.e. it is a leaf node
if (flag == 1)
{
// Weight of the current node
let x = weight[node];
// If the weight is a palindrome
if (isPalindrome(x))
cnt += 1;
}
}
for (let i = 0; i < graph.length;i++)
graph[i] = [];
// Weights of the node
weight[1] = "ab" ;
weight[2] = "abca" ;
weight[3] = "axxa" ;
weight[4] = "geeks" ;
weight[5] = "aba" ;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write(cnt);
</script> |
Output:
2
Time Complexity: O(N) where N is the number of nodes in the tree.
Auxiliary Space: O(1) as constant space is considered.