Product of nodes at k-th level in a tree represented as string
Last Updated :
04 Aug, 2022
Given an integer ‘K’ and a binary tree in string format. Every node of a tree has value in range from 0 to 9. We need to find product of elements at K-th level from root. The root is at level 0.
Note : Tree is given in the form: (node value(left subtree)(right subtree))
Examples:
Input : tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
k = 2
Output : 72
Its tree representation is shown below
Elements at level k = 2 are 6, 4, 1, 3
sum of the digits of these elements = 6 * 4 * 1 * 3 = 72
Input : tree = "(8(3(2()())(6(5()())()))(5(10()())(7(13()())())))"
k = 3
Output : 15
Elements at level k = 3 are 5, 1 and 3
sum of digits of these elements = 5 * 1 * 3 = 15
Approach :
- Input ‘tree’ in string format and level k
- Initialize level = -1 and product = 1
- for each character ‘ch’ in ‘tree’
3.1 if ch == ‘(‘ then
–> level++
3.2 else if ch == ‘)’ then
–> level–
3.3 else
if level == k then
product = product * (ch-‘0’)
- 4. Print product
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int productAtKthLevel(string tree, int k)
{
int level = -1;
int product = 1;
int n = tree.length();
for ( int i = 0; i < n; i++) {
if (tree[i] == '(' )
level++;
else if (tree[i] == ')' )
level--;
else {
if (level == k)
product *= (tree[i] - '0' );
}
}
return product;
}
int main()
{
string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" ;
int k = 2;
cout << productAtKthLevel(tree, k);
return 0;
}
|
Java
class GFG
{
static int productAtKthLevel(String tree, int k)
{
int level = - 1 ;
int product = 1 ;
int n = tree.length();
for ( int i = 0 ; i < n; i++)
{
if (tree.charAt(i) == '(' )
level++;
else if (tree.charAt(i) == ')' )
level--;
else
{
if (level == k)
product *= (tree.charAt(i) - '0' );
}
}
return product;
}
public static void main(String[] args)
{
String tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" ;
int k = 2 ;
System.out.println(productAtKthLevel(tree, k));
}
}
|
Python3
def productAtKthLevel(tree, k):
level = - 1
product = 1
n = len (tree)
for i in range ( 0 , n):
if (tree[i] = = '(' ):
level + = 1
elif (tree[i] = = ')' ):
level - = 1
else :
if (level = = k):
product * = ( int (tree[i]) - int ( '0' ))
return product
tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))"
k = 2
print (productAtKthLevel(tree, k))
|
C#
using System;
class GFG
{
static int productAtKthLevel( string tree,
int k)
{
int level = -1;
int product = 1;
int n = tree.Length;
for ( int i = 0; i < n; i++)
{
if (tree[i] == '(' )
level++;
else if (tree[i] == ')' )
level--;
else
{
if (level == k)
product *= (tree[i] - '0' );
}
}
return product;
}
static void Main()
{
string tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" ;
int k = 2;
Console.WriteLine(productAtKthLevel(tree, k));
}
}
|
PHP
<?php
function productAtKthLevel( $tree , $k )
{
$level = -1;
$product = 1;
$n = strlen ( $tree );
for ( $i = 0; $i < $n ; $i ++)
{
if ( $tree [ $i ] == '(' )
$level ++;
else if ( $tree [ $i ] == ')' )
$level --;
else
{
if ( $level == $k )
$product *= (ord( $tree [ $i ]) -
ord( '0' ));
}
}
return $product ;
}
$tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" ;
$k = 2;
echo productAtKthLevel( $tree , $k );
?>
|
Javascript
<script>
function productAtKthLevel(tree, k)
{
let level = -1;
let product = 1;
let n = tree.length;
for (let i = 0; i < n; i++)
{
if (tree[i] == '(' )
level++;
else if (tree[i] == ')' )
level--;
else
{
if (level == k)
product *= (tree[i].charCodeAt() - '0' .charCodeAt());
}
}
return product;
}
let tree = "(0(5(6()())(4()(9()())))(7(1()())(3()())))" ;
let k = 2;
document.write(productAtKthLevel(tree, k));
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
Please suggest if someone has a better solution which is more efficient in terms of space and time.
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