Number of Paths of Weight W in a K-ary tree
Last Updated :
12 Sep, 2022
Given a K-ary tree, where each node is having K children and each edge has some weight. All the edges i.e. K, that goes from a particular node to all its children have weights in ascending order 1, 2, 3, …, K. Find the number of paths having total weight as W (sum of all edge weights in the path) starting from root and containing atleast one edge of weight atleast M.
Examples:
Input : W = 3, K = 3, M = 2
Output : 3
Explanation : One path can be (1 + 2), second can be (2 + 1) and third is 3.
Input : W = 4, K = 3, M = 2
Output : 6
Approach:
This problem can be solved using dynamic programming approach. The idea is to maintain two states, one for the current weight to be required and other one for a boolean variable which denotes that the current path has included an edge of weight atleast M or not. Iterate over all possible edge weights i.e. K and recursively solve for the weight W – i for 1 ≤ i ≤ K. If the current edge weight is more than or equal to M, set the boolean variable as 1 for the next call.
Below is the implementation of above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int solve( int dp[][2], int wt, int K, int M,
int used)
{
if (wt < 0)
return 0;
if (wt == 0) {
if (used)
return 1;
return 0;
}
if (dp[wt][used] != -1)
return dp[wt][used];
int ans = 0;
for ( int i = 1; i <= K; i++) {
if (i >= M)
ans += solve(dp, wt - i,
K, M, used | 1);
else
ans += solve(dp, wt - i,
K, M, used);
}
return dp[wt][used] = ans;
}
int main()
{
int W = 3, K = 3, M = 2;
int dp[W + 1][2];
memset (dp, -1, sizeof (dp));
cout << solve(dp, W, K, M, 0) << endl;
return 0;
}
|
Java
class GFG
{
public static int solve( int [][] dp, int wt,
int K, int M, int used)
{
if (wt < 0 )
{
return 0 ;
}
if (wt == 0 )
{
if (used == 1 )
{
return 1 ;
}
return 0 ;
}
if (dp[wt][used] != - 1 )
{
return dp[wt][used];
}
int ans = 0 ;
for ( int i = 1 ; i <= K; i++)
{
if (i >= M)
{
ans += solve(dp, wt - i,
K, M, used | 1 );
}
else
{
ans += solve(dp, wt - i,
K, M, used);
}
}
return dp[wt][used] = ans;
}
public static void main(String[] args)
{
int W = 3 , K = 3 , M = 2 ;
int [][] dp = new int [W + 1 ][ 2 ];
for ( int i = 0 ; i < W + 1 ; i++)
{
for ( int j = 0 ; j < 2 ; j++)
{
dp[i][j] = - 1 ;
}
}
System.out.print(solve(dp, W, K, M, 0 ) + "\n" );
}
}
|
Python3
import numpy as np
def solve(dp, wt, K, M, used) :
if (wt < 0 ) :
return 0
if (wt = = 0 ) :
if (used) :
return 1
return 0
if (dp[wt][used] ! = - 1 ) :
return dp[wt][used]
ans = 0
for i in range ( 1 , K + 1 ) :
if (i > = M) :
ans + = solve(dp, wt - i,
K, M, used | 1 )
else :
ans + = solve(dp, wt - i,
K, M, used)
dp[wt][used] = ans
return ans
if __name__ = = "__main__" :
W = 3
K = 3
M = 2
dp = np.ones((W + 1 , 2 ));
dp = - 1 * dp
print (solve(dp, W, K, M, 0 ))
|
C#
using System;
class GFG
{
public static int solve( int [,] dp, int wt, int K, int M, int used)
{
if (wt < 0)
return 0;
if (wt == 0) {
if (used == 1)
return 1;
return 0;
}
if (dp[wt,used] != -1)
return dp[wt,used];
int ans = 0;
for ( int i = 1; i <= K; i++) {
if (i >= M)
ans += solve(dp, wt - i,
K, M, used | 1);
else
ans += solve(dp, wt - i,
K, M, used);
}
return dp[wt,used] = ans;
}
static void Main()
{
int W = 3, K = 3, M = 2;
int [,] dp = new int [W + 1,2];
for ( int i = 0;i < W + 1; i++)
for ( int j = 0; j < 2; j++)
dp[i,j] = -1;
Console.Write(solve(dp, W, K, M, 0) + "\n" );
}
}
|
Javascript
<script>
function solve(dp, wt, K, M, used)
{
if (wt < 0)
{
return 0;
}
if (wt == 0)
{
if (used == 1)
{
return 1;
}
return 0;
}
if (dp[wt][used] != -1)
{
return dp[wt][used];
}
let ans = 0;
for (let i = 1; i <= K; i++)
{
if (i >= M)
{
ans += solve(dp, wt - i,
K, M, used | 1);
}
else
{
ans += solve(dp, wt - i,
K, M, used);
}
}
return dp[wt][used] = ans;
}
let W = 3, K = 3, M = 2;
let dp = new Array(W + 1);
for (let i = 0; i < W + 1; i++)
{
dp[i] = new Array(2);
for (let j = 0; j < 2; j++)
{
dp[i][j] = -1;
}
}
document.write(solve(dp, W, K, M, 0) + "</br>" );
</script>
|
Time Complexity: O(W * K)
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