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Count unique paths with given sum in an N-ary Tree

  • Difficulty Level : Medium
  • Last Updated : 01 Oct, 2021

Given an integer X and integer N, the task is to find the number of unique paths starting from the root in N-ary tree such that the sum of all these paths is equal to X. The N-ary tree satisfies the following conditions:

  • All the nodes have N children and the weight of each edge is distinct and lies in the range [1, N].
  • The tree is extended up to infinity.

Examples:

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Input: N = 3, X = 2



Output: 2
Explanation: the two paths having path sum equal to 2 are {1, 1} and {2}.

Input: N = 3, X = 6
Output: 24

Naive Approach: The simplest approach is to recursively find all possible ways to obtain path sum equal to X and print the count obtained. 

Time Complexity: O(N * X)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Follow the steps below to solve the problem:

  • Initialize a dp[] array which for every ith index, stores the count of paths adding up to i.
  • For every vertex, iterate form strong>1 to min(X, N), i.e. all possible values of its children and find the number of paths possible with given sum from each node considered.
  • Add all the paths made using the edges 1 to N and check if count is already computed or not. If already computed, return the value. Otherwise, recursively count the number of paths with sum equal to current value by considering all possible ways to extend the tree from the current vertex.
  • Update the dp[] array and return the count obtained.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = (int)1e9 + 7;
 
// Function for counting total
// no of paths possible with
// the sum is equal to X
ll findTotalPath(int X, int n,
                 vector<int>& dp)
{
 
    // If the path of the sum
    // from the root to current
    // node is stored in sum
    if (X == 0) {
        return 1;
    }
    ll ans = 0;
 
    // If already computed
    if (dp[X] != -1) {
        return dp[X];
    }
 
    // Count different no of paths
    // using all possible ways
    for (int i = 1; i <= min(X, n); ++i) {
 
        ans += findTotalPath(
                   X - i, n, dp)
               % mod;
        ans %= mod;
    }
 
    // Return total no of paths
    return dp[X] = ans;
}
 
// Driver Code
int main()
{
 
    int n = 3, X = 2;
 
    // Stores the number of ways
    // to obtains sums 0 to X
    vector<int> dp(X + 1, -1);
 
    // Function call
    cout << findTotalPath(X, n, dp);
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
static int mod = (int)1e9 + 7;
 
// Function for counting total
// no of paths possible with
// the sum is equal to X
static int findTotalPath(int X, int n,
                         ArrayList<Integer> dp)
{
     
    // If the path of the sum
    // from the root to current
    // node is stored in sum
    if (X == 0)
    {
        return 1;
    }
    int ans = 0;
 
    // If already computed
    if (dp.get(X) != -1)
    {
        return dp.get(X);
    }
 
    // Count different no of paths
    // using all possible ways
    for(int i = 1; i <= Math.min(X, n); ++i)
    {
        ans += findTotalPath(X - i, n, dp) % mod;
        ans %= mod;
    }
 
    // Return total no of paths
    dp.set(X, ans);
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 3, X = 2;
     
    // Stores the number of ways
    // to obtains sums 0 to X
    ArrayList<Integer> dp = new ArrayList<Integer>(
        Collections.nCopies(X + 1, -1));
 
    // Function call
    System.out.print(findTotalPath(X, n, dp));
}
}
 
// This code is contributed by akhilsaini

Python3




# Python3 program for the above approach
mod = int(1e9 + 7)
 
# Function for counting total
# no of paths possible with
# the sum is equal to X
def findTotalPath(X, n, dp):
     
  # If the path of the sum
  # from the root to current
  # node is stored in sum
  if (X == 0):
    return 1
     
  ans = 0
 
  # If already computed
  if (dp[X] != -1):
    return dp[X]
     
  # Count different no of paths
  # using all possible ways
  for i in range(1, min(X, n) + 1):
    ans = ans + findTotalPath(X - i, n, dp) % mod;
    ans %= mod;
 
  # Return total no of paths
  dp[X] = ans
  return ans
 
# Driver Code
if __name__ == '__main__':
     
  n = 3
  X = 2
 
  # Stores the number of ways
  # to obtains sums 0 to X
  dp = [-1] * (X + 1)
 
  # Function call
  print(findTotalPath(X, n, dp))
   
# This code is contributed by akhilsaini

C#




// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
 
static int mod = (int)1e9 + 7;
 
// Function for counting total
// no of paths possible with
// the sum is equal to X
static int findTotalPath(int X, int n, int[] dp)
{
     
    // If the path of the sum
    // from the root to current
    // node is stored in sum
    if (X == 0)
    {
        return 1;
    }
    int ans = 0;
 
    // If already computed
    if (dp[X] != -1)
    {
        return dp[X];
    }
 
    // Count different no of paths
    // using all possible ways
    for(int i = 1; i <= Math.Min(X, n); ++i)
    {
        ans += findTotalPath(X - i, n, dp) % mod;
        ans %= mod;
    }
 
    // Return total no of paths
    dp[X] = ans;
    return ans;
}
 
// Driver Code
public static void Main()
{
    int n = 3, X = 2;
     
    // Stores the number of ways
    // to obtains sums 0 to X
    int[] dp = new int[X + 1];
    Array.Fill(dp, -1);
 
    // Function call
    Console.WriteLine(findTotalPath(X, n, dp));
}
}
 
// This code is contributed by akhilsaini

Javascript




<script>
 
// Javascript program for the above approach
var mod = 1000000007;
 
// Function for counting total
// no of paths possible with
// the sum is equal to X
function findTotalPath(X, n, dp)
{
 
    // If the path of the sum
    // from the root to current
    // node is stored in sum
    if (X == 0) {
        return 1;
    }
    var ans = 0;
 
    // If already computed
    if (dp[X] != -1) {
        return dp[X];
    }
 
    // Count different no of paths
    // using all possible ways
    for (var i = 1; i <= Math.min(X, n); ++i) {
 
        ans += findTotalPath(
                   X - i, n, dp)
               % mod;
        ans %= mod;
    }
 
    // Return total no of paths
    return dp[X] = ans;
}
 
// Driver Code
var n = 3, X = 2;
// Stores the number of ways
// to obtains sums 0 to X
var dp = Array(X + 1).fill(-1);
// Function call
document.write( findTotalPath(X, n, dp));
 
</script>
Output: 
2

 

Time Complexity: O(min (N, X))
Auxiliary Space: O(X)

 




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