Count unique paths with given sum in an N-ary Tree
Last Updated :
15 Jul, 2025
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:
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 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 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# 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
<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>
Time Complexity: O(min (N, X))
Auxiliary Space: O(X)
Another approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a Dp array to store the solution of the subproblems.
- Initialize the Dp with base cases
- Fill up the Dp iteratively and get the current value from previous computation.
- Return the final solution stored in dp[X].
Implementation :
C++
// C++ program to count total
// no of paths possible with
// the sum is equal to X
#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)
{
// Stores the number of ways
// to obtains sums 0 to X
vector<int> dp(X + 1);
// Base case
dp[0] = 1;
// Fill the DP table iteratively
for (int i = 1; i <= X; ++i) {
for (int j = 1; j <= min(i, n); ++j) {
dp[i] += dp[i - j];
dp[i] %= mod;
}
}
// Return total no of paths
return dp[X];
}
// Driver Code
int main()
{
int n = 3, X = 2;
// Function call
cout << findTotalPath(X, n);
}
// this code is contributed by bhardwajji
import java.util.*;
public class Main {
// Function for counting total
// no of paths possible with
// the sum is equal to X
static long findTotalPath(int X, int n) {
// Stores the number of ways
// to obtains sums 0 to X
int mod = (int)1e9 + 7;
int[] dp = new int[X + 1];
// Base case
dp[0] = 1;
// Fill the DP table iteratively
for (int i = 1; i <= X; ++i) {
for (int j = 1; j <= Math.min(i, n); ++j) {
dp[i] += dp[i - j];
dp[i] %= mod;
}
}
// Return total no of paths
return dp[X];
}
// Driver Code
public static void main(String[] args) {
int n = 3, X = 2;
// Function call
System.out.println(findTotalPath(X, n));
}
}
# Python program to count total
# no of paths possible with
# the sum is equal to X
# Function for counting total
# no of paths possible with
# the sum is equal to X
def findTotalPath(X, n):
# Stores the number of ways
# to obtains sums 0 to X
dp = [0] * (X + 1)
# Base case
dp[0] = 1
# Fill the DP table iteratively
for i in range(1, X + 1):
for j in range(1, min(i, n) + 1):
dp[i] += dp[i - j]
dp[i] %= int(1e9 + 7)
# Return total no of paths
return dp[X]
# Driver Code
if __name__ == "__main__":
n = 3
X = 2
# Function call
print(findTotalPath(X, n))
// Function for counting total
// no of paths possible with
// the sum is equal to X
function findTotalPath(X, n)
{
// Stores the number of ways
// to obtains sums 0 to X
const mod = 1e9 + 7;
const dp = new Array(X + 1).fill(0);
// Base case
dp[0] = 1;
// Fill the DP table iteratively
for (let i = 1; i <= X; ++i) {
for (let j = 1; j <= Math.min(i, n); ++j) {
dp[i] += dp[i - j];
dp[i] %= mod;
}
}
// Return total no of paths
return dp[X];
}
// Driver Code
const n = 3, X = 2;
// Function call
console.log(findTotalPath(X, n));
using System;
public class Program
{
public static int mod = (int)1e9 + 7;
public static long FindTotalPath(int X, int n)
{
// Stores the number of ways
// to obtains sums 0 to X
int[] dp = new int[X + 1];
// Base case
dp[0] = 1;
// Fill the DP table iteratively
for (int i = 1; i <= X; ++i)
{
for (int j = 1; j <= Math.Min(i, n); ++j)
{
dp[i] += dp[i - j];
dp[i] %= mod;
}
}
// Return total no of paths
return dp[X];
}
public static void Main()
{
int n = 3, X = 2;
// Function call
Console.WriteLine(FindTotalPath(X, n));
}
}
//THIS CODE IS CONTRIBUTED BY ZAID
Output:
2
Time Complexity: O( X * min (N, X))
Auxiliary Space: O(X)
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