Count number of paths whose weight is exactly X and has at-least one edge of weight M

Given an infinite tree and three numbers N, M, and X which has exactly N child from every node. Every edge has a weight of 1, 2, 3, 4..N. The task is to find the count of paths whose weight is exactly X and has a minimum of one edge of weight M in it.

The diagram above shows a tree shown till level-3 and N = 3.

Examples:

Input: N = 3, M = 2, X = 3
Output: 2
The path 1-2 and 2-1 in the image above 

Input:  N = 2, M = 1, X = 4
Output:  4 


Approach: The problem can be solved using Dynamic Programming and memoization. We will use a top-down approach to solve this problem. Recur starting from the root with sum initially as X, and recursively traverse all paths possible( which is from 1 to N). If the node is equal to M, then the second parameter becomes true, else it stays the same which has been passed in the previous call. Store the value in a DP[][] table to avoid visiting same states twice.

Below is the implementation of the above approach.

C++

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// C++ program to count the number of paths
#include <bits/stdc++.h>
using namespace std;
#define max 4
#define c 2
  
// Function to find the number of paths
int countPaths(int sum, int get, int m, int n, int dp[])
{
  
    // If the summation is more than X
    if (sum < 0)
        return 0;
  
    // If exactly X weights have reached
    if (sum == 0)
        return get;
  
    // Already visited
    if (dp[sum][get] != -1)
        return dp[sum][get];
  
    // Count paths
    int res = 0;
  
    // Traverse in all paths
    for (int i = 1; i <= n; i++) {
  
        // If the edge weight is M
        if (i == m)
            res += countPaths(sum - i, 1, m, n, dp);
        else // Edge's weight is not M
            res += countPaths(sum - i, get, m, n, dp);
    }
  
    dp[sum][get] = res;
  
    return dp[sum][get];
}
  
// Driver Code
int main()
{
    int n = 3, m = 2, x = 3;
  
    int dp[max + 1];
  
    // Initialized the DP array with -1
    for (int i = 0; i <= max; i++)
        for (int j = 0; j < 2; j++)
            dp[i][j] = -1;
  
    // Function to count paths
    cout << countPaths(x, 0, m, n, dp);
}

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Java

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// Java program to count the number of paths
  
public class GFG{
  
    static int max = 4 ;
    static int  c = 2 ;
      
    // Function to find the number of paths
    static int countPaths(int sum, int get, int m, int n, int dp[][])
    {
      
        // If the summation is more than X
        if (sum < 0)
            return 0;
      
        // If exactly X weights have reached
        if (sum == 0)
            return get;
      
        // Already visited
        if (dp[sum][get] != -1)
            return dp[sum][get];
      
        // Count paths
        int res = 0;
      
        // Traverse in all paths
        for (int i = 1; i <= n; i++) {
      
            // If the edge weight is M
            if (i == m)
                res += countPaths(sum - i, 1, m, n, dp);
            else // Edge's weight is not M
                res += countPaths(sum - i, get, m, n, dp);
        }
      
        dp[sum][get] = res;
      
        return dp[sum][get];
    }
      
    // Driver Code
    public static void main(String []args)
    {
        int n = 3, m = 2, x = 3;
      
        int dp[][] = new int[max + 1][2];
      
        // Initialized the DP array with -1
        for (int i = 0; i <= max; i++)
            for (int j = 0; j < 2; j++)
                dp[i][j] = -1;
      
        // Function to count paths
        System.out.println(countPaths(x, 0, m, n, dp));
    }
    // This code is contributed by Ryuga
}

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Python3

# Python3 program to count the number of paths
Max = 4
c = 2

# Function to find the number of paths
def countPaths(Sum, get, m, n, dp):

# If the Summation is more than X
if (Sum < 0): return 0 # If exactly X weights have reached if (Sum == 0): return get # Already visited if (dp[Sum][get] != -1): return dp[Sum][get] # Count paths res = 0 # Traverse in all paths for i in range(1, n + 1): # If the edge weight is M if (i == m): res += countPaths(Sum - i, 1, m, n, dp) else: # Edge's weight is not M res += countPaths(Sum - i, get, m, n, dp) dp[Sum][get] = res return dp[Sum][get] # Driver Code n = 3 m = 2 x = 3 dp = [[-1 for i in range(2)] for i in range(Max + 1)] # Initialized the DP array with -1 for i in range(Max + 1): for j in range(2): dp[i][j] = -1 # Function to count paths print(countPaths(x, 0, m, n, dp)) # This code is contributed by Mohit kumar 29 [tabbyending]

Output:

2


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Striver(underscore)79 at Codechef and codeforces D

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Improved By : Ryuga