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

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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 the 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 the same states twice.

Below is the implementation of the above approach.

C++

// 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);
}

Java

// 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
}

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

C#

// C# program to count the number of paths
using System;
 
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()
    {
        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
        Console.WriteLine(countPaths(x, 0, m, n, dp));
    }
}
 
// This code is contributed by Akanksha Rai

PHP

<?php 
 
// PHP program to count the number of paths
 
$max = 4;
$c = 2;
 
// Function to find the number of paths
function countPaths($sum, $get, $m, $n, &$dp)
{
    global $max,$c;
    // 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 = 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
 
    $n = 3;
    $m = 2;
    $x = 3;
 
    $dp = array_fill(0,$max + 1,NULL);
 
    // Initialized the DP array with -1
    for ($i = 0; $i <= $max; $i++)
        for ($j = 0; $j < 2; $j++)
            $dp[$i][$j] = -1;
 
    // Function to count paths
    echo countPaths($x, 0, $m, $n, $dp);
     
    // This code is contributed by ChitraNayal
?>

Javascript

<script>
 
// Javascript program to count the number of paths
let max = 4;
let c = 2;
 
// Function to find the number of paths
function 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
    let res = 0;
   
    // Traverse in all paths
    for(let i = 1; i <= n; i++) 
    {
         
        // If the edge weight is M
        if (i == m)
            res += countPaths(sum - i, 1, 
                              m, n, dp);
             
        // Edge's weight is not M
        else
            res += countPaths(sum - i, get,
                              m, n, dp);
    }
    dp[sum][get] = res;
   
    return dp[sum][get];
}
 
// Driver Code
let  n = 3, m = 2, x = 3;
let dp = new Array(max + 1);
   
// Initialized the DP array with -1
for(let i = 0; i <= max; i++)
{
    dp[i] = new Array(2)
    for(let j = 0; j < 2; j++)
        dp[i][j] = -1;
} 
 
// Function to count paths
document.write(countPaths(x, 0, m, n, dp));
 
// This code is contributed by avanitrachhadiya2155
     
</script>

Output

Complexity Analysis:

Time Complexity: O(x*n), as we are using a loop to traverse n times and in each traversal, we are recursively calling the function again which will cost O(x). Where n is the number of children from every node and x is the total weight.
Auxiliary Space: O(x*n), as we are using extra space for the DP matrix. Where n is the number of children from every node and x is the total weight.

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