Count ways to reach the Nth station

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Given N stations and three trains A, B, and C such that train A stops at every station, train B stops at every second station, and train C stops at every third station, the task is to find the number of ways to reach the N^th station can be reached from station 1.

Examples:

Input: N = 4
Output: 3
Explanation:
Below are the ways of reaching station 4 from station 1 as follows:

Take train A at station 1 and continue to station 4 using only train A as (A ? A ? A ? A).

Take train B at station 1, stop at station 3 and take train A from station 3 to 4 as (B? . ? A).

Take train C from station 1 and stop at station 4(C? .)

Therefore, the total number of ways to reach station 4 from station 1 is 3.

Input: N = 15
Output: 338

Approach: The given problem can be solved using the following observations:

Train A can be used to reach station X only if train A reaches X – 1.
Train B can be used to reach station X only if train B reaches X – 2.
Train C can be used to reach station X only if train C reaches X – 3.

Therefore, from the above observations, the idea is to use the concept of Dynamic Programming. Follow the steps given below to solve the problem:

Initialize a 2D array DP such that:
- DP[i][1] stores the number of ways to reach station i using train A.
- DP[i][2] stores the number of ways to reach station i using train B.
- DP[i][3] stores the number of ways to reach station i using train C.
- DP[i][4] stores the number of ways to reach station i using trains A, B, or C.
There is only one way of reaching station 1. Therefore, update the value of DP[1][1], DP[1][2], DP[1][3], DP[1][4] as 1.
Using the above observations, update the value of each state by iterating the loop as follows:
- DP[i][1] = DP[i-1][4]
- DP[i][2] = DP[i-2][4]
- DP[i][3] = DP[i-3][4]
- DP[i][4] = DP[i][1] + DP[i][2] + DP[i][3]
After completing the above steps, print the value of DP[N][4] as the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the number of ways
// to reach Nth station
int numberOfWays(int N)
{
    // Declares the DP[] array
    int DP[N + 1][5];
 
    // Initialize dp[][] array
    memset(DP, 0, sizeof(DP));
 
    // Only 1 way to reach station 1
    DP[1][1] = 1;
    DP[1][2] = 1;
    DP[1][3] = 1;
    DP[1][4] = 1;
 
    // Find the remaining states from
    // the 2nd station
    for (int i = 2; i <= N; i++) {
 
        // If the train A is present
        // at station i - 1
        if (i - 1 > 0 && DP[i - 1][1] > 0)
            DP[i][1] = DP[i - 1][4];
 
        // If the train B is present
        // at station i-2
        if (i - 2 > 0 && DP[i - 2][2] > 0)
            DP[i][2] = DP[i - 2][4];
 
        // If train C is present at
        // station i-3
        if (i - 3 > 0 && DP[i - 3][3] > 0)
            DP[i][3] = DP[i - 3][4];
 
        // The total number of ways to
        // reach station i
        DP[i][4] = (DP[i][1] + DP[i][2]
                    + DP[i][3]);
    }
 
    // Return the total count of ways
    return DP[N][4];
}
 
// Driver Code
int main()
{
    int N = 15;
    cout << numberOfWays(N);
 
    return 0;
}

Java

// Java program for the above approach
 
class GFG {
 
    // Function to find the number of ways
    // to reach Nth station
    static int numberOfWays(int N)
    {
 
        // Declares the DP[] array
        int[][] DP = new int[N + 1][5];
 
        // Initialize dp[][] array
        for (int i = 0; i < N + 1; i++) {
            for (int j = 0; j < 5; j++) {
                DP[i][j] = 0;
            }
        }
 
        // Only 1 way to reach station 1
        DP[1][1] = 1;
        DP[1][2] = 1;
        DP[1][3] = 1;
        DP[1][4] = 1;
 
        // Find the remaining states from
        // the 2nd station
        for (int i = 2; i <= N; i++) {
 
            // If the train A is present
            // at station i - 1
            if (i - 1 > 0 && DP[i - 1][1] > 0)
                DP[i][1] = DP[i - 1][4];
 
            // If the train B is present
            // at station i-2
            if (i - 2 > 0 && DP[i - 2][2] > 0)
                DP[i][2] = DP[i - 2][4];
 
            // If train C is present at
            // station i-3
            if (i - 3 > 0 && DP[i - 3][3] > 0)
                DP[i][3] = DP[i - 3][4];
 
            // The total number of ways to
            // reach station i
            DP[i][4] = (DP[i][1] + DP[i][2] + DP[i][3]);
        }
 
        // Return the total count of ways
        return DP[N][4];
    }
 
    // Driver Code
    static public void main(String[] args)
    {
        int N = 15;
        System.out.println(numberOfWays(N));
    }
}
 
// This code is contributed by ukasp.

Python3

# Python3 program for the above approach
 
# Function to find the number of ways
# to reach Nth station
def numberOfWays(N):
     
    # Declares the DP[] array
    DP = [[0 for i in range(5)] 
             for i in range(N + 1)]
 
    # Initialize dp[][] array
    # memset(DP, 0, sizeof(DP))
     
    # Only 1 way to reach station 1
    DP[1][1] = 1
    DP[1][2] = 1
    DP[1][3] = 1
    DP[1][4] = 1
 
    # Find the remaining states from
    # the 2nd station
    for i in range(2, N + 1):
         
        # If the train A is present
        # at station i - 1
        if (i - 1 > 0 and DP[i - 1][1] > 0):
            DP[i][1] = DP[i - 1][4]
 
        # If the train B is present
        # at station i-2
        if (i - 2 > 0 and DP[i - 2][2] > 0):
            DP[i][2] = DP[i - 2][4]
 
        # If train C is present at
        # station i-3
        if (i - 3 > 0 and DP[i - 3][3] > 0):
            DP[i][3] = DP[i - 3][4]
 
        # The total number of ways to
        # reach station i
        DP[i][4] = (DP[i][1] + DP[i][2] + DP[i][3])
 
    # Return the total count of ways
    return DP[N][4]
 
# Driver Code
if __name__ == '__main__':
     
    N = 15
     
    print(numberOfWays(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the number of ways
// to reach Nth station
static int numberOfWays(int N)
{
     
    // Declares the DP[] array
    int[,] DP = new int[N + 1, 5];
 
    // Initialize dp[][] array
    for(int i = 0; i < N + 1; i++) 
    {
        for(int j = 0; j < 5; j++)
        {
            DP[i, j] = 0;
        }
    }
 
    // Only 1 way to reach station 1
    DP[1, 1] = 1;
    DP[1, 2] = 1;
    DP[1, 3] = 1;
    DP[1, 4] = 1;
 
    // Find the remaining states from
    // the 2nd station
    for(int i = 2; i <= N; i++)
    {
         
        // If the train A is present
        // at station i - 1
        if (i - 1 > 0 && DP[i - 1, 1] > 0)
            DP[i, 1] = DP[i - 1, 4];
 
        // If the train B is present
        // at station i-2
        if (i - 2 > 0 && DP[i - 2, 2] > 0)
            DP[i, 2] = DP[i - 2, 4];
 
        // If train C is present at
        // station i-3
        if (i - 3 > 0 && DP[i - 3, 3] > 0)
            DP[i, 3] = DP[i - 3, 4];
 
        // The total number of ways to
        // reach station i
        DP[i, 4] = (DP[i, 1] + DP[i, 2] + DP[i, 3]);
    }
 
    // Return the total count of ways
    return DP[N, 4];
}
 
// Driver Code
static public void Main()
{
    int N = 15;
    Console.WriteLine(numberOfWays(N));
}
}
 
// This code is contributed by Dharanendra L V.

Javascript

// Javascript program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to find the number of ways
// to reach Nth station
static int numberOfWays(int N)
{
    // Declares the DP[] array
    int DP[][] = new int[N + 1][5];
 
    // Initialize dp[][] array
    for (int i = 0; i < N +1; i++) {
        for (int j = 0; j < 5; j++) {
            DP[i][j] = 0;
        }
    }
 
    // Only 1 way to reach station 1
    DP[1][1] = 1;
    DP[1][2] = 1;
    DP[1][3] = 1;
    DP[1][4] = 1;
 
    // Find the remaining states from
    // the 2nd station
    for (int i = 2; i <= N; i++) {
 
        // If the train A is present
        // at station i - 1
        if (i - 1 > 0 && DP[i - 1][1] > 0)
            DP[i][1] = DP[i - 1][4];
 
        // If the train B is present
        // at station i-2
        if (i - 2 > 0 && DP[i - 2][2] > 0)
            DP[i][2] = DP[i - 2][4];
 
        // If train C is present at
        // station i-3
        if (i - 3 > 0 && DP[i - 3][3] > 0)
            DP[i][3] = DP[i - 3][4];
 
        // The total number of ways to
        // reach station i
        DP[i][4] = (DP[i][1] + DP[i][2]
                    + DP[i][3]);
    }
 
    // Return the total count of ways
    return DP[N][4];
}
 
// Driver Code
public static void main(String[] args)
{
     
    int N = 15;
    System.out.print(numberOfWays(N));
}
}
 
// This code is contributed by code_hunt.

Output:

Time Complexity: O(N)
Auxiliary Space: O(N)

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