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Railway Station | TCS CodeVita 2020
  • Difficulty Level : Easy
  • Last Updated : 16 Mar, 2021
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Given an integer N, representing the number of stations lying between the source and the destination. There are three trains available from every station and their stoppage patterns are as follows:

  • Train 1: Stops at every station
  • Train 2: Stops at every alternate station
  • Train 3: Stops at every third station

The task is to find the number of ways to reach the destination from the source using any possible combination of trains.

Examples:

Input: N = 2 
Output: 4
Explanation: 
Four possible ways exists to travel from source to destination with 2 stations in between:
Train 1 (from source) -> Train 1 (from station 1) -> Train 1(from station 2) -> Destination
Train 2 (from source) -> Train 1 (from station 2) -> Destination
Train 1 (from source) -> Train 2 (from station 1) -> Destination
Train 3 (from source) -> Destination

Input: N = 0
Output: 1
Explanation: No station is present in between the source and destination. Therefore, there is only one way to travel, i.e.
Train 1(from source) -> Destination



Approach: The main idea to solve the problem is to use Recursion with Memoization to solve this problem. The recurrence relation is as follows:

 F(N) = F(N – 1) + F(N – 2) + F(N – 3),

where,
F(N – 1) counts ways to travel upto (N – 1)th station.
F(N – 2) counts ways to travel upto (N – 2)th station.
F(N – 3) counts ways to travel upto (N – 3)th station.

Follow the steps below to solve the problem:

  1. Initialize an array dp[] for memorization. Set all indices to -1 initially.
  2. Define a recursive function findWays() to calculate the number of ways to reach the Nth station.
  3. Following base cases are required to be considered:
    • For x < 0 return 0.
    • For x = 0, return 1.
    • For x = 1, return 2.
    • For x = 2, return 4.
  4. If the current state, say x, is already evaluated i.e. dp[x] is not equal to -1, simply return the evaluated value.
  5. Otherwise, calculate findWays(x – 1), findWays(x – 2) and findWays(x – 3) recursively and store their sum in dp[x].
  6. Return dp[x].

Below is the implementation of the above approach:

C++14




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Dp table for memoization
int dp[100000];
 
// Function to count the number
// of ways to N-th station
int findWays(int x)
{
    // Base Cases
    if (x < 0)
        return 0;
 
    if (x == 0)
        return 1;
 
    if (x == 1)
        return 2;
 
    if (x == 2)
        return 4;
 
    // If current state is
    // already evaluated
    if (dp[x] != -1)
        return dp[x];
 
    // Recursive calls
 
    // Count ways in which
    // train 1 can be chosen
    int count = findWays(x - 1);
 
    // Count ways in which
    // train 2 can be chosen
    count += findWays(x - 2);
 
    // Count ways in which
    // train 3 can be chosen
    count += findWays(x - 3);
 
    // Store the current state
    dp[x] = count;
 
    // Return the number of ways
    return dp[x];
}
 
// Driver Code
int main()
{
 
    // Given Input
    int n = 4;
 
    // Initialize DP table with -1
    memset(dp, -1, sizeof(dp));
 
    // Function call to count
    // the number of ways to
    // reach the n-th station
    cout << findWays(n) << "\n";
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Dp table for memoization
  static int dp[] = new int[100000];
 
  // Function to count the number
  // of ways to N-th station
  static int findWays(int x)
  {
 
    // Base Cases
    if (x < 0)
      return 0;
 
    if (x == 0)
      return 1;
 
    if (x == 1)
      return 2;
 
    if (x == 2)
      return 4;
 
    // If current state is
    // already evaluated
    if (dp[x] != -1)
      return dp[x];
 
    // Recursive calls
 
    // Count ways in which
    // train 1 can be chosen
    int count = findWays(x - 1);
 
    // Count ways in which
    // train 2 can be chosen
    count += findWays(x - 2);
 
    // Count ways in which
    // train 3 can be chosen
    count += findWays(x - 3);
 
    // Store the current state
    dp[x] = count;
 
    // Return the number of ways
    return dp[x];
  }
 
  // Driven Code
  public static void main(String[] args)
  {
 
    // Given Input
    int n = 4;
 
    // Initialize DP table with -1
    Arrays.fill(dp, -1);
 
    // Function call to count
    // the number of ways to
    // reach the n-th station
    System.out.print(findWays(n));
  }
}
 
// This code is contributed by splevel62.

Python3




# Python3 program of the above approach
 
# Dp table for memoization
dp = [-1 for i in range(100000)]
 
# Function to count the number
# of ways to N-th station
def findWays(x):
     
    # Base Cases
    if (x < 0):
        return 0
 
    if (x == 0):
        return 1
 
    if (x == 1):
        return 2
 
    if (x == 2):
        return 4
 
    # If current state is
    # already evaluated
    if (dp[x] != -1):
        return dp[x]
 
    # Recursive calls
 
    # Count ways in which
    # train 1 can be chosen
    count = findWays(x - 1)
 
    # Count ways in which
    # train 2 can be chosen
    count += findWays(x - 2)
 
    # Count ways in which
    # train 3 can be chosen
    count += findWays(x - 3)
 
    # Store the current state
    dp[x] = count
 
    # Return the number of ways
    return dp[x]
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    n = 4
     
    # Function call to count
    # the number of ways to
    # reach the n-th station
    print(findWays(n))
 
# This code is contributed by SURENDRA_GANGWAR
Output: 
13

 

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

 

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