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# Railway Station | TCS CodeVita 2020

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 ``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`

## C#

 `// C# program to implement above approach``using` `System;``using` `System.Collections.Generic;` `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];``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args){` `    ``// Given Input``    ``int` `n = 4;` `    ``// Initialize DP table with -1``    ``for``(``int` `i = 0 ; i < dp.Length ; i++){``      ``dp[i] = -1;``    ``}` `    ``// Function call to count``    ``// the number of ways to``    ``// reach the n-th station``    ``Console.Write(findWays(n) + ``"\n"``);` `  ``}``}` `// This code is contributed by subhamgoyal2014.`

## Javascript

 ``

Output

`13`

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

Efficient 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 + memorization(top-down) because memorization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Take the integer N as input.
• Initialize a DP table of size N+1 to store the number of ways to reach each station.
• Base cases: set the initial values of the DP table for the first three stations.
• Traverse the stations from 3 to N and for each station:
a. Calculate the number of ways to reach the current station by considering the number of ways to reach the two previous stations.
b. Update the DP table with the new number of ways to reach the current station.
• Return the number of ways to reach the N-th station from the DP table.

Implementation :

## C++

 `// C++ program of the above approach``#include ``using` `namespace` `std;` `// Function to count the number``// of ways to N-th station``int` `findWays(``int` `x)``{``    ``// initialize Dp``    ``vector<``int``> dp(x + 1, 0);` `    ``// Base Cases``    ``dp[0] = 1;``    ``dp[1] = 2;``    ``dp[2] = 4;` `    ``// initialize count``    ``// Count ways in which``    ``// train 1 can be chosen``    ``int` `count = dp[2];` `    ``for` `(``int` `i = 3; i <= x; i++) {``        ``// Count ways in which``        ``// train 2 can be chosen``        ``count += dp[i - 2];``        ``// Count ways in which``        ``// train 3 can be chosen``        ``count += dp[i - 3];``        ``// Store the current state``        ``dp[i] = count;``    ``}` `    ``// Return the number of ways``    ``return` `dp[x];``}` `// Driver Code``int` `main()``{` `    ``// Given Input``    ``int` `n = 2;` `    ``// Function call to count``    ``// the number of ways to``    ``// reach the n-th station``    ``cout << findWays(n) << ``"\n"``;``}``// this code is contributed by bhardwajji`

## Java

 `// Java program of the above approach``import` `java.util.*;` `public` `class` `Main``{``  ` `// Function to count the number``// of ways to N-th station``public` `static` `int` `findWays(``int` `x)``{``  ` `// initialize Dp``int``[] dp = ``new` `int``[x + ``1``];``      ``// Base Cases``    ``dp[``0``] = ``1``;``    ``dp[``1``] = ``2``;``    ``dp[``2``] = ``4``;` `    ``// initialize count``    ``// Count ways in which``    ``// train 1 can be chosen``    ``int` `count = dp[``2``];` `    ``for` `(``int` `i = ``3``; i <= x; i++)``    ``{``      ` `        ``// Count ways in which``        ``// train 2 can be chosen``        ``count += dp[i - ``2``];``      ` `        ``// Count ways in which``        ``// train 3 can be chosen``        ``count += dp[i - ``3``];``      ` `        ``// Store the current state``        ``dp[i] = count;``    ``}` `    ``// Return the number of ways``    ``return` `dp[x];``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ` `    ``// Given Input``    ``int` `n = ``2``;` `    ``// Function call to count``    ``// the number of ways to``    ``// reach the n-th station``    ``System.out.println(findWays(n));``}``}`

## Python3

 `# Function to count the number``# of ways to N-th station``def` `findWays(x):``    ``# initialize Dp``    ``dp ``=` `[``0``] ``*` `(x ``+` `1``)` `    ``# Base Cases``    ``dp[``0``] ``=` `1``    ``dp[``1``] ``=` `2``    ``dp[``2``] ``=` `4` `    ``# initialize count``    ``# Count ways in which``    ``# train 1 can be chosen``    ``count ``=` `dp[``2``]` `    ``for` `i ``in` `range``(``3``, x ``+` `1``):``        ``# Count ways in which``        ``# train 2 can be chosen``        ``count ``+``=` `dp[i ``-` `2``]``        ``# Count ways in which``        ``# train 3 can be chosen``        ``count ``+``=` `dp[i ``-` `3``]``        ``# Store the current state``        ``dp[i] ``=` `count` `    ``# Return the number of ways``    ``return` `dp[x]` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``# Given Input``    ``n ``=` `2` `    ``# Function call to count``    ``# the number of ways to``    ``# reach the n-th station``    ``print``(findWays(n))`

## C#

 `using` `System;` `class` `Solution``{` `  ``// Function to count the number``  ``// of ways to N-th station``  ``public` `int` `findWays(``int` `x)``  ``{` `    ``// initialize Dp``    ``int``[] dp = ``new` `int``[x + 1];` `    ``// Base Cases``    ``dp[0] = 1;``    ``dp[1] = 2;``    ``dp[2] = 4;` `    ``// initialize count``    ``// Count ways in which``    ``// train 1 can be chosen``    ``int` `count = dp[2];` `    ``for` `(``int` `i = 3; i <= x; i++)``    ``{` `      ``// Count ways in which``      ``// train 2 can be chosen``      ``count += dp[i - 2];` `      ``// Count ways in which``      ``// train 3 can be chosen``      ``count += dp[i - 3];` `      ``// Store the current state``      ``dp[i] = count;``      ``count = 0;``    ``}` `    ``// Return the number of ways``    ``return` `dp[x];``  ``}``}` `// Driver Code``public` `class` `Program {``  ``static` `void` `Main(``string``[] args)``  ``{``    ``Solution obj = ``new` `Solution();` `    ``// Given Input``    ``int` `n = 2;` `    ``// Function call to count``    ``// the number of ways to``    ``// reach the n-th station``    ``Console.WriteLine(obj.findWays(n));``  ``}``}` `// This code is contributed by usr_dtewbxkn77n`

## Javascript

 `// JavaScript program of the above approach``function` `findWays(x) {``    ``// initialize Dp``    ``let dp = ``new` `Array(x + 1).fill(0);` `    ``// Base Cases``    ``dp[0] = 1;``    ``dp[1] = 2;``    ``dp[2] = 4;` `    ``// initialize count``    ``// Count ways in which``    ``// train 1 can be chosen``    ``let count = dp[2];` `    ``for` `(let i = 3; i <= x; i++)``    ``{``    ` `        ``// Count ways in which``        ``// train 2 can be chosen``        ``count += dp[i - 2];``        ` `        ``// Count ways in which``        ``// train 3 can be chosen``        ``count += dp[i - 3];``        ` `        ``// Store the current state``        ``dp[i] = count;``    ``}` `    ``// Return the number of ways``    ``return` `dp[x];``}` `// Driver Code``let n = 2;` `// Function call to count``// the number of ways to``// reach the n-th station``console.log(findWays(n));` `// This code is contributed by sarojmcy2e`

Output

`4`

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

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