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 Nth 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 approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to find the number of ways// to reach Nth stationint numberOfWays(int N){    // Declares the DP[] array    int DP[N + 1][5];Â
    // Initialize dp[][] array    memset(DP, 0, sizeof(DP));Â
    // Only 1 way to reach station 1    DP[1][1] = 1;    DP[1][2] = 1;    DP[1][3] = 1;    DP[1][4] = 1;Â
    // Find the remaining states from    // the 2nd station    for (int i = 2; i <= N; i++) {Â
        // 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];Â
        // 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];Â
        // 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];Â
        // The total number of ways to        // reach station i        DP[i][4] = (DP[i][1] + DP[i][2]                    + DP[i][3]);    }Â
    // Return the total count of ways    return DP[N][4];}Â
// Driver Codeint main(){Â Â Â Â int N = 15;Â Â Â Â cout << numberOfWays(N);Â
    return 0;} |
Java
// Java program for the above approachÂ
class GFG {Â
    // Function to find the number of ways    // to reach Nth station    static int numberOfWays(int N)    {Â
        // Declares the DP[] array        int[][] DP = new int[N + 1][5];Â
        // Initialize dp[][] array        for (int i = 0; i < N + 1; i++) {            for (int j = 0; j < 5; j++) {                DP[i][j] = 0;            }        }Â
        // Only 1 way to reach station 1        DP[1][1] = 1;        DP[1][2] = 1;        DP[1][3] = 1;        DP[1][4] = 1;Â
        // Find the remaining states from        // the 2nd station        for (int i = 2; i <= N; i++) {Â
            // 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];Â
            // 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];Â
            // 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];Â
            // The total number of ways to            // reach station i            DP[i][4] = (DP[i][1] + DP[i][2] + DP[i][3]);        }Â
        // Return the total count of ways        return DP[N][4];    }Â
    // Driver Code    static public void main(String[] args)    {        int N = 15;        System.out.println(numberOfWays(N));    }}Â
// This code is contributed by ukasp. |
Python3
# Python3 program for the above approachÂ
# Function to find the number of ways# to reach Nth stationdef numberOfWays(N):         # Declares the DP[] array    DP = [[0 for i in range(5)]              for i in range(N + 1)]Â
    # 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] = 1Â
    # 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]Â
        # 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]Â
        # 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]Â
        # The total number of ways to        # reach station i        DP[i][4] = (DP[i][1] + DP[i][2] + DP[i][3])Â
    # Return the total count of ways    return DP[N][4]Â
# Driver Codeif __name__ == '__main__':Â Â Â Â Â Â Â Â Â N = 15Â Â Â Â Â Â Â Â Â print(numberOfWays(N))Â
# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;Â
class GFG{Â
// Function to find the number of ways// to reach Nth stationstatic int numberOfWays(int N){         // Declares the DP[] array    int[,] DP = new int[N + 1, 5];Â
    // Initialize dp[][] array    for(int i = 0; i < N + 1; i++)     {        for(int j = 0; j < 5; j++)        {            DP[i, j] = 0;        }    }Â
    // Only 1 way to reach station 1    DP[1, 1] = 1;    DP[1, 2] = 1;    DP[1, 3] = 1;    DP[1, 4] = 1;Â
    // 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];Â
        // 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];Â
        // 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];Â
        // The total number of ways to        // reach station i        DP[i, 4] = (DP[i, 1] + DP[i, 2] + DP[i, 3]);    }Â
    // Return the total count of ways    return DP[N, 4];}Â
// Driver Codestatic public void Main(){Â Â Â Â int N = 15;Â Â Â Â Console.WriteLine(numberOfWays(N));}}Â
// This code is contributed by Dharanendra L V. |
Javascript
// Javascript program for the above approachimport java.io.*;import java.lang.*;import java.util.*;Â
class GFG{Â
// Function to find the number of ways// to reach Nth stationstatic int numberOfWays(int N){    // Declares the DP[] array    int DP[][] = new int[N + 1][5];Â
    // Initialize dp[][] array    for (int i = 0; i < N +1; i++) {        for (int j = 0; j < 5; j++) {            DP[i][j] = 0;        }    }Â
    // Only 1 way to reach station 1    DP[1][1] = 1;    DP[1][2] = 1;    DP[1][3] = 1;    DP[1][4] = 1;Â
    // Find the remaining states from    // the 2nd station    for (int i = 2; i <= N; i++) {Â
        // 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];Â
        // 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];Â
        // 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];Â
        // The total number of ways to        // reach station i        DP[i][4] = (DP[i][1] + DP[i][2]                    + DP[i][3]);    }Â
    // Return the total count of ways    return DP[N][4];}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â Â Â Â Â Â int N = 15;Â Â Â Â System.out.print(numberOfWays(N));}}Â
// This code is contributed by code_hunt. |
Output:Â
338
Â
Time Complexity: O(N)
Auxiliary Space: O(N)
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