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 approach#include <bits/stdc++.h>using namespace std;// Function to find the number of ways// to reach Nth stationint 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 Codeint main(){ int N = 15; cout << numberOfWays(N); return 0;} |
Java
// Java program for the above approachclass 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 stationdef 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 Codeif __name__ == '__main__': N = 15 print(numberOfWays(N))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// 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]; // 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 Codestatic public void Main(){ int N = 15; Console.WriteLine(numberOfWays(N));}}// This code is contributed by Dharanendra L V. |
Javascript
// Javascript program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG{// 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]; // 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 Codepublic static void main(String[] args){ int N = 15; System.out.print(numberOfWays(N));}}// This code is contributed by code_hunt. |
Output:
338
Time Complexity: O(N)
Auxiliary Space: O(N)
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