Given a distance N. The task is to count the total number of ways to cover the distance with 1, 2 and 3 steps.
Examples:
Input: N = 3
Output: 4
All the required ways are (1 + 1 + 1), (1 + 2), (2 + 1) and (3).
Input: N = 4
Output: 7
Approach: In previous article, a recursive and dynamic programming based approach has been discussed. Here we will reduce the space complexity. It can be observed that to calculate the number of steps to cover the distance i, only the last three states are required (i – 1, i – 2, i – 3). So, the result can be calculated using the last three states.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the count of the// total number of ways to cover the// distance with 1, 2 and 3 stepsint countWays(int n){ // Base conditions if (n == 0) return 1; if (n <= 2) return n; // To store the last three stages int f0 = 1, f1 = 1, f2 = 2, ans; // Find the numbers of steps required // to reach the distance i for (int i = 3; i <= n; i++) { ans = f0 + f1 + f2; f0 = f1; f1 = f2; f2 = ans; } // Return the required answer return ans;}// Driver codeint main(){ int n = 4; cout << countWays(n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to return the count of the// total number of ways to cover the// distance with 1, 2 and 3 stepsstatic int countWays(int n){ // Base conditions if (n == 0) return 1; if (n <= 2) return n; // To store the last three stages int f0 = 1, f1 = 1, f2 = 2; int ans=0; // Find the numbers of steps required // to reach the distance i for (int i = 3; i <= n; i++) { ans = f0 + f1 + f2; f0 = f1; f1 = f2; f2 = ans; } // Return the required answer return ans;}// Driver codepublic static void main (String[] args) { int n = 4; System.out.println (countWays(n));}}// This code is contributed by jit_t |
Python
# Python3 implementation of the approach# Function to return the count of the# total number of ways to cover the# distance with 1, 2 and 3 stepsdef countWays(n): # Base conditions if (n == 0): return 1 if (n <= 2): return n # To store the last three stages f0 = 1 f1 = 1 f2 = 2 ans = 0 # Find the numbers of steps required # to reach the distance i for i in range(3, n + 1): ans = f0 + f1 + f2 f0 = f1 f1 = f2 f2 = ans # Return the required answer return ans# Driver coden = 4print(countWays(n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the count of the// total number of ways to cover the// distance with 1, 2 and 3 stepsstatic int countWays(int n){ // Base conditions if (n == 0) return 1; if (n <= 2) return n; // To store the last three stages int f0 = 1, f1 = 1, f2 = 2; int ans = 0; // Find the numbers of steps required // to reach the distance i for (int i = 3; i <= n; i++) { ans = f0 + f1 + f2; f0 = f1; f1 = f2; f2 = ans; } // Return the required answer return ans;}// Driver codepublic static void Main(String[] args) { int n = 4; Console.WriteLine (countWays(n));}}// This code is contributed by PrinciRaj1992 |
Javascript
<script> // Javascript implementation of the approach // Function to return the count of the // total number of ways to cover the // distance with 1, 2 and 3 steps function countWays(n) { // Base conditions if (n == 0) return 1; if (n <= 2) return n; // To store the last three stages let f0 = 1, f1 = 1, f2 = 2; let ans = 0; // Find the numbers of steps required // to reach the distance i for (let i = 3; i <= n; i++) { ans = f0 + f1 + f2; f0 = f1; f1 = f2; f2 = ans; } // Return the required answer return ans; } let n = 4; document.write(countWays(n));// This code is contributed by suresh07.</script> |
Output:
7
Time Complexity: O(N)
Space Complexity O(1)
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