# Count number of ways to cover a distance | Set 2

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In previous article, a recursive and dynamic programming based approach has been discussed. Here we will reduce the space complexity. It can be onserved 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:

## CPP

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of the ` `// total number of ways to cover the ` `// distance with 1, 2 and 3 steps ` `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, 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 code ` `int` `main() ` `{ ` `    ``int` `n = 4; ` ` `  `    ``cout << countWays(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the count of the ` `// total number of ways to cover the ` `// distance with 1, 2 and 3 steps ` `static` `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 code ` `public` `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 steps ` `def` `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 code ` `n ``=` `4` ` `  `print``(countWays(n)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to return the count of the ` `// total number of ways to cover the ` `// distance with 1, 2 and 3 steps ` `static` `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 code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `n = 4; ` `    ``Console.WriteLine (countWays(n)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```7
```

Time Complexity: O(N)
Space Complexity O(1)

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