Given a distance ‘dist, count total number of ways to cover the distance with 1, 2 and 3 steps.

**Examples:**

Input:n = 3Output:4Explantion:Below are the four ways 1 step + 1 step + 1 step 1 step + 2 step 2 step + 1 step 3 stepInput:n = 4Output:7Explantion:Below are the four ways 1 step + 1 step + 1 step + 1 step 1 step + 2 step + 1 step 2 step + 1 step + 1 step 1 step + 1 step + 2 step 2 step + 2 step 3 step + 1 step 1 step + 3 step

__Recursive solution__

**Approach:**There are n stairs, and a person is allowed to next step, skip one position or skip two positions. So there are n positions. The idea is standing at the ith position the person can move by i+1, i+2, i+3 position. So a recursive function can be formed where at current index i the function is recursively called for i+1, i+2 and i+3 positions.

There is another way of forming the recursive function. To reach position i, a person has to jump either from i-1, i-2 or i-3 position where i is the starting position.

**Algorithm:**- Create a recursive function (
*count(int n)*) which takes only one parameter. - Check the base cases. If the value of n is less than 0 then return 0, and if value of n is equal to zero then return 1 as it is the starting position.
- Call the function recursively with values n-1, n-2 and n-3 and sum up the values that are returned, i.e.
*sum = count(n-1) + count(n-2) + count(n-3)*. - Return the value of
*sum*.

- Create a recursive function (
**Implementation:**

## C++

`// A naive recursive C++ program to count number of ways to cover` `// a distance with 1, 2 and 3 steps` `#include<iostream>` `using` `namespace` `std;` `// Returns count of ways to cover 'dist'` `int` `printCountRec(` `int` `dist)` `{` ` ` `// Base cases` ` ` `if` `(dist<0) ` `return` `0;` ` ` `if` `(dist==0) ` `return` `1;` ` ` `// Recur for all previous 3 and add the results` ` ` `return` `printCountRec(dist-1) +` ` ` `printCountRec(dist-2) +` ` ` `printCountRec(dist-3);` `}` `// driver program` `int` `main()` `{` ` ` `int` `dist = 4;` ` ` `cout << printCountRec(dist);` ` ` `return` `0;` `}` |

## Java

`// A naive recursive Java program to count number` `// of ways to cover a distance with 1, 2 and 3 steps` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function returns count of ways to cover 'dist'` ` ` `static` `int` `printCountRec(` `int` `dist)` ` ` `{` ` ` `// Base cases` ` ` `if` `(dist<` `0` `) ` ` ` `return` `0` `;` ` ` `if` `(dist==` `0` `) ` ` ` `return` `1` `;` ` ` ` ` `// Recur for all previous 3 and add the results` ` ` `return` `printCountRec(dist-` `1` `) +` ` ` `printCountRec(dist-` `2` `) +` ` ` `printCountRec(dist-` `3` `);` ` ` `}` ` ` ` ` `// driver program` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `dist = ` `4` `;` ` ` `System.out.println(printCountRec(dist));` ` ` `}` `}` `// This code is contributed by Pramod Kumar` |

## Python3

`# A naive recursive Python3 program` `# to count number of ways to cover` `# a distance with 1, 2 and 3 steps` `# Returns count of ways to` `# cover 'dist'` `def` `printCountRec(dist):` ` ` ` ` `# Base cases` ` ` `if` `dist < ` `0` `:` ` ` `return` `0` ` ` ` ` `if` `dist ` `=` `=` `0` `:` ` ` `return` `1` ` ` `# Recur for all previous 3 and ` ` ` `# add the results` ` ` `return` `(printCountRec(dist` `-` `1` `) ` `+` ` ` `printCountRec(dist` `-` `2` `) ` `+` ` ` `printCountRec(dist` `-` `3` `))` `# Driver code` `dist ` `=` `4` `print` `(printCountRec(dist))` `# This code is contributed by Anant Agarwal.` |

## C#

`// A naive recursive C# program to` `// count number of ways to cover a` `// distance with 1, 2 and 3 steps` `using` `System;` `class` `GFG {` ` ` ` ` `// Function returns count of` ` ` `// ways to cover 'dist'` ` ` `static` `int` `printCountRec(` `int` `dist)` ` ` `{` ` ` `// Base cases` ` ` `if` `(dist < 0)` ` ` `return` `0;` ` ` `if` `(dist == 0)` ` ` `return` `1;` ` ` `// Recur for all previous 3` ` ` `// and add the results` ` ` `return` `printCountRec(dist - 1) +` ` ` `printCountRec(dist - 2) +` ` ` `printCountRec(dist - 3);` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `dist = 4;` ` ` `Console.WriteLine(printCountRec(dist));` ` ` `}` `}` `// This code is contributed by Sam007.` |

## PHP

`<?php` `// A naive recursive PHP program to` `// count number of ways to cover` `// a distance with 1, 2 and 3 steps` `// Returns count of ways to cover 'dist'` `function` `printCountRec( ` `$dist` `)` `{` ` ` ` ` `// Base cases` ` ` `if` `(` `$dist` `<0) ` `return` `0;` ` ` `if` `(` `$dist` `==0) ` `return` `1;` ` ` `// Recur for all previous 3` ` ` `// and add the results` ` ` `return` `printCountRec(` `$dist` `- 1) +` ` ` `printCountRec(` `$dist` `- 2) +` ` ` `printCountRec(` `$dist` `- 3);` `}` ` ` `// Driver Code` ` ` `$dist` `= 4;` ` ` `echo` `printCountRec(` `$dist` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// A naive recursive javascript program to count number of ways to cover` `// a distance with 1, 2 and 3 steps` `// Returns count of ways to cover 'dist'` `function` `printCountRec(dist)` `{` ` ` `// Base cases` ` ` `if` `(dist<0) ` `return` `0;` ` ` `if` `(dist==0) ` `return` `1;` ` ` `// Recur for all previous 3 and add the results` ` ` `return` `printCountRec(dist-1) +` ` ` `printCountRec(dist-2) +` ` ` `printCountRec(dist-3);` `}` `// driver program` `var` `dist = 4;` `document.write(printCountRec(dist));` `</script>` |

**Output:**

7

**Complexity Analysis:****Time Complexity:**O(3^{n}).

The time complexity of the above solution is exponential, a close upper bound is O(3^{n}). From each state 3, a recursive function is called. So the upper bound for n states is O(3^{n}).**Space complexity:**O(1).

No extra space is required.

__Efficient solution__

**Approach:**The idea is similar, but it can be observed that there are n states but the recursive function is called 3 ^ n times. That means that some states are called repeatedly. So the idea is to store the value of states. This can be done in two ways.- The first way is to keep the recursive structure intact and just store the value in a HashMap and whenever the function is called, return the value store without computing (Top-Down Approach).
- The second way is to take an extra space of size n and start computing values of states from 1, 2 .. to n, i.e. compute values of i, i+1, i+2 and then use them to calculate the value of i+3 (Bottom-Up Approach).
- Overlapping Subproblems in Dynamic Programming.
- Optimal substructure property in Dynamic Programming.
- Dynamic Programming(DP) problems

**Algorithm:**- Create an array of size n + 1 and initialize the first 3 variables with 1, 1, 2. The base cases.
- Run a loop from 3 to n.
- For each index i, compute value of ith position as
*dp[i] = dp[i-1] + dp[i-2] + dp[i-3]*. - Print the value of dp[n], as the Count of number of ways to cover a distance.

**Implementation:**

## C++

`// A Dynamic Programming based C++ program to count number of ways` `// to cover a distance with 1, 2 and 3 steps` `#include<iostream>` `using` `namespace` `std;` `int` `printCountDP(` `int` `dist)` `{` ` ` `int` `count[dist+1];` ` ` `// Initialize base values. There is one way to cover 0 and 1` ` ` `// distances and two ways to cover 2 distance` ` ` `count[0] = 1;` ` ` `if` `(dist >= 1)` ` ` `count[1] = 1;` ` ` `if` `(dist >= 2)` ` ` `count[2] = 2;` ` ` `// Fill the count array in bottom up manner` ` ` `for` `(` `int` `i=3; i<=dist; i++)` ` ` `count[i] = count[i-1] + count[i-2] + count[i-3];` ` ` `return` `count[dist];` `}` `// driver program` `int` `main()` `{` ` ` `int` `dist = 4;` ` ` `cout << printCountDP(dist);` ` ` `return` `0;` `}` |

## Java

`// A Dynamic Programming based Java program` `// to count number of ways to cover a distance` `// with 1, 2 and 3 steps` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function returns count of ways to cover 'dist'` ` ` `static` `int` `printCountDP(` `int` `dist)` ` ` `{` ` ` `int` `[] count = ` `new` `int` `[dist+` `1` `];` ` ` ` ` `// Initialize base values. There is one way to` ` ` `// cover 0 and 1 distances and two ways to` ` ` `// cover 2 distance` ` ` `count[` `0` `] = ` `1` `;` ` ` `if` `(dist >= ` `1` `)` ` ` `count[` `1` `] = ` `1` `;` ` ` `if` `(dist >= ` `2` `)` ` ` `count[` `2` `] = ` `2` `;` ` ` ` ` `// Fill the count array in bottom up manner` ` ` `for` `(` `int` `i=` `3` `; i<=dist; i++)` ` ` `count[i] = count[i-` `1` `] + count[i-` `2` `] + count[i-` `3` `];` ` ` ` ` `return` `count[dist];` ` ` `}` ` ` ` ` `// driver program` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `dist = ` `4` `;` ` ` `System.out.println(printCountDP(dist));` ` ` `}` `}` `// This code is contributed by Pramod Kumar` |

## Python3

`# A Dynamic Programming based on Python3` `# program to count number of ways to` `# cover a distance with 1, 2 and 3 steps` `def` `printCountDP(dist):` ` ` `count ` `=` `[` `0` `] ` `*` `(dist ` `+` `1` `)` ` ` ` ` `# Initialize base values. There is` ` ` `# one way to cover 0 and 1 distances` ` ` `# and two ways to cover 2 distance` ` ` `count[` `0` `] ` `=` `1` ` ` `if` `dist >` `=` `1` `:` ` ` `count[` `1` `] ` `=` `1` ` ` `if` `dist >` `=` `2` `:` ` ` `count[` `2` `] ` `=` `2` ` ` ` ` `# Fill the count array in bottom` ` ` `# up manner` ` ` `for` `i ` `in` `range` `(` `3` `, dist ` `+` `1` `):` ` ` `count[i] ` `=` `(count[i` `-` `1` `] ` `+` ` ` `count[i` `-` `2` `] ` `+` `count[i` `-` `3` `])` ` ` ` ` `return` `count[dist];` `# driver program` `dist ` `=` `4` `;` `print` `( printCountDP(dist))` `# This code is contributed by Sam007.` |

## C#

`// A Dynamic Programming based C# program` `// to count number of ways to cover a distance` `// with 1, 2 and 3 steps` `using` `System;` `class` `GFG {` ` ` ` ` `// Function returns count of ways` ` ` `// to cover 'dist'` ` ` `static` `int` `printCountDP(` `int` `dist)` ` ` `{` ` ` `int` `[] count = ` `new` `int` `[dist + 1];` ` ` `// Initialize base values. There is one` ` ` `// way to cover 0 and 1 distances` ` ` `// and two ways to cover 2 distance` ` ` `count[0] = 1;` ` ` `count[1] = 1;` ` ` `count[2] = 2;` ` ` `// Fill the count array` ` ` `// in bottom up manner` ` ` `for` `(` `int` `i = 3; i <= dist; i++)` ` ` `count[i] = count[i - 1] +` ` ` `count[i - 2] +` ` ` `count[i - 3];` ` ` `return` `count[dist];` ` ` `}` ` ` ` ` `// Driver Code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` `int` `dist = 4;` ` ` `Console.WriteLine(printCountDP(dist));` ` ` `}` `}` `// This code is contributed by Sam007.` |

## PHP

`<?php` `// A Dynamic Programming based PHP program` `// to count number of ways to cover a` `// distance with 1, 2 and 3 steps` `function` `printCountDP( ` `$dist` `)` `{` ` ` `$count` `= ` `array` `();` ` ` `// Initialize base values. There is` ` ` `// one way to cover 0 and 1 distances` ` ` `// and two ways to cover 2 distance` ` ` `$count` `[0] = 1; ` `$count` `[1] = 1;` ` ` `$count` `[2] = 2;` ` ` `// Fill the count array` ` ` `// in bottom up manner` ` ` `for` `( ` `$i` `= 3; ` `$i` `<= ` `$dist` `; ` `$i` `++)` ` ` `$count` `[` `$i` `] = ` `$count` `[` `$i` `- 1] +` ` ` `$count` `[` `$i` `- 2] +` ` ` `$count` `[` `$i` `- 3];` ` ` `return` `$count` `[` `$dist` `];` `}` `// Driver Code` `$dist` `= 4;` `echo` `printCountDP(` `$dist` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// A Dynamic Programming based Javascript program` `// to count number of ways to cover a distance` `// with 1, 2 and 3 steps` ` ` `// Function returns count of ways` `// to cover 'dist'` `function` `printCountDP(dist)` `{` ` ` `let count = ` `new` `Array(dist + 1);` ` ` ` ` `// Initialize base values. There is one` ` ` `// way to cover 0 and 1 distances` ` ` `// and two ways to cover 2 distance` ` ` `count[0] = 1;` ` ` `if` `(dist >= 1)` ` ` `count[1] = 1;` ` ` `if` `(dist >= 2)` ` ` `count[2] = 2;` ` ` `// Fill the count array` ` ` `// in bottom up manner` ` ` `for` `(let i = 3; i <= dist; i++)` ` ` `count[i] = count[i - 1] +` ` ` `count[i - 2] +` ` ` `count[i - 3];` ` ` `return` `count[dist];` `}` `// Driver code` `let dist = 4;` `document.write(printCountDP(dist));` ` ` `// This code is contributed by divyeshrabadiya07` `</script>` |

**Output :**

7

**Complexity Analysis:****Time Complexity:**O(n).

Only one traversal of the array is needed. So Time Complexity is O(n)**Space complexity:**O(n).

To store the values in a DP O(n) extra space is needed.

__More Optimal Solution__

**Approach: **Instead of using array of size n+1 we can use array of size 3 because for calculating no of ways for a particular step we need only last 3 steps no of ways.

** Algorithm:**

- Create an array of size 3 and initialize the values for step 0,1,2 as 1,1,2 (Base cases).
- Run a loop from 3 to n(dist).
- For each index compute the value as ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3] and store its value at i%3 index of array ways. If we are computing value for index 3 then the computed value will go at index 0 because for larger indices(4 ,5,6…..) we don’t need the value of index 0.
- Return the value of ways[n%3].

## C++

`// A Dynamic Programming based C++ program to count number of ways` `#include<iostream>` `using` `namespace` `std;` ` ` `int` `printCountDP(` `int` `dist)` `{` ` ` `//Create the array of size 3.` ` ` `int` `ways[3] , n = dist;` ` ` ` ` `//Initialize the bases cases` ` ` `ways[0] = 1;` ` ` `ways[1] = 1;` ` ` `ways[2] = 2;` ` ` ` ` `//Run a loop from 3 to n` ` ` `//Bottom up approach to fill the array` ` ` `for` `(` `int` `i=3 ;i<=n ;i++)` ` ` `ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];` ` ` ` ` `return` `ways[n%3];` `}` ` ` `// driver program` `int` `main()` `{` ` ` `int` `dist = 4;` ` ` `cout << printCountDP(dist);` ` ` `return` `0;` `}` |

## Javascript

`<script>` `// A Dynamic Programming based javascript program to count number of ways` ` ` `function` `printCountDP( dist)` `{` ` ` `//Create the array of size 3.` ` ` `var` `ways= [] , n = dist;` ` ` `ways.length = 3 ;` ` ` ` ` `//Initialize the bases cases` ` ` `ways[0] = 1;` ` ` `ways[1] = 1;` ` ` `ways[2] = 2;` ` ` ` ` `//Run a loop from 3 to n` ` ` `//Bottom up approach to fill the array` ` ` `for` `(` `var` `i=3 ;i<=n ;i++)` ` ` `ways[i%3] = ways[(i-1)%3] + ways[(i-2)%3] + ways[(i-3)%3];` ` ` ` ` `return` `ways[n%3];` `}` ` ` `// driver code` ` ` `var` `dist = 4;` ` ` `document.write(printCountDP(dist));` `</script>` |

**Output : **

7

**Time Complexity : O(n)**

**Space Complexity : O(1)**

This article is contributed by Vignesh Venkatesan. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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