# Count number of ways to cover a distance

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

Examples :

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

## 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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` `?> ` |

*chevron_right*

*filter_none*

Output:

7

The time complexity of above solution is exponential, a close upper bound is O(3^{n}). If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, when we start from n = 6, we can reach 4 by subtracting one 2 times and by subtracting 2 one times. So the subproblem for 4 is called twice.

Since same suproblems are called again, this problem has Overlapping Subprolems property. So min square sum problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array count[] in bottom up manner.

Below is the implementation of above idea :

## 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, 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 program ` `int` `main() ` `{ ` ` ` `int` `dist = 4; ` ` ` `cout << printCountDP(dist); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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` `; ` ` ` `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 program ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `{ ` ` ` `int` `dist = ` `4` `; ` ` ` `System.out.println(printCountDP(dist)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Pramod Kumar ` |

*chevron_right*

*filter_none*

## 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` ` ` `count[` `1` `] ` `=` `1` ` ` `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. ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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. ` `?> ` |

*chevron_right*

*filter_none*

**Output :**

7

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

## Recommended Posts:

- Count number of ways to jump to reach end
- Count the number of ways to traverse a Matrix
- Count number of ways to partition a set into k subsets
- Count number of ways to reach destination in a Maze
- Count number of ways to reach a given score in a game
- Count number of ways to reach a given score in a Matrix
- Count number of ways to fill a "n x 4" grid using "1 x 4" tiles
- Count the number of ways to tile the floor of size n x m using 1 x m size tiles
- Print all possible ways to convert one string into another string | Edit-Distance
- Count of different ways to express N as the sum of 1, 3 and 4
- Count possible ways to construct buildings
- Count ways to reach the n'th stair
- Count ways to increase LCS length of two strings by one
- Count ways to build street under given constraints
- Count ways to reach the nth stair using step 1, 2 or 3