** Puzzle: **A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) an unlimited number of times.

** Answer:** 25000 kms

** Solution:** Divide the lifetime of the spare tire into 4 equal part i.e., 5000 and swap it at each completion of 5000 miles distance.

Let four tyres be named as A, B, C and D and spare tyre be S.

**5000 KMs:**Replace A with S. Remaining distances (A, B, C, D, S) : 15000, 15000, 15000, 15000, 20000.

**10000 KMs:**Put A back to its original position and replace B with S. Remaining distances (A, B, C, D, S) : 15000, 10000, 10000, 10000, 15000.

**15000 KMs:**Put B back to its original position and replace C with S. Remaining distances (A, B, C, D, S) : 10000, 10000, 5000, 5000, 10000.

**20000 KMs:**Put C back to its original position and replace D with S. Remaining distances (A, B, C, D, S) : 5000, 5000, 5000, 0, 5000.

**25000 KMs:**Every tyre is now worn out completely.

All tyres are used to their full strength.

**Related Amazon interview question **

There are n pencils, each having l length. Each can write 4 kilometres. After writing 4 kilometres it has l/4 length. One can join 4 pencils which are having l/4 length and can make 1 pencil. One can’t make pencil of pieces if remaining pieces are 3 or 2 or 1 in number but one can include these remaining pieces whenever needed.

Write a relation independent of l, length of the given pencil, for how much one can write from n pencils.

**Examples:**

Input:4Output:20

**Recursive Approach: **

Suppose we use 3 pencils that will 12 and generate 3 used pencils, now if the remaining pencils are greater than zero at least 1 unused pencil can be used with those 3 unused to write 4 and that will generate 1 more unused pencil. This will keep repeating.

if(n-3 >= 1){ f(n) = f(n-3) + 12 + 4 } else{ // Used pencil that cannot be used f(n) = 4 + 4 }

Below is the implementation of the above approach:

`int` `count(` `int` `n) ` `{ ` ` ` `if` `(n < 4) { ` ` ` `return` `(n * 4); ` ` ` `} ` ` ` `else` `{ ` ` ` `return` `(16 + count(n - 3)); ` ` ` `} ` `} ` |

**Mathematical Approach O(1): **

Above relation can be optimized in O(1)

// x is max no of time we can //subtract 3 without n-3 <= 3 n - 3 + x <= 3 x > (n-3)/3 i.e.n/3 - 1if it divides exactly elsen/3

Below is the implementation of the above approach:

## CPP

`// C++ program to find relation independent of l ` `// length of the given pencil, for how much one ` `// can write from n pencils. ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find no of pencils ` `int` `count(` `int` `n) ` `{ ` ` ` `int` `x = (n / 3) - 1; ` ` ` `if` `(n % 3) { ` ` ` `x++; ` ` ` `} ` ` ` `return` `(4 * x + 4 * n); ` `} ` ` ` `// Driver function ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` `cout << count(n) << endl; ` `} ` |

## Java

`// Java program to find relation independent of l ` `// length of the given pencil, for how much one ` `// can write from n pencils. ` `class` `GFG ` `{ ` ` ` `// Function to find no of pencils ` `static` `int` `count(` `int` `n) ` `{ ` ` ` `int` `x = (n / ` `3` `) - ` `1` `; ` ` ` `if` `(n % ` `3` `> ` `0` `) ` ` ` `{ ` ` ` `x++; ` ` ` `} ` ` ` `return` `(` `4` `* x + ` `4` `* n); ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `n = ` `5` `; ` ` ` `System.out.print(count(n) +` `"\n"` `); ` `} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 ` |

## Python3

`# Python3 program to find relation independent of l ` `# length of the given pencil, for how much one ` `# can write from n pencils. ` ` ` `# Function to find no of pencils ` `def` `count(n): ` ` ` `x ` `=` `(n ` `/` `/` `3` `) ` `-` `1` `; ` ` ` `if` `(n ` `%` `3` `> ` `0` `): ` ` ` `x` `+` `=` `1` `; ` ` ` ` ` `return` `(` `4` `*` `x ` `+` `4` `*` `n); ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `5` `; ` ` ` `print` `(count(n)); ` ` ` `# This code is contributed by PrinciRaj1992 ` |

## C#

`// C# program to find relation independent of l ` `// length of the given pencil, for how much one ` `// can write from n pencils. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Function to find no of pencils ` `static` `int` `count(` `int` `n) ` `{ ` ` ` `int` `x = (n / 3) - 1; ` ` ` `if` `(n % 3 > 0) ` ` ` `{ ` ` ` `x++; ` ` ` `} ` ` ` `return` `(4 * x + 4 * n); ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `n = 5; ` ` ` `Console.Write(count(n) +` `"\n"` `); ` `} ` `} ` ` ` `// This code is contributed by 29AjayKumar ` |

**Output:**

24