# Count the number of ways to tile the floor of size n x m using 1 x m size tiles

Given a floor of size n x m and tiles of size 1 x m. The problem is to count the number of ways to tile the given floor using 1 x m tiles. A tile can either be placed horizontally or vertically.
Both n and m are positive integers and 2 < = m.

Examples:

```Input : n = 2, m = 3
Output : 1
Only one combination to place
two tiles of size 1 x 3 horizontally
on the floor of size 2 x 3.

Input :  n = 4, m = 4
Output : 2
1st combination:
All tiles are placed horizontally
2nd combination:
All tiles are placed vertically.```

This problem is mainly a more generalized approach to the Tiling Problem.
Approach: For a given value of n and m, the number of ways to tile the floor can be obtained from the following relation.

```            |  1, 1 < = n < m
count(n) = |  2, n = m
| count(n-1) + count(n-m), m < n

```
 ` `
`// C++ implementation to count number of ways to `
`// tile a floor of size n x m using 1 x m tiles`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// function to count the total number of ways`
`int` `countWays(``int` `n, ``int` `m)`
`{`
`    ``// table to store values`
`    ``// of subproblems`
`    ``int` `count[n+1];`
`    ``count[0] = 0;`
`    `
`    ``// Fill the table upto value n`
`    ``for` `(``int` `i = 1; i< = n; i++)`
`    ``{`
`        ``// recurrence relation`
`        ``if` `(i > m)`
`            ``count[i] = count[i-1] + count[i-m];`
`        `
`        ``// base cases    `
`        ``else` `if` `(i < m)    `
`            ``count[i] = 1;`
`        ``// i = = m    `
`        ``else`
`            ``count[i] = 2;`
`    ``}`
`    `
`    ``// required number of ways`
`    ``return` `count[n];`
`}`
`// Driver program to test above`
`int` `main()`
`{`
`    ``int` `n = 7, m = 4;`
`    ``cout << ``"Number of ways = "`
`         ``<< countWays(n, m);`
`    ``return` `0;`
`}`

Output:

```Number of ways = 5
```

Time Complexity: O(n)
Auxiliary Space: O(n)

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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