# Painting Fence Algorithm

Given a fence with n posts and k colors, find out the number of ways of painting the fence such that at most 2 adjacent posts have the same color. Since the answer can be large return it modulo 10^9 + 7.

**Examples:**

Input :n = 2 k = 4Output :16Explanation:We have 4 colors and 2 posts.

Ways when both posts have same color : 4

Ways when both posts have diff color :4(choices for 1st post) * 3(choices for 2nd post) = 12

Input :n = 3 k = 2Output :6

The following image depicts the 6 possible ways of painting 3 posts with 2 colors:

Consider the following image in which c, c’ and c” are the respective colors of posts i, i-1, and i -2.

According to the constraint of the problem, c = c’ = c” is not possible simultaneously, so either c’ != c or c” != c or both. There are k – 1 possibility for c’ != c and k – 1 for c” != c.

diff = no of ways when color of last two posts is different same = no of ways when color of last two posts is same total ways = diff + same for n = 1 diff = k, same = 0 total = k for n = 2 diff = k * (k-1) //k choices for first post, k-1 for next same = k //k choices for common color of two posts total = k + k * (k-1) for n = 3 diff = k * (k-1)* (k-1) //(k-1) choices for the first place // k choices for the second place //(k-1) choices for the third place same = k * (k-1) * 2 // 2 is multiplied because consider two color R and B // R R B or B R R // B B R or R B B c'' != c, (k-1) choices for it Hence we deduce that, total[i] = same[i] + diff[i] same[i] = diff[i-1] diff[i] = (diff[i-1] + diff[i-2]) * (k-1) = total[i-1] * (k-1)

Below is the implementation of the problem:

## C++

`// C++ program for Painting Fence Algorithm` `// optimised version` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of ways to color k posts` `long` `countWays(` `int` `n, ` `int` `k)` `{` ` ` `long` `dp[n + 1];` ` ` `memset` `(dp, 0, ` `sizeof` `(dp));` ` ` `long` `long` `mod = 1000000007;` ` ` `dp[1] = k;` ` ` `dp[2] = k * k;` ` ` `for` `(` `int` `i = 3; i <= n; i++) {` ` ` `dp[i] = ((k - 1) * (dp[i - 1] + dp[i - 2])) % mod;` ` ` `}` ` ` `return` `dp[n];` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3, k = 2;` ` ` `cout << countWays(n, k) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program for Painting Fence Algorithm` `import` `java.util.*;` `class` `GfG {` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `static` `long` `countWays(` `int` `n, ` `int` `k)` ` ` `{` ` ` `// To store results for subproblems` ` ` `long` `dp[] = ` `new` `long` `[n + ` `1` `];` ` ` `Arrays.fill(dp, ` `0` `);` ` ` `int` `mod = ` `1000000007` `;` ` ` `// There are k ways to color first post` ` ` `dp[` `1` `] = k;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `int` `same = ` `0` `, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `int` `i = ` `2` `; i <= n; i++) {` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = (` `int` `)(dp[i - ` `1` `] * (k - ` `1` `));` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `dp[i] = (same + diff) % mod;` ` ` `}` ` ` `return` `dp[n];` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `n = ` `3` `, k = ` `2` `;` ` ` `System.out.println(countWays(n, k));` ` ` `}` `}` `// This code contributed by Rajput-Ji` |

## Python3

`# Python3 program for Painting Fence Algorithm ` `# optimised version` `# Returns count of ways to color k posts ` `def` `countWays(n, k):` ` ` ` ` `dp ` `=` `[` `0` `] ` `*` `(n ` `+` `1` `)` ` ` `total ` `=` `k ` ` ` `mod ` `=` `1000000007` ` ` ` ` `dp[` `1` `] ` `=` `k` ` ` `dp[` `2` `] ` `=` `k ` `*` `k ` ` ` ` ` `for` `i ` `in` `range` `(` `3` `,n` `+` `1` `):` ` ` `dp[i] ` `=` `((k ` `-` `1` `) ` `*` `(dp[i ` `-` `1` `] ` `+` `dp[i ` `-` `2` `])) ` `%` `mod` ` ` ` ` `return` `dp[n]` ` ` `# Driver code ` `n ` `=` `3` `k ` `=` `2` `print` `(countWays(n, k))` ` ` `# This code is contributed by shubhamsingh10` |

## C#

`// C# program for Painting Fence Algorithm` `using` `System;` `public` `class` `GFG` `{` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `static` `long` `countWays(` `int` `n, ` `int` `k)` ` ` `{` ` ` `// To store results for subproblems` ` ` `long` `[] dp = ` `new` `long` `[n + 1];` ` ` `Array.Fill(dp, 0);` ` ` `int` `mod = 1000000007;` ` ` `// There are k ways to color first post` ` ` `dp[1] = k;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `int` `same = 0, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `int` `i = 2; i <= n; i++) ` ` ` `{` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = (` `int` `)(dp[i - 1] * (k - 1));` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `dp[i] = (same + diff) % mod;` ` ` `}` ` ` `return` `dp[n];` ` ` `}` ` ` `// Driver code` ` ` `static` `public` `void` `Main ()` ` ` `{` ` ` `int` `n = 3, k = 2;` ` ` `Console.WriteLine(countWays(n, k));` ` ` `}` `}` `// This code is contributed by avanitrachhadiya2155` |

## Javascript

`<script>` ` ` `// Javascript program for Painting Fence Algorithm` ` ` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `function` `countWays(n, k)` ` ` `{` ` ` `// To store results for subproblems` ` ` `let dp = ` `new` `Array(n + 1);` ` ` `dp.fill(0);` ` ` `let mod = 1000000007;` ` ` `// There are k ways to color first post` ` ` `dp[1] = k;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `let same = 0, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(let i = 2; i <= n; i++)` ` ` `{` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = (dp[i - 1] * (k - 1));` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `dp[i] = (same + diff) % mod;` ` ` `}` ` ` `return` `dp[n];` ` ` `}` ` ` ` ` `let n = 3, k = 2;` ` ` `document.write(countWays(n, k));` ` ` ` ` `// This code is contributed by divyeshrabadiya07.` `</script>` |

**Output**

6

**Time Complexity: **O(N)**Auxiliary Space: **O(N)

**Space optimization : **

We can optimize the above solution to use one variable instead of a table.

Below is the implementation of the problem:

## C++

`// C++ program for Painting Fence Algorithm` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns count of ways to color k posts` `// using k colors` `long` `countWays(` `int` `n, ` `int` `k)` `{` ` ` `// There are k ways to color first post` ` ` `long` `total = k;` ` ` `int` `mod = 1000000007;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color) and k ways to` ` ` `// not violate (different color)` ` ` `int` `same = 0, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `int` `i = 2; i <= n; i++) {` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = total * (k - 1);` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `total = (same + diff) % mod;` ` ` `}` ` ` `return` `total;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3, k = 2;` ` ` `cout << countWays(n, k) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program for Painting Fence Algorithm` `class` `GFG {` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `static` `long` `countWays(` `int` `n, ` `int` `k)` ` ` `{` ` ` `// There are k ways to color first post` ` ` `long` `total = k;` ` ` `int` `mod = ` `1000000007` `;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `int` `same = ` `0` `, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `int` `i = ` `2` `; i <= n; i++) {` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = (` `int` `)total * (k - ` `1` `);` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `total = (same + diff) % mod;` ` ` `}` ` ` `return` `total;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `n = ` `3` `, k = ` `2` `;` ` ` `System.out.println(countWays(n, k));` ` ` `}` `}` `// This code is contributed by Mukul Singh` |

## Python3

`# Python3 program for Painting ` `# Fence Algorithm ` `# Returns count of ways to color ` `# k posts using k colors ` `def` `countWays(n, k) :` ` ` `# There are k ways to color first post ` ` ` `total ` `=` `k` ` ` `mod ` `=` `1000000007` ` ` `# There are 0 ways for single post to ` ` ` `# violate (same color_ and k ways to ` ` ` `# not violate (different color) ` ` ` `same, diff ` `=` `0` `, k` ` ` `# Fill for 2 posts onwards ` ` ` `for` `i ` `in` `range` `(` `2` `, n ` `+` `1` `) :` ` ` ` ` `# Current same is same as ` ` ` `# previous diff ` ` ` `same ` `=` `diff ` ` ` `# We always have k-1 choices ` ` ` `# for next post ` ` ` `diff ` `=` `total ` `*` `(k ` `-` `1` `) ` ` ` `diff ` `=` `diff ` `%` `mod ` ` ` `# Total choices till i. ` ` ` `total ` `=` `(same ` `+` `diff) ` `%` `mod ` ` ` ` ` `return` `total` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `n, k ` `=` `3` `, ` `2` ` ` `print` `(countWays(n, k)) ` `# This code is contributed by Ryuga` |

## C#

`// C# program for Painting Fence Algorithm` `using` `System;` `class` `GFG {` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `static` `long` `countWays(` `int` `n, ` `int` `k)` ` ` `{` ` ` `// There are k ways to color first post` ` ` `long` `total = k;` ` ` `int` `mod = 1000000007;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `long` `same = 0, diff = k;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `int` `i = 2; i <= n; i++) {` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` `// We always have k-1 choices for next post` ` ` `diff = total * (k - 1);` ` ` `diff = diff % mod;` ` ` `// Total choices till i.` ` ` `total = (same + diff) % mod;` ` ` `}` ` ` `return` `total;` ` ` `}` ` ` `// Driver code` ` ` `static` `void` `Main()` ` ` `{` ` ` `int` `n = 3, k = 2;` ` ` `Console.Write(countWays(n, k));` ` ` `}` `}` `// This code is contributed by DrRoot_` |

## PHP

`<?php ` `// PHP program for Painting Fence Algorithm` `// Returns count of ways to color k ` `// posts using k colors` `function` `countWays(` `$n` `, ` `$k` `)` `{` ` ` `// There are k ways to color first post` ` ` `$total` `= ` `$k` `;` ` ` `$mod` `= 1000000007;` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `$same` `= 0;` ` ` `$diff` `= ` `$k` `;` ` ` `// Fill for 2 posts onwards` ` ` `for` `(` `$i` `= 2; ` `$i` `<= ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `// Current same is same as previous diff` ` ` `$same` `= ` `$diff` `;` ` ` `// We always have k-1 choices for next post` ` ` `$diff` `= ` `$total` `* (` `$k` `- 1);` ` ` `$diff` `= ` `$diff` `% ` `$mod` `;` ` ` `// Total choices till i.` ` ` `$total` `= (` `$same` `+ ` `$diff` `) % ` `$mod` `;` ` ` `}` ` ` `return` `$total` `;` `}` `// Driver code` `$n` `= 3;` `$k` `= 2;` `echo` `countWays(` `$n` `, ` `$k` `) . ` `"\n"` `;` `// This code is contributed by ita_c` `?>` |

## Javascript

`<script>` ` ` `// JavaScript program for Painting Fence Algorithm` ` ` ` ` `// Returns count of ways to color k posts` ` ` `// using k colors` ` ` `function` `countWays(n, k)` ` ` `{` ` ` `// There are k ways to color first post` ` ` `let total = k;` ` ` `let mod = 1000000007;` ` ` ` ` `// There are 0 ways for single post to` ` ` `// violate (same color_ and k ways to` ` ` `// not violate (different color)` ` ` `let same = 0, diff = k;` ` ` ` ` `// Fill for 2 posts onwards` ` ` `for` `(let i = 2; i <= n; i++) {` ` ` `// Current same is same as previous diff` ` ` `same = diff;` ` ` ` ` `// We always have k-1 choices for next post` ` ` `diff = total * (k - 1);` ` ` `diff = diff % mod;` ` ` ` ` `// Total choices till i.` ` ` `total = (same + diff) % mod;` ` ` `}` ` ` ` ` `return` `total;` ` ` `}` ` ` ` ` `let n = 3, k = 2;` ` ` `document.write(countWays(n, k));` ` ` `</script>` |

**Output**

6

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

This article is contributed by **Aditi Sharma**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

## Please

Loginto comment...