# 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 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 possibilities 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

**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

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