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 answer can be large return it modulo 10^9 + 7.**Examples:**

Input : n = 2 k = 4 Output : 16 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 = 2 Output : 6

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 + sum 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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**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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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_ ` |

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## 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 ` `?> ` |

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**Output:**

6

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