Paint Fence

Easy Paid Dynamic Programming

LintCode: http://www.lintcode.com/en/problem/paint-fence/

LeetCode: /lintcode-tags/#Paid

Question

There is a fence with n posts, each post can be painted with one of the k colors.
You have to paint all the posts such that no more than two adjacent fence posts have the same color.
Return the total number of ways you can paint the fence.

Notice
n and k are non-negative integers.

Example
Given n=3, k=2 return 6

      post 1,   post 2, post 3
way1    0         0       1
way2    0         1       0
way3    0         1       1
way4    1         0       0
way5    1         0       1
way6    1         1       0

Thinking

This is a dynamic programming. Dynamic programming : paint fence algorithm
diff - number of combinations with different colors,
same - number of combinations with same colors.

For n = 1:

diff = k;
same = 0;

For n = 2:

diff = k * (k - 1);
same = k;

For n = 3:

diff = (k + k * (k - 1)) * (k - 1);
same = k * (k - 1);

Final formula

diff = (diff[i - 1] + diff[i - 2]) * (k - 1)
same = diff[i - 1] 

Total ways

total[n] = diff[n] + same[n]
         = (diff[n - 1] + diff[n - 2]) * (k - 1) + diff[n - 1]
         = (diff[n - 1] + same[n - 1]) * (k - 1) + (diff[n - 2] + diff[n - 3]) * (k - 1)
         = total[n - 1] * (k - 1) + total[n - 2] * (k - 1)
         = (tatal[n - 1] + total[n - 2]) * (k - 1)

Solution

Java

public class Solution {
    /**
     * @param n non-negative integer, n posts
     * @param k non-negative integer, k colors
     * @return an integer, the total number of ways
     */
    public int numWays(int n, int k) {
        // Write your code here
        if (n <= 0 || k <= 0) {
            return 0;
        }
        if (n == 1) {
            return k;
        } else if (n == 2) {
            return k * k; // k + k * (k - 1)
        } 
        int num1 = k;
        int num2 = k * k;
        int num = 3;
        while (num <= n) {
            int tmp  = num2;
            num2 = (k - 1) * (num2 + num1);
            num1 = tmp;
            ++num;
        }
        return num2;
    }
}