Paint Fence

Question (LC.276)

There are n posts and 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. n and k are non-zero integers.

Example

(3, 2) => 6

P1 P2 P2
1  1  2
1  2  1
2  1  1
2  2  1
2  1  2
1  2  2

Analysis

We only want the number of ways not the actual solutions. So we can try DP.

Single Sequence DP

Step 1 Define optimal subproblem
fence(i, 0) = max number of ways to apply the same color at the ith post
fence(i, 1) = max # ways to apply different colors at the ith post

Step 2 Recurrence
fence(i, 0) = fence(i-1, 1)
fence(i, 1) = fence(i-1, 0) * (k-1) + fence(i-1, 1) * (k-1)

Step 3 Base cases
fence(0, 0) = 0
fence(0, 1) = 0
fence(1, 0) = k
fence(1, 1) = 0
fence(2, 0) = k
fence(2, 1) = k * (k - 1)
fence(3, 0) = fence(2, 1)
fence(3, 1) = fence(2, 0) * (k - 1) + fence(2, 1) * (k - 1)

Step 4 Topo Order
    for i from 3 to n

Stet 5 Final Answer
    fence(n, 0) + fence(n, 1)

Code

public int numWays(int n, int k) {
    // create memo table
    int adjN = n < 2 ? 2 : n;
    int[][] fence = new int[adjN + 1][2];

    // init base case
    fence[0][0] = 0;
    fence[0][1] = 0;
    fence[1][0] = k;
    fence[1][1] = 0;
    fence[2][0] = k;
    fence[2][1] = k * (k - 1);

    // topo order
    for (int i = 3; i <= n; i++) {
        fence[i][0] = fence[i-1][1];
        fence[i][1] = fence[i-1][0] * (k - 1) + fence[i-1][1] * (k - 1);
    }

    // final answer
    return fence[n][0] + fence[n][1];
}

There is only one layer of dependencies so we can compress the active states.

Code w/ State Compression

public int numWays(int n, int k) {
    // create memo table
    int[][] fence = new int[3][2];

    // init base case
    fence[0][0] = 0;
    fence[0][1] = 0;
    fence[1][0] = k;
    fence[1][1] = 0;
    fence[2][0] = k;
    fence[2][1] = k * (k - 1);

    // topo order
    for (int i = 3; i <= n; i++) {
        fence[i%3][0] = fence[(i-1)%3][1];
        fence[i%3][1] = fence[(i-1)%3][0] * (k - 1) + fence[(i-1)%3][1] * (k - 1);
    }

    // final answer
    return fence[n%3][0] + fence[n%3][1];
}

Time Complexity

#subproblems * time/subproblem = 2n * O(1) = O(n)