Walls Coloring II

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Intuition

Color n walls with k colors minimizing cost where adjacent walls must differ. DP.

Algorithm

1dp[i][c] = min cost to color walls 1..i with wall i colored c. Transition: min over all c != prev_c.

Common Pitfalls

•O(nk^2) naive. Optimize to O(nk) by tracking min and second-min of previous row.

Walls Coloring II.java

Java

// Approach: DP. dp[i][c] = min cost to paint wall i with color c, considering the k-window constraint.
// Time: O(n * colors) Space: O(n)
class Solution {
    int minCost(int[][] costs) {
        int n = costs.length;
        int k = costs[0].length;
        if (k == 1 && n == 1)
            return costs[0][0];
        if (k <= 1 || n == 0)
            return -1;

        int min = 0, secondMin = -1;
        int minVal = costs[0][0], secondMinVal = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            if (i != 0) {
                for (int j = 0; j < k; j++)
                    costs[i][j] += costs[i - 1][j];
            }

            min = 0;
            secondMin = -1;
            minVal = costs[i][0];
            secondMinVal = Integer.MAX_VALUE;

            // find min and second min
            for (int j = 1; j < k; j++) {

                if (costs[i][j] < minVal) {
                    secondMin = min;
                    secondMinVal = minVal;
                    min = j;
                    minVal = costs[i][j];
                } else if (j != min && (secondMin == -1 || costs[i][j] < secondMinVal)) {
                    secondMin = j;
                    secondMinVal = costs[i][j];
                }
            }

            for (int j = 0; j < k; j++) {
                if (j != min)
                    costs[i][j] = minVal;
                else
                    costs[i][j] = secondMinVal;
            }
        }

        return minVal;
    }
}

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