2742. Painting the Walls

Problem

You are given two 0-indexed integer arrays, cost and time, of size n representing the costs and the time taken to paint n different walls respectively. There are two painters available:

• A paid painter that paints the ith wall in time[i] units of time and takes cost[i] units of money.
• A free painter that paints any wall in 1 unit of time at a cost of 0. But the free painter can only be used if the paid painter is already occupied.

Return **the minimum amount of money required to paint the **n walls.

Example 1:

Input: cost = [1,2,3,2], time = [1,2,3,2]
Output: 3
Explanation: The walls at index 0 and 1 will be painted by the paid painter, and it will take 3 units of time; meanwhile, the free painter will paint the walls at index 2 and 3, free of cost in 2 units of time. Thus, the total cost is 1 + 2 = 3.

Example 2:

Input: cost = [2,3,4,2], time = [1,1,1,1]
Output: 4
Explanation: The walls at index 0 and 3 will be painted by the paid painter, and it will take 2 units of time; meanwhile, the free painter will paint the walls at index 1 and 2, free of cost in 2 units of time. Thus, the total cost is 2 + 2 = 4.

Constraints:

• 1 <= cost.length <= 500
• cost.length == time.length
• 1 <= cost[i] <= 106
• 1 <= time[i] <= 500

Solution (Java)

class Solution {
    public int paintWalls(int[] cost, int[] time) {
        int n = cost.length, dp[] = new int[n + 1];
        Arrays.fill(dp, (int)1e9);
        dp[0] = 0;
        for (int i = 0; i < n; ++i)
            for (int j = n; j > 0; --j)
                dp[j] = Math.min(dp[j], dp[Math.max(j - time[i] - 1, 0)] + cost[i]);
        return dp[n];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).