LeetCode 2742: Painting the Walls Solution

Master LeetCode problem 2742 (Painting the Walls), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2742. Painting the Walls

Hard

Array Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming to find the minimum cost required to paint the walls. We can maintain two dp arrays: one for the cost of painting the walls when the free painter is used and one when the paid painter is used. We iterate through the walls and update the dp arrays based on whether it is more optimal to use the free painter or the paid painter for each wall.

Initialize two dp arrays, dpPaid and dpFree, of size n+1.
Set dpPaid[0] = 0 and dpFree[0] = 0 as the base cases.
Iterate through each wall from 1 to n:
- For each wall i, update dpPaid[i] and dpFree[i] based on the minimum cost of painting the wall using the paid painter or the free painter.
The minimum cost to paint all walls will be the minimum value between dpPaid[n] and dpFree[n].

Time Complexity: O(n) where n is the number of walls. Space Complexity: O(n) for the dp arrays.

Solution Code

class Solution {
    public int minCost(int[] cost, int[] time) {
        int n = cost.length;
        int[] dpPaid = new int[n+1];
        int[] dpFree = new int[n+1];
        
        dpPaid[0] = 0;
        dpFree[0] = 0;
        
        for (int i = 1; i <= n; i++) {
            dpPaid[i] = Math.min(dpPaid[i-1] + cost[i-1], dpFree[i-1] + cost[i-1]);
            dpFree[i] = Math.min(dpPaid[i-1], dpFree[i-1]) + time[i-1];
        }
        
        return Math.min(dpPaid[n], dpFree[n]);
    }
}

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