2742. Painting the Walls

Description

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

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(n²)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number[]} cost
 * @param {number[]} time
 * @return {number}
 */
const paintWalls = function (cost, time) {
  const n = cost.length;
  const dp = Array.from({ length: n }, () => new Array(n + 1).fill(-1));

  const getPaintCost = (index, remainedWalls) => {
    if (remainedWalls <= 0) return 0;
    if (index >= n) return Number.MAX_SAFE_INTEGER;
    if (dp[index][remainedWalls] !== -1) return dp[index][remainedWalls];

    const nextRemainedWalls = remainedWalls - time[index] - 1;
    const paintCost = cost[index] + getPaintCost(index + 1, nextRemainedWalls);
    const skipCost = getPaintCost(index + 1, remainedWalls);
    const result = Math.min(paintCost, skipCost);

    dp[index][remainedWalls] = result;

    return result;
  };

  return getPaintCost(0, n);
};