Today I have done leetcode's October LeetCoding Challenge with cpp.

October LeetCoding Challenge 14

Description

Painting the Walls

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

class Solution {
public:
  int paintWalls(vector<int>& cost, vector<int>& time) {
    int n = cost.size();
    vector<int> dp(n + 1, 250000000);
    dp[0] = 0;

    for(int i = 0; i < n; ++i) {
      for(int j = n; j > 0; --j) {
        dp[j] = min(dp[j], dp[max(j - time[i] - 1, 0)] + cost[i]);
      }
    }

    return dp[n];
  }
};

// Accepted
// 2557/2557 cases passed (88 ms)
// Your runtime beats 95.12 % of cpp submissions
// Your memory usage beats 91.34 % of cpp submissions (89.6 MB)