Minimum Time to Make Rope Colorful

How it works

We can simplify the logic further by tracking the maximum removal time seen so far in the current group of same-colored balloons. For each balloon, we add the minimum of the current maximum and the current balloon's time to the result, then update the maximum. When we encounter a different color, we reset the maximum to 0.

Steps

1. Initialize res = 0 and maxi = 0.
2. For each index i:
   • If i > 0 and colors[i] != colors[i-1], reset maxi = 0.
   • Add min(maxi, neededTime[i]) to res.
   • Update maxi = max(maxi, neededTime[i]).
3. Return res.

Complexity

• Time complexity: O(n)
• Space complexity: O(1) extra space.

Things to watch for

Removing the Wrong Balloon

When encountering consecutive same-colored balloons, some may instinctively remove the first balloon encountered or remove all of them. The correct strategy is to keep the balloon with the maximum removal time and remove all others, minimizing total cost.

Forgetting to Reset State Between Groups

When processing groups of consecutive same-colored balloons, failing to reset the maximum or sum tracking variables when transitioning to a new color group leads to incorrect calculations. Each group must be processed independently.

Off-by-One Errors in Group Boundaries

When using two pointers to identify groups of consecutive balloons, it is easy to mishandle the boundary conditions, either including an extra balloon from the next group or missing the last balloon in the current group. Carefully ensure the inner loop condition correctly identifies when colors change.

from typing import List

class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        res = maxi = 0
        for i in range(len(colors)):
            if i and colors[i] != colors[i - 1]:
                maxi = 0
            res += min(maxi, neededTime[i])
            maxi = max(maxi, neededTime[i])
        return res

class Solution {
public:
    int minCost(string colors, vector<int>& neededTime) {
        int res = 0, maxi = 0;
        for (int i = 0; i < colors.size(); i++) {
            if (i > 0 && colors[i] != colors[i - 1]) {
                maxi = 0;
            }
            res += min(maxi, neededTime[i]);
            maxi = max(maxi, neededTime[i]);
        }
        return res;
    }
};