LeetCode 1578 Solution

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the ith balloon.

Alice wants the rope to be colorful. She does not want two consecutive balloons to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it colorful. You are given a 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the ith balloon from the rope.

Return the minimum time Bob needs to make the rope colorful.

Example 1:

• Input: colors = "abaac", neededTime = [1,2,3,4,5]
• Output: 3
• Explanation: In the above image, 'a' is blue, 'b' is red, and 'c' is green.
  Bob can remove the blue balloon at index 2. This takes 3 seconds.
  There are no longer two consecutive balloons of the same color. Total time = 3.

python

class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        result = 0
        flag = None  
        record = []  

        for c, t in zip(colors, neededTime):
            if c != flag:  
                if record: 
                    result += sum(record) - max(record)
                record = [t] 
                flag = c  
            else:
                record.append(t) 

        if record:  
            result += sum(record) - max(record)

        return result

The consecutive block of colors needs to be minimized to just one color and one neededTime. So if the flag is same as the front parts of colors, just append it to the record and flag continues. But if the flag differentialize part appear, sum(record) - max(record) execute and reinitialize the flag and record.

if record check whether the record list is empty or not. If it's empty, it means that the record is reinitialized and needs to be full with current colors.

As the solution search zipped list for just one time and search all not just needed part, it is greedy algorithm. And time and space complexity of this code is both $O(n)$.