Solving "Minimum time to make rope colorful"

Today we're diving into a challenging problem: Minimum time to make rope colorful.

Understanding the problem

In this challenge, we are presented to a rope with some balloons on it. Alice, the owner of the balloons, does not like when the rope has consecutive balloons of the same color.

Each balloon has a different time (in seconds) to remove.

The goal is to remove the minimum number of balloons to ensure no two consecutive balloons are of the same color.

Visualizing the challenge

Look at the image below to better understand our task:

Initial arrangement of balloons: green(1), blue(2), blue(4), red(3).

Which one do you think we need to remove?

In this setup, the best choice is to remove the second balloon (blue, 2 seconds).

For the rope to be colorful in the best time, we need to remove the balloons that takes less time.

After removal, the arrangement looks like this:

Arrangement after optimal removal: green(1), blue(4), red(3).

Approaching the solution

One of the difficulties of this problem is that the optimal solution relies on a technique we don't often use daily: greedy algorithm.

Before explaining how greedy algorithms work, let's write down how we decide which balloons we remove.

Algorithm steps

What are the necessary steps to find the minimum time to make the rope colorful?

One reason this problem is challenging is the difficulty in identifying the underlying pattern. Many problems can be solved by first considering the appropriate data structure — for instance, using Depth-First Search (DFS) for graph problems or recursion for binary tree problems. However, this particular challenge requires a different approach.

• Iterate over the balloon sequence (a string is a sequence).
• Start from the second balloon to easily compare it with the first.
• When encountering balloons of the same color, remove the one taking less time.
• Keep track of the maximum time spent on a single balloon removal.
• Sum up the time for each removal to get the total time.
• If the current balloon is a different color, reset the maximum time to the current balloon's time.

This approach highlights the importance of making local optimal choices at each step, a hallmark of greedy algorithms.

Greedy algorithm explained

Being "greedy" in this context means making the best immediate choice at each step without reconsidering previous choices. This differs from strategies like backtracking, where you might revisit past decisions.

In this problem, a "sequence" of same-colored balloons forms our local problem. By determining the minimum time to remove balloons in each sequence, we find the overall solution.

Solution

Below is the Python code that solves the problem. Pay close attention to the part of the code that checks if the balloons are the same color and how we calculate the minimum time.

class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        min_time = 0
        max_time_in_sequence = neededTime[0]

        for idx in range(1, len(colors)):
            if colors[idx] == colors[idx - 1]:
                min_time += min(max_time_in_sequence, neededTime[idx])
                max_time_in_sequence = max(max_time_in_sequence, neededTime[idx])
            else:
                max_time_in_sequence = neededTime[idx]

        return min_time

Efficiency analysis: time and space complexity

Time complexity: O(N) where N is the length of the colors string (or neededTime). This is because our solution involves a single for loop that iterates each color exactly once.

Space complexity: O(1). It is because we don't use any extra additional memory. Our solution uses a fixed amount of space, the variables min_time and max_time_in_sequence, which remain constant regardless of whether the number of colors is 1 or 100.

It's important to note that when calculating space complexity, we do not consider the space used by the input itself.

Conclusion

In this article, we learned how to solve the challenge "Minimum Time to Make Rope Colorful" and apply greedy algorithm using a greedy algorithm. We also explored how to determine when a greedy approach is the most effective strategy.

Keys points of this article include:

• Iterating through the balloon sequence starting from the second balloon, allowing for a straightforward comparison between adjacent balloons:

for idx in range(1, len(colors)):

• Calculating the minimum time for removal when encountering balloons of the same color:

min_time += min(max_time_in_sequence, neededTime[idx])
max_time_in_sequence = max(max_time_in_sequence, neededTime[idx])