#1578

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Time to Make Rope Colorful

Minimize the time Bob needs to remove balloons to make a rope colorful using dynamic programming with state transitions.

array string dynamic programming greedy

Problem Statement

Alice has a rope with n balloons, each balloon colored as per the string colors. Each balloon has a removal cost represented by the neededTime array, where each value corresponds to the time it takes to remove the corresponding balloon. Bob is tasked with making the rope colorful by ensuring no two adjacent balloons have the same color. He can remove balloons from the rope, and the objective is to minimize the total time spent removing balloons.

You are required to return the minimum time Bob needs to make the rope colorful. This can be achieved by calculating the minimum removal times while ensuring that there are no consecutive balloons of the same color, employing dynamic programming to track state transitions.

Examples

Example 1

Input: colors = "abaac", neededTime = [1,2,3,4,5]

Output: 3

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.

Example 2

Input: colors = "abc", neededTime = [1,2,3]

Output: 0

The rope is already colorful. Bob does not need to remove any balloons from the rope.

Example 3

Input: colors = "aabaa", neededTime = [1,2,3,4,1]

Output: 2

Bob will remove the balloons at indices 0 and 4. Each balloons takes 1 second to remove. There are no longer two consecutive balloons of the same color. Total time = 1 + 1 = 2.

Constraints

n == colors.length == neededTime.length
1 <= n <= 105
1 <= neededTime[i] <= 104
colors contains only lowercase English letters.

Solution Approach

State Transition Dynamic Programming

The problem can be modeled using dynamic programming where we track the running sum of removal times for repeated colors. As we traverse through the colors array, if two consecutive balloons are the same, we keep the balloon with the minimum removal time and discard the one with the higher time.

Tracking Maximum Removal Time

While processing the repeated balloons, we maintain a running total of the removal time. For each pair of consecutive balloons with the same color, the balloon with the larger removal time should be removed. The total removal time will be the sum of these removals.

Efficient Traversal

Traverse the colors and neededTime arrays once. For each repeated balloon, update the removal time, ensuring that the approach works in linear time to meet the problem's constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of balloons, as the solution involves a single traversal of the colors array. The space complexity is O(1), as we only need to maintain a few variables for the running sum and maximum removal time during the traversal.

What Interviewers Usually Probe

Candidate should understand how dynamic programming can minimize the removal time by handling repeated characters efficiently.
They should demonstrate the ability to track state transitions and handle consecutive characters with minimal time.
A successful candidate should consider space optimization, ensuring the approach runs in O(1) space.

Common Pitfalls or Variants

Common pitfalls

Forgetting to update the removal time correctly when processing consecutive balloons with the same color.
Not efficiently handling the state transition between colors, leading to a suboptimal solution.
Misunderstanding the dynamic programming approach, which may lead to excessive computation or unnecessary space usage.

Follow-up variants

The problem could involve different balloon removal constraints or require tracking the number of removals.
The same concept can be applied to more complex sequences of colors or higher constraints.
The dynamic programming solution might be adapted for cases where there are additional conditions for removing balloons, such as prioritizing certain colors.

FAQ

How do I handle repeated colors efficiently?

Track the removal time for each pair of consecutive repeated colors and always remove the balloon with the higher removal time to minimize the total time.

What is the time complexity of the solution?

The time complexity of the solution is O(n), as it requires a single pass through the input arrays to calculate the removal times.

How can I ensure that my solution meets the problem's constraints?

Ensure that your approach runs in O(n) time and uses O(1) space, as these are the expected complexities for this problem.

Can I use a greedy approach to solve this problem?

Yes, a greedy approach works in this case, as we always want to remove the balloon with the higher removal time between consecutive repeated colors.

What pattern does the problem follow?

This problem follows the 'State Transition Dynamic Programming' pattern, where you minimize the removal time by tracking repeated characters and updating removal costs efficiently.

terminal

Solution

Solution 1: Two Pointers + Greedy

We can use two pointers to point to the beginning and end of the current consecutive balloons with the same color, then calculate the total time $s$ of these consecutive balloons with the same color, as well as the maximum time $mx$. If the number of consecutive balloons with the same color is greater than $1$, we can greedily choose to keep the balloon with the maximum time and remove the other balloons with the same color, which takes time $s - mx$, and add it to the answer. Then we continue to traverse until all balloons are traversed.

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class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        ans = i = 0
        n = len(colors)
        while i < n:
            j = i
            s = mx = 0
            while j < n and colors[j] == colors[i]:
                s += neededTime[j]
                if mx < neededTime[j]:
                    mx = neededTime[j]
                j += 1
            if j - i > 1:
                ans += s - mx
            i = j
        return ans

Continue Practicing

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