To solve this problem, we can apply dynamic programming with state transitions to handle the removal of balloons. Track the removal time for consecutive repeated colors while keeping a running sum and max time to minimize the cost. This approach ensures that the time is minimized while ensuring no two consecutive balloons share the same color.
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.
How GhostInterview Helps
- GhostInterview helps by breaking down the problem into clear steps of state transitions and removal time calculations.
- It provides an optimized approach for minimizing removal time, ensuring an O(n) solution that meets the problem's constraints.
- The step-by-step guidance of GhostInterview helps refine the dynamic programming approach for handling repeated characters efficiently.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
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.
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