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Problem solution · Python

Minimum Time to Make Rope Colorful

Minimum Time to Make Rope Colorful: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
19 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Minimum Time to Make Rope Colorful, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 19 lines of Python from the credited upstream file 1578.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Time to Make Rope Colorful · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def minCost(self, colors: str, neededTime: list[int]) -> int:    ans = 0    maxNeededTime = neededTime[0]     for i in range(1, len(colors)):      if colors[i] == colors[i - 1]:        ans += min(maxNeededTime, neededTime[i])        # For each continuous group, Bob needs to remove every balloon except        # the one with the maximum `neededTime`. So, he should hold the balloon        # with the highest `neededTime` in his hand.        maxNeededTime = max(maxNeededTime, neededTime[i])      else:        # If the current balloon is different from the previous one, discard        # the balloon from the previous group and hold the new one in hand.        maxNeededTime = neededTime[i]     return ans

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