1578. Minimum Time to Make Rope Colorful LeetCode Solution

Last updated: February 24, 2025

In this guide, you will get 1578. Minimum Time to Make Rope Colorful LeetCode Solution with the best time and space complexity. The solution to Minimum Time to Make Rope Colorful problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Minimum Time to Make Rope Colorful solution in C++
Minimum Time to Make Rope Colorful solution in Java
Minimum Time to Make Rope Colorful solution in Python
Additional Resources

1578. Minimum Time to Make Rope Colorful LeetCode Solution image

Problem Statement of Minimum Time to Make Rope Colorful

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.
Example 2:

Input: colors = “abc”, neededTime = [1,2,3]
Output: 0
Explanation: 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
Explanation: 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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(1)

1578. Minimum Time to Make Rope Colorful LeetCode Solution in C++

class Solution {
 public:
  int minCost(string colors, vector<int>& neededTime) {
    int ans = 0;
    int maxNeededTime = neededTime[0];

    for (int i = 1; i < colors.length(); ++i)
      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;
  }
};
/* code provided by PROGIEZ */

1578. Minimum Time to Make Rope Colorful LeetCode Solution in Java

class Solution {
  public int minCost(String colors, int[] neededTime) {
    int ans = 0;
    int maxNeededTime = neededTime[0];

    for (int i = 1; i < colors.length(); ++i)
      if (colors.charAt(i) == colors.charAt(i - 1)) {
        ans += Math.min(maxNeededTime, neededTime[i]);
        // For each continuous group, Bob needs to remove every balloon except the one with the max
        // `neededTime`. So, he should hold the balloon with the highest `neededTime` in his hand.
        maxNeededTime = Math.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;
  }
}
// code provided by PROGIEZ

1578. Minimum Time to Make Rope Colorful LeetCode Solution in Python

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
 # code by PROGIEZ

Additional Resources

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