LeetCode 1578 - Minimum Time to Make Rope Colorful

The problem requires transforming a rope of balloons into a "colorful" rope, which means no two consecutive balloons sha

leetcodemediumarraystringdynamic-programminggreedy

LeetCode Problem 1578

Difficulty: 🟡 Medium
Topics: Array, String, Dynamic Programming, Greedy

Solution

Problem Understanding

The problem requires transforming a rope of balloons into a "colorful" rope, which means no two consecutive balloons share the same color. The input provides a string colors representing the colors of the balloons in order and an array neededTime representing the time in seconds it takes to remove each corresponding balloon. The task is to determine the minimum total time required to remove balloons such that no two consecutive balloons have the same color.

The key is that when consecutive balloons share the same color, you must remove all but one of them. The naive way might try every possible removal sequence, but we only need to keep the balloon with the highest removal time in each consecutive group to minimize the total time.

The constraints show that the number of balloons can be up to 10^5 and each removal time is at most 10^4. This suggests that an efficient linear-time solution is necessary, as any algorithm significantly worse than O(n) will be too slow.

Important edge cases include strings with no consecutive repeating colors, strings where all balloons are the same color, and strings where the optimal choice requires keeping a non-first or non-last balloon in a group.

Approaches

The brute-force approach would iterate through the string, and every time two consecutive colors match, consider removing one. However, this can lead to repeatedly recalculating which balloon to remove, especially for long consecutive runs of the same color. The complexity could degrade toward O(n²) for worst-case inputs like "aaaaa...".

The optimal approach leverages a greedy strategy: for any consecutive group of balloons with the same color, remove all except the one with the largest neededTime. The rationale is that removing the balloons with smaller times minimizes the total removal cost. This can be implemented in a single linear pass through the array, maintaining a running total of the cost.

Approach Time Complexity Space Complexity Notes
Brute Force O(n²) O(1) Iteratively removes balloons for each duplicate pair, recalculating conflicts each time
Optimal (Greedy) O(n) O(1) Keeps max-cost balloon in each group of consecutive same colors, sums other removal times

Algorithm Walkthrough

  1. Initialize total_time to 0. This will accumulate the minimum removal time.

  2. Initialize prev_time to the neededTime of the first balloon. This keeps track of the max time in the current consecutive color group.

  3. Iterate over the balloons from the second one to the end:

  4. If the current balloon's color is the same as the previous one:

  5. Add the smaller of prev_time and the current balloon's neededTime to total_time.

  6. Update prev_time to the maximum of prev_time and the current balloon's neededTime (keeping the most expensive balloon to keep).

  7. If the current balloon's color is different from the previous one, update prev_time to the current balloon's neededTime.

  8. After processing all balloons, total_time will represent the minimum time to make the rope colorful.

Why it works: This approach guarantees that in any consecutive group of the same color, the balloon with the highest removal time is preserved, and all others are removed, which minimizes total removal time. The invariant maintained is that at any step, prev_time always contains the maximum neededTime of the current color block.

Python Solution

from typing import List

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

        for i in range(1, len(colors)):
            if colors[i] == colors[i - 1]:
                total_time += min(prev_time, neededTime[i])
                prev_time = max(prev_time, neededTime[i])
            else:
                prev_time = neededTime[i]

        return total_time

Explanation: We start with the first balloon and iterate through the rest. For each consecutive duplicate, we remove the cheaper balloon and update the max cost for the current group. When we encounter a new color, we reset the previous max to the current balloon's removal time. This guarantees the total time is minimal.

Go Solution

func minCost(colors string, neededTime []int) int {
    totalTime := 0
    prevTime := neededTime[0]

    for i := 1; i < len(colors); i++ {
        if colors[i] == colors[i-1] {
            if neededTime[i] < prevTime {
                totalTime += neededTime[i]
            } else {
                totalTime += prevTime
                prevTime = neededTime[i]
            }
        } else {
            prevTime = neededTime[i]
        }
    }

    return totalTime
}

Explanation: The Go solution mirrors the Python logic. Since Go does not have dynamic typing, we use integer comparisons directly. The prevTime variable tracks the highest removal time in a consecutive group, ensuring the minimal removal cost.

Worked Examples

Example 1:

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

Index Color neededTime Action total_time prev_time
0 a 1 start 0 1
1 b 2 new color 0 2
2 a 3 new color 0 3
3 a 4 duplicate 3 4
4 c 5 new color 3 5

Output: 3

Example 2:

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

No duplicates, total_time = 0.

Example 3:

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

Index Color neededTime Action total_time prev_time
0 a 1 start 0 1
1 a 2 duplicate 1 2
2 b 3 new color 1 3
3 a 4 new color 1 4
4 a 1 duplicate 2 4

Output: 2

Complexity Analysis

Measure Complexity Explanation
Time O(n) Single pass through the colors array, comparing each balloon with its predecessor
Space O(1) Only constant extra space for total_time and prev_time is used

The algorithm efficiently handles the maximum input size because it avoids nested loops and uses only a few variables.

Test Cases

# Provided examples
assert Solution().minCost("abaac", [1,2,3,4,5]) == 3  # example 1
assert Solution().minCost("abc", [1,2,3]) == 0       # example 2
assert Solution().minCost("aabaa", [1,2,3,4,1]) == 2 # example 3

# Edge cases
assert Solution().minCost("a", [5]) == 0             # single balloon
assert Solution().minCost("aa", [1,2]) == 1         # two same-color balloons
assert Solution().minCost("aaa", [3,1,2]) == 3      # multiple duplicates, keep max
assert Solution().minCost("ababab", [1,2,1,2,1,2]) == 0 # alternating colors
assert Solution().minCost("bbbbbb", [1,2,3,4,5,6]) == 15 # all same color
Test Why
"abaac", [1,2,3,4,5] Normal case with one duplicate group
"abc", [1,2,3] No duplicates, should return 0
"aabaa", [1,2,3,4,1] Duplicates at start and end, multiple removal
"a", [5] Single balloon, no removal
"aa", [1,2] Two balloons same color, remove the cheaper
"aaa", [3,1,2] Three same-color balloons, keep max cost
"ababab", [1,2,1,2,1,2] Alternating colors, no removal
"bbbbbb", [1,2,3,4,5,6] All same color, sum all except max

Edge Cases

Single Balloon: If colors contains only one balloon, no removal is needed. The algorithm handles this because the loop never executes and total_time remains 0.

All Balloons Same Color: For inputs like "aaaaa", the algorithm correctly sums