1578 - Minimum Time to Make Rope Colorful

Difficulty: Medium | Pattern: Greedy | Company tags: Amazon, Bloomberg

Problem Statement

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the i-th balloon.

Alice wants the rope to be colorful. Two consecutive balloons with the same color are not allowed. You are given a 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the i-th balloon.

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: Remove the 2nd "a" (cost 3) → "abaac" → "abac" (no wait, "abac" still has 'a')
Actually: colors[3]='a' costs 4, but we remove the cheaper one at index 2 (cost=3).

Example 2:

Input: colors = "abc", neededTime = [1,2,3]
Output: 0  (already colorful)

Example 3:

Input: colors = "aabaa", neededTime = [1,2,3,4,1]
Output: 2  (remove first 'a' at cost 1, last 'a' at cost 1)

Approach: Greedy (Keep Max, Remove Rest) — O(n)

Key insight: For a group of consecutive same-colored balloons, we must remove all but one. Keeping the most expensive one minimizes cost. Total cost = group sum - group max.

def minCost(colors: str, neededTime: list[int]) -> int:
    total_cost = 0
    i = 0
    n = len(colors)
    
    while i < n:
        j = i
        group_sum = 0
        group_max = 0
        
        # Gather all balloons of the same color
        while j < n and colors[j] == colors[i]:
            group_sum += neededTime[j]
            group_max = max(group_max, neededTime[j])
            j += 1
        
        total_cost += group_sum - group_max
        i = j
    
    return total_cost

Algorithm Flow

Simplified Two-Pointer

def minCost(colors: str, neededTime: list[int]) -> int:
    cost = 0
    for i in range(1, len(colors)):
        if colors[i] == colors[i-1]:
            cost += min(neededTime[i], neededTime[i-1])
            neededTime[i] = max(neededTime[i], neededTime[i-1])
    return cost

This propagates the max forward — each time we encounter a consecutive duplicate, we "remove" the cheaper one.

Dry Run

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

Groups: [1,2] for "aa" → sum=3, max=2, cost=1; [3] for "b" → cost=0; [4,1] for "aa" → sum=5, max=4, cost=1

Total: 1+0+1 = 2 ✓

Edge Cases

• No consecutive duplicates → return 0
• All same color → remove all but the most expensive
• Single balloon → return 0

Complexity

• Time: O(n)
• Space: O(1)

Key Terms

Term	Definition
Greedy algorithm	Making the locally optimal choice (keep the costliest balloon) at each step to reach a globally optimal result.
Consecutive group	A maximal run of adjacent balloons sharing the same color, processed as one unit.
Two-pointer propagation	Carrying the max cost forward through a run instead of re-scanning it, collapsing group logic into a single pass.
Group sum minus group max	The core cost formula: removing all but the most expensive balloon in a duplicate run.

FAQ

Q: Can this be solved without extra space? A: Yes. Both the grouping approach and the two-pointer approach use O(1) extra space; the two-pointer version even reuses neededTime in place.

Q: What if the input is empty or has one balloon? A: With 0 or 1 balloons there are no consecutive duplicates possible, so the answer is 0 — both approaches return this naturally since the loop body never triggers a removal.

Q: Why keep the max instead of just removing the first duplicate found? A: Removing anything other than the most expensive balloon in a group leaves that expensive one to still need eventual removal, or wastes cost; keeping the max always minimizes the sum of removed costs for that group.

Q: How would this change if colors could repeat non-consecutively (any two same-colored balloons, not just adjacent) were disallowed? A: The problem would become fundamentally different — you'd need to remove all but one occurrence of each color globally, which is a distinct (and harder) selection problem, not a simple linear scan.

Q: Does the order of processing groups matter? A: No. Each group's cost is independent of the others, so a single left-to-right pass suffices — there's no need to sort or revisit earlier groups.

Quick Revision

• Problem: remove minimum total time so no two adjacent balloons share a color.
• Pattern: greedy, processed per consecutive same-color group.
• Rule: in each duplicate run, keep the balloon with the largest neededTime, remove the rest.
• Cost per group = sum(group) - max(group).
• Two-pointer variant: compare colors[i] to colors[i-1]; on match add min(times) to cost and let neededTime[i] inherit the max.
• Time: O(n) single pass; Space: O(1).
• No duplicates anywhere → answer is 0.
• All balloons same color → remove all but the globally most expensive one.
• Contrast with brute force (trying removal subsets): greedy avoids exponential blowup by exploiting that each group is independent.

Related Problems

• 135 - Candy — another greedy problem processing runs/neighbors under a local constraint.
• 42 - Trapping Rain Water — shares the "track a running max while scanning" technique.
• 2216 - Minimum Deletions to Make Array Beautiful — same core idea of removing minimum elements to break adjacent equality.