Minimum Time to Make Rope Colorful

Minimum Time to Make Rope Colorful

What is this problem about?

The Minimum Time to Make Rope Colorful problem gives you a string of balloon colors and a list of removal times for each balloon. Two adjacent balloons of the same color are considered "ugly." You can remove balloons at a cost equal to their removal time. Find the minimum total cost to eliminate all consecutive duplicate colors. This Minimum Time to Make Rope Colorful coding problem combines greedy group-based processing with string scanning.

Why is this asked in interviews?

Microsoft, Amazon, Google, and Bloomberg ask this because it tests clean greedy thinking: within each run of identical consecutive colors, you must keep at most one balloon (the one with the maximum removal time) and remove all others. The array, string, greedy, and dynamic programming interview pattern applies, and the greedy solution is elegant enough that interviewers use it to distinguish candidates who overthink (using DP) from those who find the simple pattern.

Algorithmic pattern used

Greedy group scanning. Traverse the string, grouping consecutive identical characters. For each group of length > 1, compute the sum of all removal times in the group and subtract the maximum removal time. The result is the cost to reduce the group to one balloon (keep the most expensive one, remove all others). Sum these costs across all groups.

Example explanation

Colors: "aabbbaa", times: [1, 2, 3, 1, 2, 3, 4].

• Group "aa" at indices 0-1: sum=3, max=2. Cost = 3-2 = 1.
• Group "bbb" at indices 2-4: sum=6, max=3. Cost = 6-3 = 3.
• Group "aa" at indices 5-6: sum=7, max=4. Cost = 7-4 = 3. Total minimum cost = 1+3+3 = 7.

Common mistakes candidates make

• Using DP unnecessarily (the greedy solution is O(n) and equally correct).
• Forgetting to subtract the maximum (keeping the most expensive one minimizes cost).
• Treating non-adjacent duplicates as groups (only consecutive identical colors matter).
• Double-counting removal costs when groups overlap.

Interview preparation tip

When adjacent duplicates need to be reduced to one element, and you have a cost per removal, the greedy approach is almost always: keep the "best" (most expensive/valuable) one in each group and remove the rest. This avoids expensive DP derivation when a simpler scan suffices. Practice scanning for groups of identical consecutive elements — it's a common preprocessing step in string problems and can usually be done with a simple while-loop or by comparing s[i] == s[i+1].