Problem Description

You are given a string colors where each character represents the color of a balloon in a rope, and an array neededTime where neededTime[i] is the time needed to remove the i-th balloon.

The goal is to make the rope "colorful," which means no two adjacent balloons have the same color. To achieve this, you can remove balloons, and the cost of removing the i-th balloon is neededTime[i]. Your task is to find the minimum total time required to make the rope colorful.

Constraints:

• Only adjacent balloons of the same color need to be separated.
• You must choose which balloons to remove in a way that minimizes the total time cost.
• Each balloon can only be removed once.
• There is always at least one valid way to make the rope colorful.

Thought Process

First, let's understand the problem intuitively. If two or more adjacent balloons have the same color, we must remove some of them to ensure only one remains in that group. The naive approach is to try all combinations of removals, but that's inefficient.

Instead, for each group of consecutive balloons with the same color, we only need to keep one balloon (the one with the highest neededTime cost, since removing cheaper ones costs less). So, for each group, we remove all but the most "expensive" balloon to minimize the total removal time.

This leads us from a brute-force approach to a greedy one: always keep the balloon with the highest neededTime in each group of consecutive same-colored balloons.

Solution Approach

The solution uses a greedy algorithm, processing the rope from left to right:

• Initialize a variable to keep track of the total minimum time required (totalTime).
• Iterate through the colors string, tracking groups of consecutive balloons with the same color.
• For each group, sum up all neededTime values and find the maximum in that group.
• The minimum time to remove balloons in that group is the sum minus the maximum (i.e., remove all but the most expensive balloon).
• Add this value to totalTime and move to the next group.

Example Walkthrough

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

Steps:

1. The first group is "aa" (positions 0 and 1). The neededTime are 1 and 2. Remove the one with smaller cost (remove index 0, cost 1).
2. Next, "b" at position 2 is alone, so no removal needed.
3. Next, "aa" at positions 3 and 4. neededTime are 4 and 1. Remove the one with smaller cost (remove index 4, cost 1).

Time and Space Complexity

Brute-force approach: Try all combinations of removals for adjacent duplicates. This is exponential in time, O(2^n), and impractical for large inputs.

Optimized (greedy) approach:

• Time Complexity: O(n), where n is the length of colors. We only need a single pass through the string.
• Space Complexity: O(1), since we only use a few variables to track sums and maximums for each group.

Summary

The problem asks us to minimize the time to make a rope colorful by removing adjacent same-colored balloons. The key insight is that for each group of consecutive same-colored balloons, we should keep the one with the highest removal cost and remove the rest. This greedy approach leads to an efficient O(n) solution that is both simple and elegant.

Code Implementation

class Solution:
    def minCost(self, colors: str, neededTime: list[int]) -> int:
        totalTime = 0
        n = len(colors)
        i = 0
        while i < n:
            sumTime = neededTime[i]
            maxTime = neededTime[i]
            j = i + 1
            while j < n and colors[j] == colors[i]:
                sumTime += neededTime[j]
                maxTime = max(maxTime, neededTime[j])
                j += 1
            totalTime += sumTime - maxTime
            i = j
        return totalTime
      

class Solution {
public:
    int minCost(string colors, vector<int>& neededTime) {
        int totalTime = 0, n = colors.size(), i = 0;
        while (i < n) {
            int sumTime = neededTime[i], maxTime = neededTime[i];
            int j = i + 1;
            while (j < n && colors[j] == colors[i]) {
                sumTime += neededTime[j];
                maxTime = max(maxTime, neededTime[j]);
                j++;
            }
            totalTime += sumTime - maxTime;
            i = j;
        }
        return totalTime;
    }
};
      

class Solution {
    public int minCost(String colors, int[] neededTime) {
        int totalTime = 0, n = colors.length(), i = 0;
        while (i < n) {
            int sumTime = neededTime[i];
            int maxTime = neededTime[i];
            int j = i + 1;
            while (j < n && colors.charAt(j) == colors.charAt(i)) {
                sumTime += neededTime[j];
                maxTime = Math.max(maxTime, neededTime[j]);
                j++;
            }
            totalTime += sumTime - maxTime;
            i = j;
        }
        return totalTime;
    }
}
      

var minCost = function(colors, neededTime) {
    let totalTime = 0, n = colors.length, i = 0;
    while (i < n) {
        let sumTime = neededTime[i];
        let maxTime = neededTime[i];
        let j = i + 1;
        while (j < n && colors[j] === colors[i]) {
            sumTime += neededTime[j];
            maxTime = Math.max(maxTime, neededTime[j]);
            j++;
        }
        totalTime += sumTime - maxTime;
        i = j;
    }
    return totalTime;
};