Minimum Time to Make Rope Colorful

class Solution {
public:
    int minimumCost(string s, vector<int>& timeRequired) {
        int summedCost = 0, maxTimeInGroup = 0;
        
        for (int j = 0; j < s.length(); ++j) {
            if (j > 0 && s[j] != s[j - 1]) {
                maxTimeInGroup = 0;
            }
            
            summedCost += min(maxTimeInGroup, timeRequired[j]);
            maxTimeInGroup = max(maxTimeInGroup, timeRequired[j]);
        }
        
        return summedCost;
    }
};

This C++ solution tackles the problem of finding the minimum cost to ensure all characters in a string s representing colors on a rope are unique with the least expense. The costs associated with modifying the colors are defined in the vector<int> timeRequired. The approach is based on a single-pass, greedy algorithm which iterates through the string while keeping track of costs:

• Initialize summedCost to accumulate the total minimum cost.
• maxTimeInGroup helps to manage the maximum cost within a group of consecutive same-character sequences.
• Iterate through the string s:
  • If the current character is different from the preceding one, reset maxTimeInGroup. This indicates the start of a new group.
  • Add the minimum of maxTimeInGroup or the current character's time from timeRequired to summedCost. This ensures that the smallest possible modification cost is used for consecutive similar characters.
  • Update maxTimeInGroup to be the maximum of its current value or the current character's cost, preparing for potential further characters in the same group.

The key insight of this solution is to greedily select the lesser cost during sequences of similar characters using min(maxTimeInGroup, timeRequired[j]), aiming to achieve minimal transformation cost for a colorful rope.

In scenarios where characters repeat consecutively, the function cleverly chooses to add the lesser costs first, leading to optimal savings, and adjusts totals only when characters change, ensuring that you accrue the minimum necessary cost to break a sequence of identical characters.

The function ultimately returns the calculated summedCost which represents the minimized expenditure required to make the rope colorful. This solution efficiently handles the problem in linear time relative to the length of the string, making it suitable for large inputs.

class Solution {
    public int minimumTime(String inputColors, int[] timeRequired) {
        int accumTime = 0, maxTimePerGroup = 0;
        for (int index = 0; index < inputColors.length(); ++index) {
            if (index > 0 && inputColors.charAt(index) != inputColors.charAt(index - 1)) {
                maxTimePerGroup = 0;
            }
            accumTime += Math.min(maxTimePerGroup, timeRequired[index]);
            maxTimePerGroup = Math.max(maxTimePerGroup, timeRequired[index]);
        }
        return accumTime;
    }
}

To solve the "Minimum Time to Make Rope Colorful" problem in Java, follow these steps outlined in the provided solution:

1. Initialize two integer variables, accumTime to store the cumulative minimum time required and maxTimePerGroup to hold the maximum time of the current consecutive group of balloons with the same color.

2. Iterate through the string inputColors using a for loop with index ranging from 0 to the length of inputColors.

3. Inside the loop, check if the current character at the position index is different from the previous character (inputColors.charAt(index) != inputColors.charAt(index - 1)). If true, reset maxTimePerGroup to 0.

4. Accumulate the minimal cost so far to accumTime by adding the smaller value between maxTimePerGroup and the current time required (timeRequired[index]).

5. Update maxTimePerGroup to the larger value between the current maxTimePerGroup and timeRequired[index].

6. Once the loop is complete, return accumTime which now holds the minimum time required.

This approach ensures efficient computation by maintaining a running total of the necessary time to modify the balloon colors without completely repainting adjacent balloons of the same color that involve higher cost times. The choice of maintaining maximum time within a group helps in optimizing the decision of which balloon's time to take as representative for continuous same-color segments.

var calculateMinCost = function(s, costs) {
    let totalCost = 0, maxCostInGroup = 0;
    for (let i = 0; i < s.length; ++i) {
        if (i > 0 && s[i] !== s[i - 1]) {
            maxCostInGroup = 0;
        }
        totalCost += Math.min(maxCostInGroup, costs[i]);
        maxCostInGroup = Math.max(maxCostInGroup, costs[i]);
    }
    return totalCost;
};

This solution addresses the problem of calculating the minimum cost required to make a string of colored balloons (represented as characters) non-adjacent by removing the minimal necessary colored balloons based on a given cost array for each balloon.

• The JavaScript function calculateMinCost takes two arguments: s (the string of balloons) and costs (the array representing the removal cost of each balloon).
• Initialize totalCost to zero, which will store the cumulative minimum cost to remove necessary balloons.
• maxCostInGroup is used to keep track of the maximum removal cost encountered so far for consecutive balloons of the same color.
• Iterate through the string s using a for loop:
  • If the current balloon is different from the previous balloon, reset maxCostInGroup to zero, indicating the start of a new group of identical balloons.
  • Update totalCost by adding the smaller value between maxCostInGroup (which is initially zero) and the current balloon's cost. This represents choosing the least expensive option to maintain non-adjacent balloons of the same color.
  • Update maxCostInGroup to be the maximum of its current value and the current balloon's cost. This step is crucial as it sets up maxCostInGroup for the comparison in the next iteration of the loop or for potentially the next group of same-colored balloons.
• Return totalCost, which by the end of the loop contains the total minimum cost to make the balloons non-adjacent by removing some balloons based on the given costs.

class Solution:
    def minimizeCost(self, hues: str, times: List[int]) -> int:
        accumulated_time = 0
        max_current_time = 0
        
        for index in range(len(hues)):
            if index > 0 and hues[index] != hues[index - 1]:
                max_current_time = 0
            
            accumulated_time += min(max_current_time, times[index])
            max_current_time = max(max_current_time, times[index])
                
        return accumulated_time

This solution focuses on determining the minimum cost to ensure no two adjacent balloons in a rope have the same color. To achieve this, the approach utilizes dynamic comparison and conditional calculations.

The provided code defines a solution class with a method called minimizeCost. This function processes two inputs: hues representing a string of colors for the balloons and times representing an array of integers where each element corresponds to the cost of painting each balloon.

Approach:

1. Initialize accumulated_time to store the total minimal cost.
2. Initialize max_current_time to keep track of the maximum time spent on the last balloon of the same hue.

As the loop iterates through each balloon:

1. Check whether the current balloon has the same hue as the previous one.
2. If the hues differ, reset max_current_time to zero. This indicates a color change where no cost calculation is needed for removal of the previous hue continuity.
3. Add the smaller value between max_current_time and the current balloon’s time to the accumulated_time. This accounts for the already considered high cost of consecutive balloons of the same hue and adds lesser values consequently.
4. Update max_current_time for the next iteration by comparing it with the current time.

Finally, return the accumulated_time which now contains the minimized cost to make the rope colorful with no adjacent balloons sharing the same color. This solution efficiently handles the accumulation and resetting of the time costs associated with painting the balloons different colors.