Introduction

In this blog post, we will explore how to solve the LeetCode problem "1578" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 ith balloon. Alice wants the rope to be colorful. She does not want two consecutive balloons to be of the same color, so she asks Bob for help. Bob can remove some balloons from the rope to make it colorful. You are given a 0-indexed integer array neededTime where neededTime[i] is the time (in seconds) that Bob needs to remove the ith balloon from the rope. 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: In the above image, 'a' is blue, 'b' is red, and 'c' is green. Bob can remove the blue balloon at index 2. This takes 3 seconds. There are no longer two consecutive balloons of the same color. Total time = 3. Example 2: Input: colors = "abc", neededTime = [1,2,3] Output: 0 Explanation: The rope is already colorful. Bob does not need to remove any balloons from the rope. Example 3: Input: colors = "aabaa", neededTime = [1,2,3,4,1] Output: 2 Explanation: Bob will remove the balloons at indices 0 and 4. Each balloons takes 1 second to remove. There are no longer two consecutive balloons of the same color. Total time = 1 + 1 = 2. Constraints: n == colors.length == neededTime.length 1 <= n <= 105 1 <= neededTime[i] <= 104 colors contains only lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Identify Consecutive Groups: Iterate through the colors string, identifying consecutive groups of balloons with the same color.
• Minimize Removal Time within Groups: For each group of consecutive balloons with the same color, find the minimum total neededTime to remove balloons such that only one balloon remains in the group. This is done by summing the neededTime of all balloons in the group and subtracting the maximum neededTime within the group.

• Accumulate Total Time: Sum the minimum removal time for each group to obtain the total minimum time required.

• Runtime Complexity: O(n) - We iterate through the input string once. Storage Complexity: O(1) - We use a constant amount of extra space.

Code

    def minCost(colors: str, neededTime: list[int]) -> int:
    """
    Calculates the minimum time needed to make the rope colorful by removing balloons.

    Args:
        colors: A string representing the colors of the balloons.
        neededTime: A list of integers representing the time needed to remove each balloon.

    Returns:
        The minimum time needed to make the rope colorful.
    """

    total_time = 0
    i = 0
    while i < len(colors):
        j = i
        group_time = 0
        max_time = 0
        while j < len(colors) and colors[i] == colors[j]:
            group_time += neededTime[j]
            max_time = max(max_time, neededTime[j])
            j += 1

        total_time += (group_time - max_time)
        i = j

    return total_time