Minimum Time to Make Rope Colorful

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ArraysStringsGreedy Algorithms

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.

For example:

If colors = "abaac", and neededTime = [1,2,3,4,5], the expected output is 3. Bob can remove the blue balloon at index 2, taking 3 seconds. There are no longer two consecutive balloons of the same color.
If colors = "abc", and neededTime = [1,2,3], the expected output is 0. The rope is already colorful.
If colors = "aabaa", and neededTime = [1,2,3,4,1], the expected output is 2. Bob will remove the balloons at indices 0 and 4. Total time = 1 + 1 = 2.

How would you approach this problem? Consider both naive and optimal solutions. What are the time and space complexities of your solutions? Can you write the code for the optimal solution? What are some edge cases to consider?

Minimum Time to Make Rope Colorful

Medium

Google

1 view

Topics:

ArraysStringsGreedy Algorithms

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.

Return the minimum time Bob needs to make the rope colorful.

For example:

If colors = "abaac", and neededTime = [1,2,3,4,5], the expected output is 3. Bob can remove the blue balloon at index 2, taking 3 seconds. There are no longer two consecutive balloons of the same color.
If colors = "abc", and neededTime = [1,2,3], the expected output is 0. The rope is already colorful.
If colors = "aabaa", and neededTime = [1,2,3,4,1], the expected output is 2. Bob will remove the balloons at indices 0 and 4. Total time = 1 + 1 = 2.

Solution

Naive Solution (Brute Force)

One straightforward approach is to try all possible combinations of balloon removals and check if the remaining balloons are colorful. This involves generating all subsets of indices to remove, removing the balloons at those indices, and then verifying if the remaining sequence is colorful. If colorful, we can keep track of the minimum time to achieve this. However, this method is highly inefficient due to the exponential number of subsets to consider.

Code (Python)

def min_cost_to_make_colorful_brute_force(colors, neededTime):
    n = len(colors)
    min_time = float('inf')

    for i in range(1 << n):
        time = 0
        new_colors = ""
        for j in range(n):
            if not (i & (1 << j)):
                new_colors += colors[j]
            else:
                time += neededTime[j]

        is_colorful = True
        for k in range(len(new_colors) - 1):
            if new_colors[k] == new_colors[k+1]:
                is_colorful = False
                break

        if is_colorful:
            min_time = min(min_time, time)

    return min_time if min_time != float('inf') else 0

Time Complexity

O(2ⁿ * n), where n is the number of balloons. We iterate through all possible subsets (2ⁿ), and for each subset, we might iterate through the entire array (n) to check if it's colorful.

Space Complexity

O(n), primarily due to the creation of new_colors string in the worst-case scenario where no balloons are removed.

Optimal Solution (Greedy)

A more efficient approach is to use a greedy algorithm. We iterate through the balloons, and whenever we find a sequence of balloons with the same color, we remove all but the one that takes the longest time to remove. This ensures that we minimize the total removal time while making the rope colorful.

Algorithm

Initialize total_time to 0.
Iterate through the colors string from left to right.
If colors[i] is the same as colors[i+1], we have a sequence of the same color.
Find the sum of removal times for this sequence and the maximum removal time.
Add (sum - max) to the total_time.
Move to the next different color, repeat.

Code (Python)

def min_cost_to_make_colorful(colors, neededTime):
    total_time = 0
    i = 0
    while i < len(colors) - 1:
        if colors[i] == colors[i + 1]:
            j = i
            current_sum = 0
            current_max = 0
            while j < len(colors) and colors[i] == colors[j]:
                current_sum += neededTime[j]
                current_max = max(current_max, neededTime[j])
                j += 1
            total_time += current_sum - current_max
            i = j
        else:
            i += 1
    return total_time

Time Complexity

O(n), where n is the number of balloons. We iterate through the balloons at most once.

Space Complexity

O(1). We use a constant amount of extra space.

Edge Cases

Empty input: The algorithm handles the empty input gracefully, returning 0, since no removals are needed.
All balloons have different colors: The algorithm correctly identifies that no removals are needed, and the time taken is 0.
All balloons have the same color: The algorithm correctly removes all but the one with the maximum removal time.
Input with mixed colors and sequences of same colors: The algorithm handles the mixed cases efficiently, removing the minimum time required to make the rope colorful.

Solution

Naive Solution (Brute Force)

Code (Python)

def min_cost_to_make_colorful_brute_force(colors, neededTime):
    n = len(colors)
    min_time = float('inf')

    for i in range(1 << n):
        time = 0
        new_colors = ""
        for j in range(n):
            if not (i & (1 << j)):
                new_colors += colors[j]
            else:
                time += neededTime[j]

        is_colorful = True
        for k in range(len(new_colors) - 1):
            if new_colors[k] == new_colors[k+1]:
                is_colorful = False
                break

        if is_colorful:
            min_time = min(min_time, time)

    return min_time if min_time != float('inf') else 0

Time Complexity

O(2ⁿ * n), where n is the number of balloons. We iterate through all possible subsets (2ⁿ), and for each subset, we might iterate through the entire array (n) to check if it's colorful.

Space Complexity

O(n), primarily due to the creation of new_colors string in the worst-case scenario where no balloons are removed.

Optimal Solution (Greedy)

Algorithm

Initialize total_time to 0.
Iterate through the colors string from left to right.
If colors[i] is the same as colors[i+1], we have a sequence of the same color.
Find the sum of removal times for this sequence and the maximum removal time.
Add (sum - max) to the total_time.
Move to the next different color, repeat.

Code (Python)

def min_cost_to_make_colorful(colors, neededTime):
    total_time = 0
    i = 0
    while i < len(colors) - 1:
        if colors[i] == colors[i + 1]:
            j = i
            current_sum = 0
            current_max = 0
            while j < len(colors) and colors[i] == colors[j]:
                current_sum += neededTime[j]
                current_max = max(current_max, neededTime[j])
                j += 1
            total_time += current_sum - current_max
            i = j
        else:
            i += 1
    return total_time

Time Complexity

O(n), where n is the number of balloons. We iterate through the balloons at most once.

Space Complexity

O(1). We use a constant amount of extra space.

Edge Cases

Empty input: The algorithm handles the empty input gracefully, returning 0, since no removals are needed.
All balloons have different colors: The algorithm correctly identifies that no removals are needed, and the time taken is 0.
All balloons have the same color: The algorithm correctly removes all but the one with the maximum removal time.
Input with mixed colors and sequences of same colors: The algorithm handles the mixed cases efficiently, removing the minimum time required to make the rope colorful.