Minimum Time to Make Rope Colorful

#850 Most AskedMedium

Minimum Time to Make Rope Colorful #850 Most Asked

Medium

3 views

Topics:

ArraysGreedy AlgorithmsTwo Pointers

Alice has n balloons arranged on a rope. You are given a 0-indexed string colors where colors[i] is the color of the i^th 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 i^th 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible lengths of the `colors` string and the `neededTime` array? What is the maximum value for any element in the `neededTime` array?
Can the `colors` string be empty or null? If so, what should the function return?
Is the length of the `colors` string always equal to the length of the `neededTime` array?
What characters are allowed in the `colors` string? Can I assume it only contains lowercase English letters?
If all balloons in a consecutive group of the same color need to be removed, should I remove all but the one with the maximum `neededTime`, or is there another rule for determining which balloons to keep?

Brute Force Solution

Approach

The brute force strategy for this rope problem involves exploring all possible ways to remove balloons to ensure no adjacent balloons have the same color. It essentially tries every single combination of removals to find the one with the minimum total cost.

Here's how the algorithm would work step-by-step:

Start by considering all possible subsets of balloons to remove.
For each subset, remove those balloons from the rope.
Check if the remaining balloons now satisfy the condition that no two adjacent balloons have the same color.
If the condition is satisfied, calculate the total cost of removing that particular subset of balloons.
Keep track of the minimum cost found so far across all subsets that satisfy the color condition.
After checking all possible subsets, the minimum cost you've recorded is the answer.

Code Implementation

def min_cost_to_make_rope_colorful_brute_force(colors, needed_time):
    number_of_balloons = len(colors)
    minimum_cost = float('inf')

    for i in range(2 ** number_of_balloons):
        subset_to_remove = []
        removal_cost = 0

        for balloon_index in range(number_of_balloons):
            if (i >> balloon_index) & 1:
                subset_to_remove.append(balloon_index)
                removal_cost += needed_time[balloon_index]

        # Create a new rope with the specified balloons removed.
        new_colors = ""
        new_needed_time = []
        for balloon_index in range(number_of_balloons):
            if balloon_index not in subset_to_remove:
                new_colors += colors[balloon_index]
                new_needed_time.append(needed_time[balloon_index])

        # Check if the resulting rope is colorful.
        is_colorful = True
        for balloon_index in range(len(new_colors) - 1):
            if new_colors[balloon_index] == new_colors[balloon_index + 1]:
                is_colorful = False
                break

        # Update the minimum cost if the rope is colorful.
        if is_colorful:
            minimum_cost = min(minimum_cost, removal_cost)

    return minimum_cost

def main():
    colors = "abaac"
    needed_time = [1, 2, 3, 4, 5]
    result = min_cost_to_make_rope_colorful_brute_force(colors, needed_time)
    print(f"Minimum cost: {result}")

    colors = "abc"
    needed_time = [1, 2, 3]
    result = min_cost_to_make_rope_colorful_brute_force(colors, needed_time)
    print(f"Minimum cost: {result}")

    colors = "aabaa"
    needed_time = [1, 2, 3, 4, 1]
    result = min_cost_to_make_rope_colorful_brute_force(colors, needed_time)
    print(f"Minimum cost: {result}")

if __name__ == "__main__":
    main()

Big(O) Analysis

Time Complexity

O(2^n * n) – The brute force approach examines all possible subsets of balloons. For n balloons, there are 2^n possible subsets. For each subset, we need to verify if the remaining balloons satisfy the color condition. This verification involves iterating through the remaining balloons, which in the worst case takes O(n) time. Thus, for each subset, we perform O(n) operations to check for the colorful rope condition and calculate cost. Therefore, the overall time complexity is O(2^n * n).

Space Complexity

O(N) – The brute force algorithm explores all possible subsets of balloons to remove. In the worst-case scenario, where the initial rope already satisfies the condition, it might still create a copy of the balloons array to check if removing each individual balloon leads to a better result. This temporary copy of the balloons, in the worst case, will store N elements where N is the number of balloons. Therefore, the space complexity is O(N).

Optimal Solution

Approach

The goal is to minimize the total cost of removing balloons to avoid having adjacent balloons of the same color. We can achieve this by iterating through the balloons and, whenever we find adjacent balloons of the same color, keeping the balloon with the higher cost and removing the other.

Here's how the algorithm would work step-by-step:

Start by looking at the first balloon and the balloon next to it.
If the two balloons are the same color, compare their costs.
Remove the balloon with the lower cost, adding its cost to a running total. Keep the balloon with the higher cost.
If the two balloons are different colors, move on to the next balloon.
Repeat steps 2-4 until you have checked all balloons in the rope.
The running total is the minimum cost to make the rope colorful.

Code Implementation

def min_cost_to_make_rope_colorful(colors, needed_time):
    total_cost = 0
    current_index = 0

    while current_index < len(colors) - 1:

        #If adjacent balloons have the same color, compare costs
        if colors[current_index] == colors[current_index + 1]:

            #Remove the balloon with the lower cost.
            if needed_time[current_index] < needed_time[current_index + 1]:
                total_cost += needed_time[current_index]
                current_index += 1
            else:
                total_cost += needed_time[current_index + 1]

                #Advance the index after 'removing' the balloon
                needed_time[current_index + 1] = needed_time[current_index]
                current_index += 1
        else:
            current_index += 1

    return total_cost

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the balloons in the rope once, comparing each balloon to its adjacent balloon. The number of comparisons is directly proportional to the number of balloons, n. Therefore, the time complexity is linear, or O(n).

Space Complexity

O(1) – The algorithm primarily uses a running total to accumulate the cost, and it keeps track of adjacent balloons using a fixed number of variables for comparison. No auxiliary data structures dependent on the input size (N, representing the number of balloons) are created or used during the process. Therefore, the space complexity is constant, regardless of the input size N.

Edge Cases

Case	How to Handle
Empty colors string or null colors array	Return 0 immediately, as there's no rope to make colorful, hence no cost.
Empty needed time array or null needed time array	Return 0 immediately, as there's no time associated with the rope.
Colors string and needed time array have different lengths	Throw an IllegalArgumentException or return -1 to indicate invalid input.
Single character string	Return 0 immediately, because a single-character string is already colorful.
All characters in colors are the same	Iterate through neededTime and sum all but the maximum value in neededTime.
neededTime array contains zero values	The algorithm should handle zero values for time without issue; it simply adds zero to the total cost when removing a balloon with zero time.
Large input size for colors and neededTime to consider performance implications	Ensure the solution has O(n) time complexity to handle large inputs efficiently.
Integer overflow in neededTime summation	Use long data type to store accumulated cost to prevent overflow.

Empty colors string or null colors array

How to Handle:

Return 0 immediately, as there's no rope to make colorful, hence no cost.

Empty needed time array or null needed time array

How to Handle:

Return 0 immediately, as there's no time associated with the rope.

Colors string and needed time array have different lengths

How to Handle:

Throw an IllegalArgumentException or return -1 to indicate invalid input.

Single character string

How to Handle:

Return 0 immediately, because a single-character string is already colorful.

All characters in colors are the same

How to Handle:

Iterate through neededTime and sum all but the maximum value in neededTime.

neededTime array contains zero values

How to Handle:

The algorithm should handle zero values for time without issue; it simply adds zero to the total cost when removing a balloon with zero time.

Large input size for colors and neededTime to consider performance implications

How to Handle:

Ensure the solution has O(n) time complexity to handle large inputs efficiently.

Integer overflow in neededTime summation

How to Handle:

Use long data type to store accumulated cost to prevent overflow.