Problem Description

Alice has a rope with n balloons attached to it. Each balloon has a color represented by a character in the string colors, where colors[i] is the color of the balloon at position i.

Alice wants to make the rope "colorful" by ensuring no two consecutive balloons have the same color. To achieve this, she asks Bob to remove some balloons. Each balloon takes a certain amount of time to remove, given in the array neededTime, where neededTime[i] is the time in seconds required to remove the balloon at position i.

The task is to find the minimum total time Bob needs to spend removing balloons to make the rope colorful.

For example:

• If colors = "abaac" and neededTime = [1,2,3,4,5], Bob needs to remove balloons to eliminate consecutive duplicates
• The consecutive 'a's at positions 2 and 3 require removing one of them (remove the one with less time)
• The goal is to keep the balloon with the maximum removal time in each group of consecutive same-colored balloons and remove all others

The solution iterates through the string, identifies groups of consecutive balloons with the same color, and for each group, removes all balloons except the one with the maximum removal time (as it's most efficient to keep the hardest-to-remove balloon). The total removal time is calculated by summing the removal times of all balloons in the group minus the maximum time.

Intuition

When we have consecutive balloons of the same color, we need to remove some of them to break the sequence. The key insight is that we want to minimize the total removal time, so we should keep the balloon that takes the longest time to remove and remove all the others.

Think of it this way: if we have three red balloons in a row with removal times [3, 7, 2], we must remove at least two of them to avoid having consecutive same-colored balloons. Which one should we keep? The one with time 7, because removing it would be most expensive. So we remove the balloons with times 3 and 2, giving us a total removal time of 3 + 2 = 5.

This leads to a greedy strategy:

1. Find each group of consecutive balloons with the same color
2. Within each group, calculate the sum of all removal times
3. Keep the balloon with the maximum removal time (don't actually remove it)
4. The cost for this group is: (sum of all times) - (maximum time)

Why does this work? Because in any group of k consecutive same-colored balloons, we must remove exactly k-1 balloons to make them non-consecutive. To minimize cost, we remove the k-1 balloons with the smallest removal times, which is equivalent to keeping the one with the maximum removal time.

The algorithm processes the string in one pass, identifying consecutive groups and applying this logic to each group. When a group has only one balloon, no removal is needed, so we only accumulate cost when the group size is greater than 1.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation uses a two-pointer technique to identify and process groups of consecutive balloons with the same color.

Algorithm Steps:

1. Initialize variables:

   • ans = 0: accumulates the total minimum removal time
   • i = 0: starting pointer for the current group
   • n = len(colors): total number of balloons

2. Process each group of consecutive same-colored balloons:

   while i < n:
       j = i
       s = mx = 0

   • j: ending pointer for the current group
   • s: sum of all removal times in the current group
   • mx: maximum removal time in the current group

3. Expand the group while colors match:

   while j < n and colors[j] == colors[i]:
       s += neededTime[j]
       if mx < neededTime[j]:
           mx = neededTime[j]
       j += 1

   • Keep moving j forward as long as the color matches colors[i]
   • Accumulate the sum of removal times in s
   • Track the maximum removal time in mx

4. Calculate the cost for this group:

   if j - i > 1:
       ans += s - mx

   • If the group has more than one balloon (j - i > 1), we need to remove some
   • The minimum cost is the sum of all times minus the maximum time
   • This effectively removes all balloons except the one with the highest removal time

5. Move to the next group:

   i = j

   • Set i to j to start processing the next group

Time Complexity: O(n) where n is the length of the colors string. Each balloon is visited exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for variables.

The beauty of this solution is that it processes the entire string in a single pass, making it highly efficient while maintaining the greedy principle of keeping the most expensive balloon in each consecutive group.

Example Walkthrough

Let's walk through the algorithm with colors = "aabbcc" and neededTime = [1, 3, 2, 5, 4, 2].

Initial State:

• ans = 0 (total removal time)
• i = 0 (start of current group)
• n = 6 (total balloons)

Group 1: Process balloons at positions 0-1 (both 'a')

• Start with j = 0, s = 0, mx = 0
• Position 0: color is 'a', time is 1
  • s = 0 + 1 = 1, mx = max(0, 1) = 1, j = 1
• Position 1: color is 'a', time is 3
  • s = 1 + 3 = 4, mx = max(1, 3) = 3, j = 2
• Position 2: color is 'b' (different), stop expanding
• Group size: j - i = 2 - 0 = 2 (need removal)
• Cost: s - mx = 4 - 3 = 1 (remove balloon with time 1, keep balloon with time 3)
• ans = 0 + 1 = 1
• Move to next group: i = 2

Group 2: Process balloons at positions 2-3 (both 'b')

• Start with j = 2, s = 0, mx = 0
• Position 2: color is 'b', time is 2
  • s = 0 + 2 = 2, mx = max(0, 2) = 2, j = 3
• Position 3: color is 'b', time is 5
  • s = 2 + 5 = 7, mx = max(2, 5) = 5, j = 4
• Position 4: color is 'c' (different), stop expanding
• Group size: j - i = 4 - 2 = 2 (need removal)
• Cost: s - mx = 7 - 5 = 2 (remove balloon with time 2, keep balloon with time 5)
• ans = 1 + 2 = 3
• Move to next group: i = 4

Group 3: Process balloons at positions 4-5 (both 'c')

• Start with j = 4, s = 0, mx = 0
• Position 4: color is 'c', time is 4
  • s = 0 + 4 = 4, mx = max(0, 4) = 4, j = 5
• Position 5: color is 'c', time is 2
  • s = 4 + 2 = 6, mx = max(4, 2) = 4, j = 6
• Position 6: out of bounds, stop expanding
• Group size: j - i = 6 - 4 = 2 (need removal)
• Cost: s - mx = 6 - 4 = 2 (remove balloon with time 2, keep balloon with time 4)
• ans = 3 + 2 = 5
• Move to next group: i = 6 (terminates loop)

Final Result: Total minimum time = 5

The algorithm removed:

• Balloon at position 0 (time 1) from the 'a' group
• Balloon at position 2 (time 2) from the 'b' group
• Balloon at position 5 (time 2) from the 'c' group

This leaves us with balloons at positions 1, 3, and 4, creating the colorful rope "abc" with no consecutive same colors.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the colors string.

The algorithm uses a two-pointer approach with variables i and j. While there are nested while loops, each element in the array is visited exactly once:

• The outer while loop iterates through starting positions
• The inner while loop advances j to find consecutive balloons of the same color
• After the inner loop completes, i is set to j, ensuring no element is revisited
• Each element is processed once for comparison and sum calculation

Therefore, despite the nested structure, the total number of operations is linear with respect to the input size.

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variables ans, i, j, n, s, and mx are all scalar values
• No additional data structures are created
• The space usage does not scale with the input size

The algorithm modifies only local variables and returns a single integer, making it an in-place solution with constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Error in Group Detection

A common mistake is incorrectly handling the boundary conditions when identifying groups of consecutive same-colored balloons. Developers might accidentally include balloons from different groups or miss the last balloon in a group.

Incorrect Implementation:

# Wrong: May miss processing the last group properly
while current_index < n - 1:  # Wrong boundary
    if colors[current_index] == colors[current_index + 1]:
        # Process group

Solution: Use the two-pointer approach correctly with proper boundary checks:

while current_index < n:
    group_start = current_index
    while current_index < n and colors[current_index] == colors[group_start]:
        # Process group
        current_index += 1

2. Forgetting to Handle Single Balloon Groups

Some implementations might incorrectly add removal costs even for groups with only one balloon, where no removal is necessary.

Incorrect Implementation:

# Wrong: Always adds to total_cost without checking group size
total_cost += group_sum - max_time  # This could subtract when group size is 1

Solution: Only add removal costs when the group has more than one balloon:

if current_index - group_start > 1:  # Only process groups with 2+ balloons
    total_cost += group_sum - max_time

3. Using the Wrong Greedy Strategy

A critical mistake is removing the balloon with the maximum removal time instead of keeping it. The optimal strategy is to keep the most expensive balloon (highest removal time) and remove all others.

Incorrect Implementation:

# Wrong: Removes the most expensive balloon
total_cost += max_time  # This is suboptimal

Solution: Keep the most expensive balloon and remove all others:

total_cost += group_sum - max_time  # Remove all except the maximum

4. Inefficient Group Processing with Nested Loops

Using nested loops or repeatedly scanning for groups can lead to O(n²) time complexity.

Incorrect Implementation:

# Wrong: Inefficient O(n²) approach
for i in range(n):
    if i > 0 and colors[i] == colors[i-1]:
        for j in range(i, n):
            if colors[j] != colors[i]:
                break
        # Process group from i to j

Solution: Use a single-pass approach with two pointers for O(n) complexity:

while current_index < n:
    group_start = current_index
    # Process entire group in one go
    while current_index < n and colors[current_index] == colors[group_start]:
        current_index += 1

5. Not Updating the Maximum Value Correctly

When tracking the maximum removal time in a group, developers might initialize or update it incorrectly.

Incorrect Implementation:

# Wrong: Using built-in max on each iteration (less efficient)
max_time = max(neededTime[group_start:current_index])  # Creates unnecessary sublist

Solution: Track the maximum value incrementally during the single pass:

max_time = 0
while current_index < n and colors[current_index] == colors[group_start]:
    if max_time < neededTime[current_index]:
        max_time = neededTime[current_index]
    current_index += 1