Minimum Time to Make Rope Colorful - Practice Coding Problems

Minimum Time to Make Rope Colorful - Problem

Alice's Colorful Balloon Rope Challenge

Alice has arranged n balloons on a rope for a party, but there's a problem! She doesn't want any two consecutive balloons to have the same color - it just doesn't look festive enough.

Here's where Bob comes in to help. Bob can remove balloons from the rope, but each balloon takes a different amount of time to remove. You're given:

• A string colors where colors[i] represents the color of the i-th balloon
• An array neededTime where neededTime[i] is the time (in seconds) Bob needs to remove the i-th balloon

Goal: Find the minimum time Bob needs to make the rope colorful by ensuring no two consecutive balloons have the same color.

Example: If we have balloons "aabaa" with times [1,2,3,4,1], Bob should remove the balloons that take less time from each group of consecutive same-colored balloons.

Input & Output

example_1.py — Basic Case

$ Input: colors = "aabaa", neededTime = [1,2,3,4,1]

› Output: 2

💡 Note: We have two groups of consecutive same-colored balloons: "aa" at positions 0-1 and "aa" at positions 3-4. For the first group, we remove the balloon with time 1 (keep the one with time 2). For the second group, we remove the balloon with time 1 (keep the one with time 4). Total cost: 1 + 1 = 2.

example_2.py — All Same Color

$ Input: colors = "abc", neededTime = [1,2,3]

› Output: 0

💡 Note: No two consecutive balloons have the same color, so the rope is already colorful. No balloons need to be removed, so the cost is 0.

example_3.py — Long Consecutive Group

$ Input: colors = "aaaaaa", neededTime = [1,5,3,7,2,4]

› Output: 18

💡 Note: All balloons have the same color, so we need to remove all but one. We keep the balloon with maximum time (7) and remove all others. Total cost: 1 + 5 + 3 + 2 + 4 = 15. Wait, let me recalculate: we keep 7, remove 1+5+3+2+4 = 15.

Constraints

n == colors.length == neededTime.length
1 ≤ n ≤ 10⁵
1 ≤ neededTime[i] ≤ 10⁴
colors contains only lowercase English letters

Visualization

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Understanding the Visualization

Identify the Problem

Consecutive balloons of same color make the rope non-colorful

Find Groups

Locate all consecutive groups of same-colored balloons

Apply Greedy Strategy

Keep the most expensive balloon from each group, remove others

Calculate Total Cost

Sum up all removal costs for optimal solution

Key Takeaway

🎯 Key Insight: For each group of consecutive same-colored balloons, we must remove all but one. Greedily keeping the most expensive balloon from each group minimizes the total removal cost.

Asked in

a Amazon 45 G Google 32 ⊞ Microsoft 28 f Meta 19

The optimal solution uses a greedy approach that processes consecutive same-colored balloon groups. For each group, we calculate the total removal cost and keep only the balloon with maximum removal time, removing all others. This works because in any group of consecutive identical balloons, we must remove all but one, and keeping the most expensive minimizes the total cost. Time complexity: O(n), Space complexity: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Combinations)	O(2^n * n)	O(n)	Try all possible ways to remove balloons and find the minimum cost
Greedy Approach (Optimal)	O(n)	O(1)	Process consecutive same-colored groups, keeping only the most expensive balloon

Brute Force (Try All Combinations) — Algorithm Steps

Generate all possible combinations of balloons to remove (2^n possibilities)
For each combination, simulate removing those balloons
Check if the resulting rope has no consecutive same-colored balloons
Track the minimum time among all valid combinations

Visualization

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Step-by-Step Walkthrough

Generate All Combinations

Create all 2^n possible ways to remove balloons

Test Each Combination

For each combination, check if result is colorful

Track Minimum Cost

Keep track of the minimum cost among valid solutions

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <limits.h>

int minCost(char* colors, int* neededTime, int neededTimeSize) {
    int n = strlen(colors);
    int minTime = INT_MAX;
    
    // Try all possible removal combinations
    for (int mask = 0; mask < (1 << n); mask++) {
        char remainingColors[1001] = {0};
        int removalCost = 0;
        int remainingCount = 0;
        
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) {  // Remove this balloon
                removalCost += neededTime[i];
            } else {  // Keep this balloon
                remainingColors[remainingCount++] = colors[i];
            }
        }
        
        // Check if remaining rope is colorful
        int isColorful = 1;
        for (int i = 1; i < remainingCount; i++) {
            if (remainingColors[i] == remainingColors[i-1]) {
                isColorful = 0;
                break;
            }
        }
        
        if (isColorful && removalCost < minTime) {
            minTime = removalCost;
        }
    }
    
    return minTime == INT_MAX ? 0 : minTime;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n * n)

2^n combinations to try, each taking O(n) time to validate

⚠ Quadratic Growth

Space Complexity

O(n)

Space to store current combination and validation state

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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