Minimum Time to Make Rope Colorful solution and explanation
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The problem
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.length1 <= n <= 1051 <= neededTime[i] <= 104colorscontains only lowercase English letters.
Approach overview
Problem Overview: You receive a string colors where each character represents a balloon color and an array neededTime where neededTime[i] is the time required to remove the i-th balloon. Adjacent balloons cannot share the same color in the final rope. Remove balloons with minimum total time so that no two neighbors have identical colors.
Approach 1: Two Pointer Greedy (O(n) time, O(1) space)
This problem becomes simple once you notice that conflicts only occur inside groups of consecutive identical colors. For every such group, you must remove all balloons except the one with the highest removal time. Use two pointers (or a running group tracker) to iterate through the string and detect segments where colors[i] == colors[i-1]. Within each segment, accumulate the total removal time while keeping track of the maximum cost balloon to keep. The minimum removal time for the group equals sum - max. This greedy decision works because keeping the most expensive balloon minimizes the total cost of removals. The algorithm scans the array once, making it O(n) time and O(1) extra space. This technique is common in greedy problems on arrays and strings where local optimal choices lead to a global optimum.
Approach 2: Dynamic Programming with State Compression (O(n) time, O(1) space)
The same logic can also be expressed with dynamic programming. Define a state that tracks the maximum removal cost balloon kept in the current color segment and the minimum cost accumulated so far. When the current balloon has a different color from the previous one, the state resets because no conflict exists. When the colors match, a decision is required: remove the current balloon or remove the previously kept balloon. The cheaper option is added to the total cost while the state keeps the balloon with the larger removal time for future comparisons. State compression reduces the DP storage to a few variables since each step depends only on the previous state. The resulting algorithm still runs in O(n) time with O(1) space but frames the solution as a decision process, which some candidates find easier when reasoning about transitions.
Recommended for interviews: The greedy two-pointer approach is what interviewers typically expect. It demonstrates the ability to recognize that each same-color segment can be solved independently and that keeping the maximum-cost balloon minimizes removals. Explaining the segment insight clearly usually matters more than the code itself. Mentioning the dynamic programming interpretation shows deeper understanding of state transitions and optimization.
Approach 1: Two Pointer Approach
This is a greedy technique using two pointers. Iterate over the string of balloon colors. For each sequence of consecutive balloons of the same color, use one pointer to find the maximum time required to remove a balloon (among those consecutive balloons). Accumulate the total removal time by subtracting this maximum time from the sum of all removal times for the sequence.
class Solution:
def minCost(self, colors: str, neededTime: List[int]) -> int:
totalCost = 0
for i in range(1, len(colors)):
if colors[i] == colors[i - 1]:
totalCost += min(neededTime[i], neededTime[i - 1])
if neededTime[i] < neededTime[i - 1]:
neededTime[i] = neededTime[i - 1]
return totalCost
# Example usage:
sol = Solution()
print(sol.minCost("aabaa", [1, 2, 3, 4, 1])) # Output: 2public class Solution {
public int minCost(String colors, int[] neededTime) {
int totalCost = 0;
for (int i = 1; i < colors.length(); i++) {
if (colors.charAt(i) == colors.charAt(i - 1)) {
totalCost += Math.min(neededTime[i], neededTime[i - 1]);
if (neededTime[i] < neededTime[i - 1])
neededTime[i] = neededTime[i - 1];
}
}
return totalCost;
}
public static void main(String[] args) {
Solution sol = new Solution();
int[] neededTime = {1, 2, 3, 4, 1};
System.out.println(sol.minCost("aabaa", neededTime));
}
}#include <iostream>
#include <string>
#include <vector>
using namespace std;
int minCost(string colors, vector<int>& neededTime) {
int result = 0;
for (int i = 1; i < colors.size(); i++) {
if (colors[i] == colors[i - 1]) {
result += min(neededTime[i], neededTime[i - 1]);
if (neededTime[i] < neededTime[i - 1])
neededTime[i] = neededTime[i - 1];
}
}
return result;
}
int main() {
vector<int> neededTime = {1, 2, 3, 4, 1};
cout << minCost("aabaa", neededTime) << endl;
return 0;
}function minCost(colors, neededTime) {
let totalCost = 0;
for (let i = 1; i < colors.length; i++) {
if (colors[i] === colors[i - 1]) {
totalCost += Math.min(neededTime[i], neededTime[i - 1]);
if (neededTime[i] < neededTime[i - 1])
neededTime[i] = neededTime[i - 1];
}
}
return totalCost;
}
console.log(minCost("aabaa", [1, 2, 3, 4, 1])); // Output: 2#include <stdio.h>
#include <string.h>
int minTime(char *colors, int *neededTime, int n) {
int totalCost = 0;
for (int i = 1; i < n; i++) {
if (colors[i] == colors[i - 1]) {
totalCost += neededTime[i] < neededTime[i-1] ? neededTime[i] : neededTime[i-1];
if (neededTime[i] < neededTime[i-1])
neededTime[i] = neededTime[i-1];
}
}
return totalCost;
}
int main() {
char colors[] = "aabaa";
int neededTime[] = {1, 2, 3, 4, 1};
int n = strlen(colors);
printf("%d\n", minTime(colors, neededTime, n));
return 0;
}using System;
namespace BalloonColors {
class Solution {
public int MinCost(string colors, int[] neededTime) {
int totalCost = 0;
for (int i = 1; i < colors.Length; i++) {
if (colors[i] == colors[i - 1]) {
totalCost += Math.Min(neededTime[i], neededTime[i - 1]);
if (neededTime[i] < neededTime[i - 1])
neededTime[i] = neededTime[i - 1];
}
}
return totalCost;
}
static void Main(string[] args) {
var solution = new Solution();
int[] neededTime = {1, 2, 3, 4, 1};
Console.WriteLine(solution.MinCost("aabaa", neededTime));
}
}
}The Python solution iterates over the balloon colors with a for-loop. It keeps track of any consecutive balloons of the same color by adding the cost of removing one of them to the total cost. The needed time of the retained balloon is updated to the max of the two, ensuring future iterations are accurate.
Time Complexity: O(n), where n is the length of the input.
Space Complexity: O(1), additional space usage is minimal.
Approach 2: Dynamic Programming with State Compression
This approach uses dynamic programming (DP) with state compression to minimize the time required to make all the balloons non-consecutive in terms of color. The solution leverages optimization by compressing the state to monitor the necessary reduction within nested loops over sequences of consecutive duplicates.
class Solution:
def minCost(self, colors: str, neededTime: List[int]) -> int:
dp = [0, 0]
for i in range(len(colors)):
cur, prev = i % 2, (i - 1) % 2
dp[cur] = neededTime[i] + dp[prev]
if i > 0 and colors[i] == colors[i - 1]:
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1]
return dp[(len(colors) - 1) % 2]
# Example usage:
sol = Solution()
print(sol.minCost("aabaa", [1, 2, 3, 4, 1])) # Output: 2public class Solution {
public int minCost(String colors, int[] neededTime) {
int[] dp = new int[2];
for (int i = 0; i < colors.length(); i++) {
int cur = i % 2;
int prev = 1 - cur;
dp[cur] = neededTime[i] + dp[prev];
if (i > 0 && colors.charAt(i) == colors.charAt(i - 1)) {
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1];
}
}
return dp[(colors.length() - 1) % 2];
}
public static void main(String[] args) {
Solution sol = new Solution();
int[] neededTime = {1, 2, 3, 4, 1};
System.out.println(sol.minCost("aabaa", neededTime));
}
}#include <iostream>
#include <vector>
using namespace std;
int minCost(string colors, vector<int>& neededTime) {
int dp[2] = {0, 0};
for (size_t i = 0; i < colors.size(); ++i) {
int cur = i % 2;
int prev = 1 - cur;
dp[cur] = neededTime[i] + dp[prev];
if (i > 0 && colors[i] == colors[i - 1]) {
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1];
}
}
return dp[(colors.size() - 1) % 2];
}
int main() {
vector<int> neededTime = {1, 2, 3, 4, 1};
cout << minCost("aabaa", neededTime) << endl;
return 0;
}function minCost(colors, neededTime) {
let dp = [0, 0];
for (let i = 0; i < colors.length; i++) {
let cur = i % 2, prev = (i - 1) % 2;
dp[cur] = neededTime[i] + dp[prev];
if (i > 0 && colors[i] === colors[i - 1]) {
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1];
}
}
return dp[(colors.length - 1) % 2];
}
console.log(minCost("aabaa", [1, 2, 3, 4, 1])); // Output: 2#include <stdio.h>
#include <string.h>
#include <limits.h>
int minTime(char *colors, int *neededTime, int n) {
int dp[2] = {0, 0};
for (int i = 0; i < n; ++i) {
int cur = i % 2;
int prev = 1 - cur;
dp[cur] = neededTime[i] + dp[prev];
if (i > 0 && colors[i] == colors[i - 1]) {
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1];
}
}
return dp[(n - 1) % 2];
}
int main() {
char colors[] = "aabaa";
int neededTime[] = {1, 2, 3, 4, 1};
int n = strlen(colors);
printf("%d\n", minTime(colors, neededTime, n));
return 0;
}using System;
namespace BalloonColors {
class Solution {
public int MinCost(string colors, int[] neededTime) {
int[] dp = new int[2];
for (int i = 0; i < colors.Length; i++) {
int cur = i % 2;
int prev = 1 - cur;
dp[cur] = neededTime[i] + dp[prev];
if (i > 0 && colors[i] == colors[i - 1]) {
dp[cur] = dp[prev] + neededTime[i] - neededTime[i - 1];
}
}
return dp[(colors.Length - 1) % 2];
}
static void Main(string[] args) {
var solution = new Solution();
int[] neededTime = {1, 2, 3, 4, 1};
Console.WriteLine(solution.MinCost("aabaa", neededTime));
}
}
}Python's concise version applying a dynamic program approach to efficiently store state with minimum balloon removal time by employing compressed states.
Time Complexity: O(n), operating through all elements.
Space Complexity: O(1), balances memory via state adjustments.
Approach 3: Two Pointers + Greedy
We can use two pointers to point to the beginning and end of the current consecutive balloons with the same color, then calculate the total time s of these consecutive balloons with the same color, as well as the maximum time mx. If the number of consecutive balloons with the same color is greater than 1, we can greedily choose to keep the balloon with the maximum time and remove the other balloons with the same color, which takes time s - mx, and add it to the answer. Then we continue to traverse until all balloons are traversed.
The time complexity is O(n) and the space complexity is O(1), where n is the number of balloons.
function minCost(colors: string, neededTime: number[]): number {
let ans = 0;
const n = neededTime.length;
for (let i = 0, j = 0; i < n; i = j) {
j = i;
let [s, mx] = [0, 0];
while (j < n && colors[j] === colors[i]) {
s += neededTime[j];
mx = Math.max(mx, neededTime[j]);
++j;
}
if (j - i > 1) {
ans += s - mx;
}
}
return ans;
}func minCost(colors string, neededTime []int) (ans int) {
n := len(colors)
for i, j := 0, 0; i < n; i = j {
j = i
s, mx := 0, 0
for j < n && colors[j] == colors[i] {
s += neededTime[j]
mx = max(mx, neededTime[j])
j++
}
if j-i > 1 {
ans += s - mx
}
}
return
}impl Solution {
pub fn min_cost(colors: String, needed_time: Vec<i32>) -> i32 {
let n = needed_time.len();
let mut ans = 0;
let bytes = colors.as_bytes();
let mut i = 0;
while i < n {
let mut j = i;
let mut s = 0;
let mut mx = 0;
while j < n && bytes[j] == bytes[i] {
s += needed_time[j];
mx = mx.max(needed_time[j]);
j += 1;
}
if j - i > 1 {
ans += s - mx;
}
i = j;
}
ans
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Two Pointer Approach | Time Complexity: O(n), where n is the length of the input. |
| Dynamic Programming with State Compression | Time Complexity: O(n), operating through all elements. |
| Two Pointers + Greedy | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Two Pointer Greedy | O(n) | O(1) | Best general solution. Detect consecutive color groups and keep the maximum cost balloon. |
| Dynamic Programming (State Compression) | O(n) | O(1) | Useful when reasoning about decision transitions between removing current vs previous balloon. |
Frequently asked questions
Is Minimum Time to Make Rope Colorful easy or hard?
Minimum Time to Make Rope Colorful is classified as a Medium difficulty problem. The implementation is straightforward once you recognize that each group of identical adjacent colors should keep the balloon with the maximum removal cost while removing the rest.
Minimum Time to Make Rope Colorful Python/Java solution
Both Python and Java implementations follow the same greedy logic. Iterate through the colors string, track the maximum time in the current color segment, and add the smaller removal time whenever two adjacent colors match. This produces an O(n) solution with constant space.
How to solve Minimum Time to Make Rope Colorful in O(n)?
Traverse the string once while comparing each balloon with the previous one. If the colors match, add the smaller removal time to the answer and keep the balloon with the larger cost as the active candidate. This greedy rule ensures the cheapest balloons are removed while preserving the most expensive balloon in each same-color segment.
What is the best approach for Minimum Time to Make Rope Colorful?
The greedy two-pointer approach is the optimal solution. Iterate through the balloons and group consecutive balloons with the same color. For each group, keep the balloon with the maximum removal time and remove the rest. This ensures the minimum total removal cost while maintaining different adjacent colors.
Is Minimum Time to Make Rope Colorful asked at Google/Amazon/Meta?
Minimum Time to Make Rope Colorful appears in interviews at companies that test greedy reasoning and array processing, including Amazon and Google-style interview rounds. The problem checks whether candidates can identify local optimal decisions within consecutive segments.
What data structure is used in Minimum Time to Make Rope Colorful?
The solution primarily uses arrays and string traversal. No advanced data structures are required. The algorithm relies on greedy comparisons and simple variables to track the maximum removal time within a consecutive color group.
What is the time complexity of Minimum Time to Make Rope Colorful?
The optimal solution runs in O(n) time where n is the number of balloons. The algorithm performs a single linear scan through the colors array and processes each balloon once. Space complexity is O(1) because only a few variables are needed to track the current group and maximum removal time.