LeetCode #1578 — MEDIUM

Minimum Time to Make Rope Colorful

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Greedy

Example 1

"abaac"
[1,2,3,4,5]

Example 2

"abc"
[1,2,3]

Example 3

"aabaa"
[1,2,3,4,1]

Step 02

Core Insight

What unlocks the optimal approach

Maintain the running sum and max value for repeated letters.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1578: Minimum Time to Make Rope Colorful
class Solution {
    public int minCost(String colors, int[] neededTime) {
        int ans = 0;
        int n = neededTime.length;
        for (int i = 0, j = 0; i < n; i = j) {
            j = i;
            int s = 0, mx = 0;
            while (j < n && colors.charAt(j) == colors.charAt(i)) {
                s += neededTime[j];
                mx = Math.max(mx, neededTime[j]);
                ++j;
            }
            if (j - i > 1) {
                ans += s - mx;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1578: Minimum Time to Make Rope Colorful
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
}

# Accepted solution for LeetCode #1578: Minimum Time to Make Rope Colorful
class Solution:
    def minCost(self, colors: str, neededTime: List[int]) -> int:
        ans = i = 0
        n = len(colors)
        while i < n:
            j = i
            s = mx = 0
            while j < n and colors[j] == colors[i]:
                s += neededTime[j]
                if mx < neededTime[j]:
                    mx = neededTime[j]
                j += 1
            if j - i > 1:
                ans += s - mx
            i = j
        return ans

// Accepted solution for LeetCode #1578: Minimum Time to Make Rope Colorful
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
    }
}

// Accepted solution for LeetCode #1578: Minimum Time to Make Rope Colorful
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.