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Hint Alice has n balloons arranged on a rope. You ...

Criado em: 3 de novembro de 2025

Respondido usando GPT-5 Thinking por Chat01

Pergunta

Hint
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.

Pensamento

Providing a hint for the problem

The user wants a hint for the "Minimum Time to Make Rope Colorful" problem, not a full solution. I can offer some tiered hints: start by grouping consecutive equal-color segments and removing all but the maximum needed time from each group. Another suggestion: Use a sliding window approach where you update the time as you iterate, keeping track of the maximum.

Crafting the hints

Since they asked for a "Hint," I’ll break it down progressively:

  • Hint 1: Look for contiguous blocks and leave exactly one per block.
  • Hint 2: To minimize the cost, keep the highest removal time in each block, removing

Thought for 9s

Resposta

Here are a few graded hints—stop whenever you’ve got it:

Hint 1:
Look at runs of consecutive balloons with the same color. Within any such run, after removals there must be exactly one balloon left.

Hint 2:
For a run of equal colors, to minimize time you should keep the balloon with the largest neededTime and remove all the others. So the cost for a run is sum(neededTime in the run) - max(neededTime in the run).

Hint 3 (one-pass trick):
You don’t need to pre-build runs. Walk left→right; whenever colors[i] == colors[i-1], add min(neededTime[i], neededTime[i-1]) to the answer and treat the remaining “kept” time as max(neededTime[i], neededTime[i-1]) for the next comparison (i.e., carry the bigger time forward). This collapses each run on the fly.

Tiny example:
colors = "aabaa", neededTime = [1,2,3,4,1]
Runs: "aa" (1,2) → keep 2 (pay 1); "b" → 0; "aa" (4,1) → keep 4 (pay 1).
Total = 1 + 0 + 1 = 2.

Pergunta

Hint
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.

give c++ solution

Pensamento

Creating C++ solution

We need to provide a concise C++ solution for the algorithm, implementing it efficiently with a time complexity of O(n). The function should be int minCost(string colors, vector<int>& neededTime), handling edge cases and large inputs safely with long long for possible sums. Using a single pass is essential to minimize complexity. I’ll ensure no math errors while keeping the explanation brief. The main focus is to iterate through and accumulate the minimum cost. I'll provide the code snippet next.

Providing C++ solution

Here’s the C

Thought for 9s

Resposta

Here’s a clean O(n) one-pass solution. We keep the most expensive balloon in each same-color run and delete the rest.

cpp
#include <bits/stdc++.h>
using namespace std;

long long minCost(string colors, const vector<int>& neededTime) {
    const int n = (int)colors.size();
    if (n == 0) return 0;

    long long ans = 0;
    int keep = neededTime[0]; // max time kept in the current same-color run

    for (int i = 1; i < n; ++i) {
        if (colors[i] == colors[i - 1]) {
            // Remove the cheaper of the two; keep the higher cost for future comparisons
            ans += min(keep, neededTime[i]);
            keep = max(keep, neededTime[i]);
        } else {
            // New color run starts
            keep = neededTime[i];
        }
    }
    return ans;
}

// Example usage
int main() {
    string colors = "aabaa";
    vector<int> neededTime = {1, 2, 3, 4, 1};
    cout << minCost(colors, neededTime) << "\n"; // Output: 2
    return 0;
}

Why this works

  • For any run of identical colors, you must leave exactly one balloon.
  • Minimizing time = delete all but the one with the largest neededTime in that run.
  • The loop merges runs on the fly by always adding the smaller of the current pair and carrying forward the larger as the “kept” one.

Complexity: O(n) time, O(1) extra space.

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