3296. Minimum Number of Seconds to Make Mountain Height Zero

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3296. Minimum Number of Seconds to Make Mountain Height Zero

Description

You are given an integer mountainHeight denoting the height of a mountain.

You are also given an integer array workerTimes representing the work time of workers in seconds .

The workers work simultaneously to reduce the height of the mountain. For worker i:

  • To decrease the mountain’s height by x, it takes workerTimes[i] + workerTimes[i] * 2 + ... + workerTimes[i] * x seconds. For example:

  • To reduce the height of the mountain by 1, it takes workerTimes[i] seconds.

  • To reduce the height of the mountain by 2, it takes workerTimes[i] + workerTimes[i] * 2 seconds, and so on.

Return an integer representing the minimum number of seconds required for the workers to make the height of the mountain 0.

Example 1:

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Input: mountainHeight = 4, workerTimes = [2,1,1]

Output: 3

Explanation:

One way the height of the mountain can be reduced to 0 is:

  • Worker 0 reduces the height by 1, taking workerTimes[0] = 2 seconds.
  • Worker 1 reduces the height by 2, taking workerTimes[1] + workerTimes[1] * 2 = 3 seconds.
  • Worker 2 reduces the height by 1, taking workerTimes[2] = 1 second.

Since they work simultaneously, the minimum time needed is max(2, 3, 1) = 3 seconds.

Example 2:

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Input: mountainHeight = 10, workerTimes = [3,2,2,4]

Output: 12

Explanation:

  • Worker 0 reduces the height by 2, taking workerTimes[0] + workerTimes[0] * 2 = 9 seconds.
  • Worker 1 reduces the height by 3, taking workerTimes[1] + workerTimes[1] * 2 + workerTimes[1] * 3 = 12 seconds.
  • Worker 2 reduces the height by 3, taking workerTimes[2] + workerTimes[2] * 2 + workerTimes[2] * 3 = 12 seconds.
  • Worker 3 reduces the height by 2, taking workerTimes[3] + workerTimes[3] * 2 = 12 seconds.

The number of seconds needed is max(9, 12, 12, 12) = 12 seconds.

Example 3:

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Input: mountainHeight = 5, workerTimes = [1]

Output: 15

Explanation:

There is only one worker in this example, so the answer is workerTimes[0] + workerTimes[0] * 2 + workerTimes[0] * 3 + workerTimes[0] * 4 + workerTimes[0] * 5 = 15.

Constraints:

  • 1 <= mountainHeight <= 10^5
  • 1 <= workerTimes.length <= 10^4
  • 1 <= workerTimes[i] <= 10^6

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    // interpret the problem:
    // n workers, divide the mountainHeight to n sets
    // each set with size h takes t = h * (h + 1) / 2 seconds
    // find the min of max(t * worktime)
    long long minNumberOfSeconds(int mountainHeight, vector<int>& workerTimes) {
        // use priority queue:
        // first, insert all workerTimes into the priority queue,
        // the structure of the priority queue <t, 1, t0)
        // when we pop the item with minmum value out, we then insert a new value updated tri back
        priority_queue<vector<long long>, vector<vector<long long>>, greater<vector<long long>>> pq;
        for (int t : workerTimes) {
            pq.push({t, 1, t});
        }
        long long res = 0;
        while (mountainHeight > 0) {
            auto tri = pq.top();
            pq.pop();
            res = max(res, tri[0]);
            long long n = tri[1] + 1, t = tri[2], t1 = (n + 1) * n / 2 * t;
            pq.push({t1, n, t});
            mountainHeight--;
        }
        return res;
    }
};