2136. Earliest Possible Day of Full Bloom

Description

You have n flower seeds. Every seed must be planted first before it can begin to grow, then bloom. Planting a seed takes time and so does the growth of a seed. You are given two 0-indexed integer arrays plantTime and growTime, of length n each:

• plantTime[i] is the number of full days it takes you to plant the ith seed. Every day, you can work on planting exactly one seed. You do not have to work on planting the same seed on consecutive days, but the planting of a seed is not complete until you have worked plantTime[i] days on planting it in total.
• growTime[i] is the number of full days it takes the ith seed to grow after being completely planted. After the last day of its growth, the flower blooms and stays bloomed forever.

From the beginning of day 0, you can plant the seeds in any order.

Return the earliest possible day where all seeds are blooming.

 

Example 1:

Input: plantTime = [1,4,3], growTime = [2,3,1]
Output: 9
Explanation: The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms.
One optimal way is:
On day 0, plant the 0th seed. The seed grows for 2 full days and blooms on day 3.
On days 1, 2, 3, and 4, plant the 1st seed. The seed grows for 3 full days and blooms on day 8.
On days 5, 6, and 7, plant the 2nd seed. The seed grows for 1 full day and blooms on day 9.
Thus, on day 9, all the seeds are blooming.

Example 2:

Input: plantTime = [1,2,3,2], growTime = [2,1,2,1]
Output: 9
Explanation: The grayed out pots represent planting days, colored pots represent growing days, and the flower represents the day it blooms.
One optimal way is:
On day 1, plant the 0th seed. The seed grows for 2 full days and blooms on day 4.
On days 0 and 3, plant the 1st seed. The seed grows for 1 full day and blooms on day 5.
On days 2, 4, and 5, plant the 2nd seed. The seed grows for 2 full days and blooms on day 8.
On days 6 and 7, plant the 3rd seed. The seed grows for 1 full day and blooms on day 9.
Thus, on day 9, all the seeds are blooming.

Example 3:

Input: plantTime = [1], growTime = [1]
Output: 2
Explanation: On day 0, plant the 0th seed. The seed grows for 1 full day and blooms on day 2.
Thus, on day 2, all the seeds are blooming.

 

Constraints:

• n == plantTime.length == growTime.length
• 1 <= n <= 105
• 1 <= plantTime[i], growTime[i] <= 104

Solutions

Solution 1: Greedy + Sorting

According to the problem description, we know that only one seed can be planted per day. Therefore, regardless of the planting order, the sum of the planting times for all seeds is always equal to $\sum_{i=0}^{n-1} plantTime[i]$. To make all seeds bloom as soon as possible, we should prioritize planting the seeds with the longest growth time. Hence, we can sort all seeds by their growth time in descending order and then plant them in sequence.

The time complexity is $O(n \log n)$, and the space complexity is $O(n)$, where $n$ is the number of seeds.

Python3JavaC++GoTypeScriptRust

class Solution:
    def earliestFullBloom(self, plantTime: List[int], growTime: List[int]) -> int:
        ans = t = 0
        for pt, gt in sorted(zip(plantTime, growTime), key=lambda x: -x[1]):
            t += pt
            ans = max(ans, t + gt)
        return ans

class Solution {
    public int earliestFullBloom(int[] plantTime, int[] growTime) {
        int n = plantTime.length;
        Integer[] idx = new Integer[n];
        for (int i = 0; i < n; i++) {
            idx[i] = i;
        }
        Arrays.sort(idx, (i, j) -> growTime[j] - growTime[i]);
        int ans = 0, t = 0;
        for (int i : idx) {
            t += plantTime[i];
            ans = Math.max(ans, t + growTime[i]);
        }
        return ans;
    }
}

class Solution {
public:
    int earliestFullBloom(vector<int>& plantTime, vector<int>& growTime) {
        int n = plantTime.size();
        vector<int> idx(n);
        iota(idx.begin(), idx.end(), 0);
        sort(idx.begin(), idx.end(), [&](int i, int j) { return growTime[j] < growTime[i]; });
        int ans = 0, t = 0;
        for (int i : idx) {
            t += plantTime[i];
            ans = max(ans, t + growTime[i]);
        }
        return ans;
    }
};

func earliestFullBloom(plantTime []int, growTime []int) (ans int) {
    n := len(plantTime)
    idx := make([]int, n)
    for i := range idx {
        idx[i] = i
    }
    sort.Slice(idx, func(i, j int) bool { return growTime[idx[j]] < growTime[idx[i]] })
    t := 0
    for _, i := range idx {
        t += plantTime[i]
        ans = max(ans, t+growTime[i])
    }
    return
}

function earliestFullBloom(plantTime: number[], growTime: number[]): number {
    const n = plantTime.length;
    const idx: number[] = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => growTime[j] - growTime[i]);
    let [ans, t] = [0, 0];
    for (const i of idx) {
        t += plantTime[i];
        ans = Math.max(ans, t + growTime[i]);
    }
    return ans;
}

impl Solution {
    pub fn earliest_full_bloom(plant_time: Vec<i32>, grow_time: Vec<i32>) -> i32 {
        let mut idx: Vec<usize> = (0..plant_time.len()).collect();
        idx.sort_by_key(|&i| -&grow_time[i]);
        let mut ans = 0;
        let mut t = 0;
        for &i in &idx {
            t += plant_time[i];
            ans = ans.max(t + grow_time[i]);
        }
        ans
    }
}

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