1482. Minimum Number of Days to Make m Bouquets

[Tcdian]

1482. Minimum Number of Days to Make m Bouquets

给你一个整数数组 bloomDay，以及两个整数 m 和 k 。

现需要制作 m 束花。制作花束时，需要使用花园中 相邻的 k 朵花 。

花园中有 n 朵花，第 i 朵花会在 bloomDay[i] 时盛开，恰好 可以用于 一束 花中。

请你返回从花园中摘 m 束花需要等待的最少的天数。如果不能摘到 m 束花则返回-1 。

Example 1

Input: bloomDay = [1,10,3,10,2], m = 3, k = 1
Output: 3
Explanation: Let's see what happened in the first three days. x means flower bloomed and _ means flower didn't bloom in the garden.
We need 3 bouquets each should contain 1 flower.
After day 1: [x, _, _, _, _]   // we can only make one bouquet.
After day 2: [x, _, _, _, x]   // we can only make two bouquets.
After day 3: [x, _, x, _, x]   // we can make 3 bouquets. The answer is 3.

Example 2

Input: bloomDay = [1,10,3,10,2], m = 3, k = 2
Output: -1
Explanation: We need 3 bouquets each has 2 flowers, that means we need 6 flowers. We only have 5 flowers so it is impossible to get the needed bouquets and we return -1.

Example 3

Input: bloomDay = [7,7,7,7,12,7,7], m = 2, k = 3
Output: 12
Explanation: We need 2 bouquets each should have 3 flowers.
Here's the garden after the 7 and 12 days:
After day 7: [x, x, x, x, _, x, x]
We can make one bouquet of the first three flowers that bloomed. We cannot make another bouquet from the last three flowers that bloomed because they are not adjacent.
After day 12: [x, x, x, x, x, x, x]
It is obvious that we can make two bouquets in different ways.

Example 4

Input: bloomDay = [1000000000,1000000000], m = 1, k = 1
Output: 1000000000
Explanation: You need to wait 1000000000 days to have a flower ready for a bouquet.

Example 5

Input: bloomDay = [1,10,2,9,3,8,4,7,5,6], m = 4, k = 2
Output: 9

Note

• bloomDay.length == n
• 1 <= n <= 10^5
• 1 <= bloomDay[i] <= 10^9
• 1 <= m <= 10^6
• 1 <= k <= n

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[]} bloomDay
 * @param {number} m
 * @param {number} k
 * @return {number}
 */
var minDays = function(bloomDay, m, k) {
    if (bloomDay.length < m * k) {
        return -1;
    }
    let left = 0;
    let right = 10**9;
    while(left < right) {
        const mid = (left + right) >> 1;
        let bouquets = 0;
        let flowers = 0;
        for (let i = 0; i < bloomDay.length; i++) {
            if (bloomDay[i] <= mid) {
                flowers += 1;
            } else {
                flowers = 0;
            }
            if (flowers === k) {
                bouquets += 1;
                flowers = 0;
            }
        }
        if (bouquets < m) {
            left = mid+1;
        } else {
            right = mid;
        }
    }
    return left;
};

• TypeScript Solution

function minDays(bloomDay: number[], m: number, k: number): number {
    if (bloomDay.length < m * k) {
        return -1;
    }
    let left = 0;
    let right = 10**9;
    while(left < right) {
        const mid = (left + right) >> 1;
        let bouquets = 0;
        let flowers = 0;
        for (let i = 0; i < bloomDay.length; i++) {
            if (bloomDay[i] <= mid) {
                flowers += 1;
            } else {
                flowers = 0;
            }
            if (flowers === k) {
                bouquets += 1;
                flowers = 0;
            }
        }
        if (bouquets < m) {
            left = mid+1;
        } else {
            right = mid;
        }
    }
    return left;
};