# 1482. Minimum Number of Days to Make m Bouquets

## LeetCode [1482. Minimum Number of Days to Make m Bouquets](https://github.com/txfs19260817/LeetCode-Go/blob/master/solutions/title/README.md)

### Description

Given an integer array `bloomDay`, an integer `m` and an integer `k`.

We need to make `m` bouquets. To make a bouquet, you need to use `k` **adjacent flowers** from the garden.

The garden consists of `n` flowers, the `ith` flower will bloom in the `bloomDay[i]` and then can be used in **exactly one** bouquet.

Return *the minimum number of days* you need to wait to be able to make `m` bouquets from the garden. If it is impossible to make `m` bouquets return **-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
```

**Constraints:**

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

### Tags

Array, Binary Search

### Solution

Based on the hint that "If we can make `m` or more bouquets at day `x`, then we can still make `m` or more bouquets at day `y, y > x`.", we can perform the binary search strategy between the interval `[min(bloomDay), max(bloomDay)]`. We check if we can make at least `m` bouquets within `mid = (start+end)/2` days. If we can, assign `mid` to `end`; otherwise let `start = mid`. After this loop, we also need to check if either `start` days or `end` days can make m bouquets. If both cannot, return -1.

### Complexity

* Time complexity: $$O(n\log(\max(bloomDay)))$$
* Space complexity: $$O(1)$$

### Code

```go
func minDays(bloomDay []int, m int, k int) int {
	start, end := minmaxInSlice(bloomDay)
	for start+1 < end {
		mid := start + (end-start)/2
		if check(bloomDay, m, k, mid) {
			end = mid
		} else {
			start = mid
		}
	}
	if check(bloomDay, m, k, start) {
		return start
	}
	if check(bloomDay, m, k, end) {
		return end
	}
	return -1
}

func minmaxInSlice(a []int) (mini, maxi int) {
	mini = a[0]
	for _, n := range a {
		if n > maxi {
			maxi = n
		}
		if n < mini {
			mini = n
		}
	}
	return
}

func check(bloomDay []int, m, k, days int) bool {
	copyK := k
	for _, d := range bloomDay {
		if d <= days {
			copyK--
			if copyK == 0 {
				m--
				if m <= 0 {
					return true
				}
				copyK = k
			}
		} else {
			copyK = k
		}
	}
	return false
}
```


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