Friday, May 25, 2018

LeetCode 683 - K Empty Slots


http://zxi.mytechroad.com/blog/simulation/leetcode-683-k-empty-slots/
There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in Ndays. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.
Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.
For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.
Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.
If there isn't such day, output -1.
Example 1:
Input: 
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.
Example 2:
Input: 
flowers: [1,2,3]
k: 1
Output: -1
Note:
  1. The given array will be in the range [1, 20000].
花园中有N个槽,每次在槽中种一朵花。给定种花的顺序flowers,flowers[i] = x表示第i天,在第x个槽种下一朵花。
另外给定数字k,求flowers中是否存在某一天,满足相隔k距离的两个端点恰好各有一朵花,而这两朵花之间的k个槽都没有花。

X.

最优解法,时间复杂度为O(n),空间复杂度为O(n),思路是用额外空间记录一下1 ~ n个花槽,是第几天开花。再从头遍历,看看每隔K天两个花槽之间有没有更早的天数,没有则更新result,有则从天数最早的index向后走直到right走到最后。

class Solution {
    public int kEmptySlots(int[] flowers, int k) {
        int[] days = new int[flowers.length];
        for (int i = 0; i < flowers.length; i++) days[flowers[i] - 1] = i + 1;

        int left = 0, right = k + 1, result = Integer.MAX_VALUE;
        for (int i = 0; right < days.length; i++) {
            if (days[i] < days[left] || days[i] <= days[right]) {
                if (i == right)
                    result = Math.min(result, Math.max(days[left], days[right]));
                left = i;
                right = k + 1 + i;
            }
        }

        return (result == Integer.MAX_VALUE) ? -1 : result;
    }
}

BST/Buckets

https://blog.csdn.net/katrina95/article/details/79070941

用treemap – 内部排序的hashmap,时间复杂度是O(nlogn),空间复杂度是O(n)。思路是遍历数组,每次都将花槽编号放进treeset,同时每次都检查到这天为止离此花槽编号最近的花槽编号,如果两个编号之间刚好相差k,那么就返回这时的天数。

    public int kEmptySlots(int[] flowers, int k) {
        int n = flowers.length;
        if (n == 1 && k == 0) return 1;
        TreeSet<Integer> sort = new TreeSet<>();
        for (int i = 0; i < n; ++i) {
            sort.add(flowers[i]);
            Integer min = sort.lower(flowers[i]);
            Integer max = sort.higher(flowers[i]);
            if (min != null && flowers[i] - min == k + 1) return i + 1;
            if (max != null && max - flowers[i] == k + 1) return i + 1;
        }
        return -1;
    }
X. Brute Force

用hashset,从头遍历数组,每次找相距k的花槽编号在不在hashset里,在的话检查直接有没有开花,没有则可以返回,这样的时间复杂度是O(kn),空间复杂度是O(n),但是LeetCode里TLE了,不过这里还是放出来吧,拓展一下思路。

public int kEmptySlots(int[] flowers, int k) {
        if (flowers.length == 0) return -1;
        HashSet<Integer> set = new HashSet<>();
        for (int i = 0; i < flowers.length; i++) {
            set.add(flowers[i]);
            int pre = flowers[i] - k - 1, next = flowers[i] + k + 1, tag = 0;
            if (pre >= 1 && set.contains(pre)) {
                for (int j = pre + 1; j < pre + 1 + k; j++) {
                    if (set.contains(j)) {
                        tag = 1;
                        break;
                    }
                }
                if (tag == 0) return i + 1;
            }
            tag = 0;
            if (next <= flowers.length && set.contains(next)) {
                for (int j = flowers[i] + 1; j < next; j++) {
                    while (set.contains(j)) {
                        tag = 1;
                        break;
                    }
                }
                if (tag == 0) return i + 1;
            }
        }
        return -1;
    }

    public int kEmptySlots(int[] flowers, int k) {
        int n = flowers.length;
        if (n == 0 || k >= n) return -1;
        int[] f = new int[n + 1];
        
        for (int i = 0; i < n; ++i)
            if (IsValid(flowers[i], k, n, f))
                return i + 1;
        
        return -1;
    }
    
    private boolean IsValid(int x, int k, int n, int[] f) {
        f[x] = 1;
        if (x + k + 1 <= n && f[x + k + 1] == 1) {
            boolean valid = true;
            for (int i = 1; i <= k; ++i)
                if (f[x + i] == 1) {
                    valid = false;
                    break;
                }
            if (valid) return true;
        }
        if (x - k - 1 > 0 && f[x - k - 1] == 1) {
            for (int i = 1; i <= k; ++i)
                if (f[x - i] == 1) return false;
            return true;
        }
        return false;
    }


X.
树状数组(Fenwick Tree) 树状数组ft[k]存储前k个槽一共有多少朵花,则区间[m, n]的花朵总数 = ft[n] - ft[m - 1] 利用该数据结构,遍历flowers即可求解。

http://bookshadow.com/weblog/2017/09/24/leetcode-k-empty-slots/
def kEmptySlots(self, flowers, k): """ :type flowers: List[int] :type k: int :rtype: int """ maxn = max(flowers) nums = [0] * (maxn + 1) ft = FenwickTree(maxn) for i, v in enumerate(flowers): ft.add(v, 1) nums[v] = 1 if v >= k and ft.sum(v) - ft.sum(v - k - 2) == 2 and nums[v - k - 1]: return i + 1 if v + k + 1<= maxn and ft.sum(v + k + 1) - ft.sum(v - 1) == 2 and nums[v + k + 1]: return i + 1 return -1 class FenwickTree(object): def __init__(self, n): self.n = n self.sums = [0] * (n + 1) def add(self, x, val): while x <= self.n: self.sums[x] += val x += self.lowbit(x) def lowbit(self, x): return x & -x def sum(self, x): res = 0 while x > 0: res += self.sums[x] x -= self.lowbit(x) return res




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