Question

683. K Empty Slots

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Description

You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are turned off. We turn on exactly one bulb every day until all bulbs are on after n days.

You are given an array bulbs of length n where bulbs[i] = x means that on the (i+1)th day, we will turn on the bulb at position x where i is 0-indexed and x is 1-indexed.

Given an integer k, return _the minimum day number such that there exists two turned on bulbs that have exactly k bulbs between them that are all turned off. If there isn't such day, return -1._

 

Example 1:

Input: bulbs = [1,3,2], k = 1
Output: 2
Explanation:
On the first day: bulbs[0] = 1, first bulb is turned on: [1,0,0]
On the second day: bulbs[1] = 3, third bulb is turned on: [1,0,1]
On the third day: bulbs[2] = 2, second bulb is turned on: [1,1,1]
We return 2 because on the second day, there were two on bulbs with one off bulb between them.

Example 2:

Input: bulbs = [1,2,3], k = 1
Output: -1

 

Constraints:

• n == bulbs.length
• 1 <= n <= 2 * 104
• 1 <= bulbs[i] <= n
• bulbs is a permutation of numbers from 1 to n.
• 0 <= k <= 2 * 104

Solutions

Solution 1: Binary Indexed Tree

We can use a Binary Indexed Tree to maintain the prefix sum of the bulbs. Every time we turn on a bulb, we update the corresponding position in the Binary Indexed Tree. Then we check if the $k$ bulbs to the left or right of the current bulb are all turned off and the $(k+1)$-th bulb is already turned on. If either of these conditions is met, we return the current day.

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

class BinaryIndexedTree:
  def __init__(self, n):
    self.n = n
    self.c = [0] * (n + 1)

  def update(self, x, delta):
    while x <= self.n:
      self.c[x] += delta
      x += x & -x

  def query(self, x):
    s = 0
    while x:
      s += self.c[x]
      x -= x & -x
    return s


class Solution:
  def kEmptySlots(self, bulbs: List[int], k: int) -> int:
    n = len(bulbs)
    tree = BinaryIndexedTree(n)
    vis = [False] * (n + 1)
    for i, x in enumerate(bulbs, 1):
      tree.update(x, 1)
      vis[x] = True
      y = x - k - 1
      if y > 0 and vis[y] and tree.query(x - 1) - tree.query(y) == 0:
        return i
      y = x + k + 1
      if y <= n and vis[y] and tree.query(y - 1) - tree.query(x) == 0:
        return i
    return -1

class Solution {
  public int kEmptySlots(int[] bulbs, int k) {
    int n = bulbs.length;
    BinaryIndexedTree tree = new BinaryIndexedTree(n);
    boolean[] vis = new boolean[n + 1];
    for (int i = 1; i <= n; ++i) {
      int x = bulbs[i - 1];
      tree.update(x, 1);
      vis[x] = true;
      int y = x - k - 1;
      if (y > 0 && vis[y] && tree.query(x - 1) - tree.query(y) == 0) {
        return i;
      }
      y = x + k + 1;
      if (y <= n && vis[y] && tree.query(y - 1) - tree.query(x) == 0) {
        return i;
      }
    }
    return -1;
  }
}

class BinaryIndexedTree {
  private int n;
  private int[] c;

  public BinaryIndexedTree(int n) {
    this.n = n;
    this.c = new int[n + 1];
  }

  public void update(int x, int delta) {
    for (; x <= n; x += x & -x) {
      c[x] += delta;
    }
  }

  public int query(int x) {
    int s = 0;
    for (; x > 0; x -= x & -x) {
      s += c[x];
    }
    return s;
  }
}

class BinaryIndexedTree {
public:
  int n;
  vector<int> c;

  BinaryIndexedTree(int _n)
    : n(_n)
    , c(_n + 1) {}

  void update(int x, int delta) {
    for (; x <= n; x += x & -x) {
      c[x] += delta;
    }
  }

  int query(int x) {
    int s = 0;
    for (; x; x -= x & -x) {
      s += c[x];
    }
    return s;
  }
};

class Solution {
public:
  int kEmptySlots(vector<int>& bulbs, int k) {
    int n = bulbs.size();
    BinaryIndexedTree* tree = new BinaryIndexedTree(n);
    bool vis[n + 1];
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; ++i) {
      int x = bulbs[i - 1];
      tree->update(x, 1);
      vis[x] = true;
      int y = x - k - 1;
      if (y > 0 && vis[y] && tree->query(x - 1) - tree->query(y) == 0) {
        return i;
      }
      y = x + k + 1;
      if (y <= n && vis[y] && tree->query(y - 1) - tree->query(x) == 0) {
        return i;
      }
    }
    return -1;
  }
};

type BinaryIndexedTree struct {
  n int
  c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
  c := make([]int, n+1)
  return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) update(x, delta int) {
  for ; x <= this.n; x += x & -x {
    this.c[x] += delta
  }
}

func (this *BinaryIndexedTree) query(x int) (s int) {
  for ; x > 0; x -= x & -x {
    s += this.c[x]
  }
  return
}

func kEmptySlots(bulbs []int, k int) int {
  n := len(bulbs)
  tree := newBinaryIndexedTree(n)
  vis := make([]bool, n+1)
  for i, x := range bulbs {
    tree.update(x, 1)
    vis[x] = true
    i++
    y := x - k - 1
    if y > 0 && vis[y] && tree.query(x-1)-tree.query(y) == 0 {
      return i
    }
    y = x + k + 1
    if y <= n && vis[y] && tree.query(y-1)-tree.query(x) == 0 {
      return i
    }
  }
  return -1
}

class BinaryIndexedTree {
  private n: number;
  private c: number[];

  constructor(n: number) {
    this.n = n;
    this.c = Array(n + 1).fill(0);
  }

  public update(x: number, delta: number) {
    for (; x <= this.n; x += x & -x) {
      this.c[x] += delta;
    }
  }

  public query(x: number): number {
    let s = 0;
    for (; x > 0; x -= x & -x) {
      s += this.c[x];
    }
    return s;
  }
}

function kEmptySlots(bulbs: number[], k: number): number {
  const n = bulbs.length;
  const tree = new BinaryIndexedTree(n);
  const vis: boolean[] = Array(n + 1).fill(false);
  for (let i = 1; i <= n; ++i) {
    const x = bulbs[i - 1];
    tree.update(x, 1);
    vis[x] = true;
    let y = x - k - 1;
    if (y > 0 && vis[y] && tree.query(x - 1) - tree.query(y) === 0) {
      return i;
    }
    y = x + k + 1;
    if (y <= n && vis[y] && tree.query(y - 1) - tree.query(x) === 0) {
      return i;
    }
  }
  return -1;
}