花花酱 LeetCode 1409. Queries on a Permutation With Key

zxi — Sun, 12 Apr 2020 06:02:52 +0000

Given the array queries of positive integers between 1 and m, you have to process all queries[i] (from i=0 to i=queries.length-1) according to the following rules:

In the beginning, you have the permutation P=[1,2,3,...,m].
For the current i, find the position of queries[i] in the permutation P (indexing from 0) and then move this at the beginning of the permutation P. Notice that the position of queries[i] in P is the result for queries[i].

Return an array containing the result for the given queries.

Example 1:

Input: queries = [3,1,2,1], m = 5
Output: [2,1,2,1] 
Explanation: The queries are processed as follow: 
For i=0: queries[i]=3, P=[1,2,3,4,5], position of 3 in P is 2, then we move 3 to the beginning of P resulting in P=[3,1,2,4,5]. 
For i=1: queries[i]=1, P=[3,1,2,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,3,2,4,5]. 
For i=2: queries[i]=2, P=[1,3,2,4,5], position of 2 in P is 2, then we move 2 to the beginning of P resulting in P=[2,1,3,4,5]. 
For i=3: queries[i]=1, P=[2,1,3,4,5], position of 1 in P is 1, then we move 1 to the beginning of P resulting in P=[1,2,3,4,5]. 
Therefore, the array containing the result is [2,1,2,1].

Example 2:

Input: queries = [4,1,2,2], m = 4
Output: [3,1,2,0]

Example 3:

Input: queries = [7,5,5,8,3], m = 8
Output: [6,5,0,7,5]

Constraints:

1 <= m <= 10^3
1 <= queries.length <= m
1 <= queries[i] <= m

Solution1: Simulation + Hashtable

Use a hashtable to store the location of each key.
For each query q, use h[q] to get the index of q, for each key, if its current index is less than q, increase their indices by 1. (move right). Set h[q] to 0.

Time complexity: O(q*m)
Space complexity: O(m)

C++

class Solution {
public:
  vector processQueries(vector& queries, int m) {
    vector p(m + 1);
    iota(begin(p), end(p), -1);
    vector ans;
    for (int q : queries) {      
      ans.push_back(p[q]);
      for (int i = 1; i <= m; ++i)
        if (p[i] < p[q]) ++p[i];
      p[q] = 0;
    }      
    return ans;
  }
};

Solution 2: Fenwick Tree + HashTable

Time complexity: O(qlogm)
Space complexity: O(m)

C++

// Author: Huahua
class Fenwick {
public:
  explicit Fenwick(int n): vals_(n) {}
  
  // sum(A[1~i])
  int query(int i) const {
    int s = 0;
    while (i > 0) {
      s += vals_[i];
      i -= i & -i;
    }
    return s;
  }
  
  // A[i] += delta
  void update(int i, int delta) {
    while (i < vals_.size()) {
      vals_[i] += delta;
      i += i & -i;
    }
  }
private:
  vector vals_;
};

class Solution {
public:
  vector processQueries(vector& queries, int m) {
    Fenwick tree(m * 2  + 1);
    vector pos(m + 1);
    for (int i = 1; i <= m; ++i)     
      tree.update(pos[i] = i + m, 1);
    
    vector ans;
    for (int q : queries) {
      ans.push_back(tree.query(pos[q] - 1));
      tree.update(pos[q], -1); // set to 0.      
      tree.update(pos[q] = m--, 1); // move to the front.
    }
    return ans;
  }
};

Python3

class Fenwick:
  def __init__(self, n):
    self.n = n
    self.val = [0] * n
  
  def query(self, i):
    s = 0
    while i > 0:
      s += self.val[i]
      i -= i & -i
    return s
  
  def update(self, i, delta):
    while i < self.n:
      self.val[i] += delta
      i += i & -i
    
class Solution:
  def processQueries(self, queries: List[int], m: int) -> List[int]:    
    pos = [i + m for i in range(m + 1)]
    tree = Fenwick(2 * m + 1)
    for i in range(1, m + 1): tree.update(i + m, 1)    
    ans = []
    for q in queries:
      ans.append(tree.query(pos[q] - 1))
      tree.update(pos[q], -1)
      pos[q] = m
      m -= 1
      tree.update(pos[q], 1)
    return ans

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花花酱 LeetCode 1409. Queries on a Permutation With Key

Solution1: Simulation + Hashtable

C++

Solution 2: Fenwick Tree + HashTable

C++

Python3