Question

1409. Queries on a Permutation With Key

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Description

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

Solutions

Solution 1

class Solution:
  def processQueries(self, queries: List[int], m: int) -> List[int]:
    p = list(range(1, m + 1))
    ans = []
    for v in queries:
      j = p.index(v)
      ans.append(j)
      p.pop(j)
      p.insert(0, v)
    return ans

class Solution {
  public int[] processQueries(int[] queries, int m) {
    List<Integer> p = new LinkedList<>();
    for (int i = 1; i <= m; ++i) {
      p.add(i);
    }
    int[] ans = new int[queries.length];
    int i = 0;
    for (int v : queries) {
      int j = p.indexOf(v);
      ans[i++] = j;
      p.remove(j);
      p.add(0, v);
    }
    return ans;
  }
}

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

func processQueries(queries []int, m int) []int {
  p := make([]int, m)
  for i := range p {
    p[i] = i + 1
  }
  ans := []int{}
  for _, v := range queries {
    j := 0
    for i := range p {
      if p[i] == v {
        j = i
        break
      }
    }
    ans = append(ans, j)
    p = append(p[:j], p[j+1:]...)
    p = append([]int{v}, p...)
  }
  return ans
}

Solution 2

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

  @staticmethod
  def lowbit(x):
    return x & -x

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

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


class Solution:
  def processQueries(self, queries: List[int], m: int) -> List[int]:
    n = len(queries)
    pos = [0] * (m + 1)
    tree = BinaryIndexedTree(m + n)
    for i in range(1, m + 1):
      pos[i] = n + i
      tree.update(n + i, 1)

    ans = []
    for i, v in enumerate(queries):
      j = pos[v]
      tree.update(j, -1)
      ans.append(tree.query(j))
      pos[v] = n - i
      tree.update(n - i, 1)
    return ans

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

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

  public void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

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

  public static int lowbit(int x) {
    return x & -x;
  }
}

class Solution {
  public int[] processQueries(int[] queries, int m) {
    int n = queries.length;
    BinaryIndexedTree tree = new BinaryIndexedTree(m + n);
    int[] pos = new int[m + 1];
    for (int i = 1; i <= m; ++i) {
      pos[i] = n + i;
      tree.update(n + i, 1);
    }
    int[] ans = new int[n];
    int k = 0;
    for (int i = 0; i < n; ++i) {
      int v = queries[i];
      int j = pos[v];
      tree.update(j, -1);
      ans[k++] = tree.query(j);
      pos[v] = n - i;
      tree.update(n - i, 1);
    }
    return ans;
  }
}

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

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

  void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

  int query(int x) {
    int s = 0;
    while (x > 0) {
      s += c[x];
      x -= lowbit(x);
    }
    return s;
  }

  int lowbit(int x) {
    return x & -x;
  }
};

class Solution {
public:
  vector<int> processQueries(vector<int>& queries, int m) {
    int n = queries.size();
    vector<int> pos(m + 1);
    BinaryIndexedTree* tree = new BinaryIndexedTree(m + n);
    for (int i = 1; i <= m; ++i) {
      pos[i] = n + i;
      tree->update(n + i, 1);
    }
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      int v = queries[i];
      int j = pos[v];
      tree->update(j, -1);
      ans.push_back(tree->query(j));
      pos[v] = n - i;
      tree->update(n - i, 1);
    }
    return ans;
  }
};

type BinaryIndexedTree struct {
  n int
  c []int
}

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

func (this *BinaryIndexedTree) lowbit(x int) int {
  return x & -x
}

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

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

func processQueries(queries []int, m int) []int {
  n := len(queries)
  pos := make([]int, m+1)
  tree := newBinaryIndexedTree(m + n)
  for i := 1; i <= m; i++ {
    pos[i] = n + i
    tree.update(n+i, 1)
  }
  ans := []int{}
  for i, v := range queries {
    j := pos[v]
    tree.update(j, -1)
    ans = append(ans, tree.query(j))
    pos[v] = n - i
    tree.update(n-i, 1)
  }
  return ans
}