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Formatted question description: https://leetcode.ca/all/1409.html

1409. Queries on a Permutation With Key (Medium)

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

Related Topics:
Array

Solution 1. Brute force

Simulate the process.

// OJ: https://leetcode.com/problems/queries-on-a-permutation-with-key/
// Time: O(MN)
// Space: O(M)
class Solution {
public:
    vector<int> processQueries(vector<int>& Q, int m) {
        int N = Q.size();
        vector<int> v(m), ans;
        for (int i = 0; i < m; ++i) v[i] = i+1;
        for (int q : Q) {
            int i = 0;
            for (; i < m && v[i] != q; ++i);
            ans.push_back(i);
            for (int j = i; j > 0; --j) v[j] = v[j - 1];
            v[0] = q;
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go

• class Solution {
      public int[] processQueries(int[] queries, int m) {
          LinkedList<Integer> list = new LinkedList<Integer>();
          for (int i = 1; i <= m; i++)
              list.add(i);
          int length = queries.length;
          int[] queryResults = new int[length];
          for (int i = 0; i < length; i++) {
              int query = queries[i];
              int index = list.indexOf(query);
              queryResults[i] = index;
              list.remove(new Integer(query));
              list.addFirst(query);
          }
          return queryResults;
      }
  }
  
  ############
  
  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;
      }
  }
  

• // OJ: https://leetcode.com/problems/queries-on-a-permutation-with-key/
  // Time: O(MN)
  // Space: O(M)
  class Solution {
  public:
      vector<int> processQueries(vector<int>& Q, int m) {
          int N = Q.size();
          vector<int> v(m), ans;
          for (int i = 0; i < m; ++i) v[i] = i+1;
          for (int q : Q) {
              int i = 0;
              for (; i < m && v[i] != q; ++i);
              ans.push_back(i);
              for (int j = i; j > 0; --j) v[j] = v[j - 1];
              v[0] = q;
          }
          return ans;
      }
  };
  

• 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:
      def processQueries(self, queries: List[int], m: int) -> List[int]:
          P = list(range(1, m + 1))
          res = []
          for q in queries:
              idx = P.index(q)
              P = [q] + P[:idx] + P[idx+1:]
              res.append(idx)
          return res
  

• 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
  }
  

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