[ Leetcode ] 1409. Queries on a Permutation With Key - Java

 

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

 

Solution

 

 
 import java.util.*;

 class Solution {
    public int[] processQueries(int[] queries, int m) {
        int[] answer = new int[queries.length];
        List<Integer> P = new ArrayList<>();
        for(int i=1; i<= m;i++){
            P.add(i);    
        }
        
        for(int i=0; i< queries.length; i++){
            int value = queries[i];
            int order = P.indexOf(value); 
            answer[i] = order;    
            P.remove(order);
            P.add(0, value);
        }
        return answer;
    }
 }
 

1. P를 만들어준다 1부터 m까지

2. 값에 대한 순서를 찾아서 answer에 넣고 맨 앞으로 꺼낸다.

 

 

++

 

 
 import java.util.*;

 class Solution {
    public int[] processQueries(int[] queries, int m) {
    	int length = queries.length;
        
        int[] answer = new int[length];
        List<Integer> P = new ArrayList<>();
        for(int i=1; i<= m;i++){
            P.add(i);    
        }
        
        for(int i=0; i< length; i++){
            int value = queries[i];
            int order = P.indexOf(value); 
            answer[i] = order;    
            P.remove(order);
            P.add(0, value);
        }
        return answer;
    }
 }
 

queries.length를 미리 length에 담으면 메모리는 좀 더 줄어든다 !