LeetCode-in-Java

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:

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:

Solution

import java.util.ArrayList;
import java.util.List;

@SuppressWarnings("java:S5413")
public class Solution {
    public int[] processQueries(int[] queries, int m) {
        int[] ans = new int[queries.length];
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            list.add(i + 1);
        }
        for (int i = 0; i < queries.length; i++) {
            int index = list.indexOf(queries[i]);
            ans[i] = index;
            list.remove(index);
            list.add(0, queries[i]);
        }
        return ans;
    }
}