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:
P=[1,2,3,...,m]
.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
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;
}
}