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.length1
) according to the following rules:
 In the beginning, you have the permutation
P=[1,2,3,...,m]
.  For the current
i
, find the position ofqueries[i]
in the permutationP
(indexing from 0) and then move this at the beginning of the permutationP.
Notice that the position ofqueries[i]
inP
is the result forqueries[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/queriesonapermutationwithkey/
// 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

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; } }

// OJ: https://leetcode.com/problems/queriesonapermutationwithkey/ // 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)) res = [] for q in queries: idx = P.index(q) P = [q] + P[:idx] + P[idx+1:] res.append(idx) return res