# 1409. Queries on a Permutation With Key

## Description

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

## Solutions

• class Solution {
public int[] processQueries(int[] queries, int m) {
for (int i = 1; i <= m; ++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);
}
return ans;
}
}

• class Solution {
public:
vector<int> processQueries(vector<int>& queries, int m) {
vector<int> p(m);
iota(p.begin(), p.end(), 1);
vector<int> ans;
for (int v : queries) {
int j = 0;
for (int i = 0; i < m; ++i) {
if (p[i] == v) {
j = i;
break;
}
}
ans.push_back(j);
p.erase(p.begin() + j);
p.insert(p.begin(), v);
}
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


• 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
}