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.&nbsp;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.

&nbsp;
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]


&nbsp;
Constraints:


	1 &lt;= m &lt;= 10^3
	1 &lt;= queries.length &lt;= m
	1 &lt;= queries[i] &lt;= m

class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        perm=[i for i in range(1,m+1)]
        res=[]
        for i in queries:
            ind=perm.index(i)
            res.append(ind)
            perm=[perm.pop(ind)]+perm
        return res
class Solution {
    public int[] processQueries(int[] queries, int m) {
        int[] results = new int[queries.length];
        ArrayList<Integer> permutations = new ArrayList<Integer>();

        // Filling the permuations array with numbers.
        for (int i = 0; i < m; i++)
            permutations.add(i+1);

        // Looping on the queries & checking their index in the permuations
        for (int i = 0; i < queries.length; i++) {
            int query = queries[i];
            for (int j = 0; j < permutations.size(); j++)
                if (permutations.get(j) == query) {
                    results[i] = j;
                    int temp = permutations.get(j);
                    permutations.remove(j);
                    permutations.add(0, temp);
                    break;
                }
        }

        return results;

    }
}
class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        
        vector<int> p;

        for(int i=0; i<m; i++) p.push_back(m-i);
        
        for(int i=0; i<queries.size(); i++)
        {
            auto it = find(p.begin(), p.end(), queries[i]);
            int j = it - p.begin();
            int tmp = p[j];
            p.erase(it);
            p.push_back(tmp);
            queries[i] = m-j-1;
        }
        return queries;
    }
};
var processQueries = function(queries, m) {
    let result = [];
    let permutation = [];
    for(let i=0; i<m; i++){
        permutation.push(i+1);
    }
    
    for(let i=0; i<queries.length; i++){
        let index = permutation.indexOf(queries[i]);
        result.push(index);
        permutation.splice(index,1);
        permutation.unshift(queries[i]);
    }
    return result;
};
Queries on a Permutation With Key