class Solution {

    // Solution @ Sergey Leschev, Belarusian State University

    // 1409. Queries on a Permutation With Key
    // 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

    func processQueries(_ queries: [Int], _ m: Int) -> [Int] {
        var orray = [Int](repeating: 0, count: m)
        var output = [Int](repeating: 0, count: queries.count)
        
        for i in 1...m { orray[i - 1] = i - 1 }
        
        for (index, query) in queries.enumerated() {
            output[index] = orray[query - 1]
            if orray[query - 1] != 0 {
                for i in 0..<orray.count where orray[i] < orray[query - 1] { orray[i] += 1 }
                orray[query - 1] = 0
            }
        }

        return output
    }
    
}