The Power of C# List Class

The solution to this problem demonstrates the power of the C# List Class. Here it is: https://leetcode.com/problems/queries-on-a-permutation-with-key/

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.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

Accepted

7,507

Submissions

9,107

As with many programming languages today, C# also provides a powerful List Class implementation which provides all the different methods needed to tackle this problem, methods such as: insert at any position, delete at any position, find the index of a given element. In few lines one can come up with a very fast running code. Source is below, cheers, and Happy Easter! ACC.

public class Solution
{
    public int[] ProcessQueries(int[] queries, int m)
    {
        List<int> list = new List<int>();

        for (int i = 1; i <= m; i++) list.Add(i);

        int[] retVal = new int[queries.Length];
        for (int i = 0; i < queries.Length; i++)
        {
            int index = list.IndexOf(queries[i]);
            retVal[i] = index;
            list.RemoveAt(index);
            list.Insert(0, queries[i]);
        }

        return retVal;
    }
}