LeetCode in Kotlin

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

Solution

class Solution {
    fun processQueries(queries: IntArray, m: Int): IntArray {
        val ans = IntArray(queries.size)
        val list: MutableList<Int> = ArrayList()
        for (i in 0 until m) {
            list.add(i + 1)
        }
        for (i in queries.indices) {
            val index = list.indexOf(queries[i])
            ans[i] = index
            list.removeAt(index)
            list.add(0, queries[i])
        }
        return ans
    }
}

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