README.md

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

Solutions (Ruby)

1. Simulation

# @param {Integer[]} queries
# @param {Integer} m
# @return {Integer[]}
def process_queries(queries, m)
    p = Array(m.downto(1))
    ret = Array.new(queries.length)

    for i in 0...queries.length
        ret[i] = m - 1 - p.index(queries[i])
        p.delete(queries[i])
        p.push(queries[i])
    end

    return ret
end

Solutions (Rust)

1. Simulation

impl Solution {
    pub fn process_queries(queries: Vec<i32>, m: i32) -> Vec<i32> {
        let mut p = (1..=m).rev().collect::<Vec<i32>>();
        let mut ret = vec![0; queries.len()];

        for i in 0..queries.len() {
            let posi = p.iter().position(|&x| x == queries[i]).unwrap();
            let x = p.remove(posi);
            p.push(x);
            ret[i] = m - 1 - posi as i32;
        }

        ret
    }
}

Solutions (Kotlin)

1. Simulation

class Solution {
    fun processQueries(queries: IntArray, m: Int): IntArray {
        var p = ArrayList<Int>((1..m).toList().reversed())
        var ret = IntArray(queries.size)

        for (i in 0 until queries.size) {
            ret[i] = m - 1 - p.indexOf(queries[i])
            p.remove(queries[i])
            p.add(queries[i])
        }

        return ret
    }
}