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1409. Queries on a Permutation With Key

Description

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

  • class Solution {
    public int[] processQueries(int[] queries, int m) {
        List<Integer> p = new LinkedList<>();
        for (int i = 1; i <= m; ++i) {
            p.add(i);
        }
        int[] ans = new int[queries.length];
        int i = 0;
        for (int v : queries) {
            int j = p.indexOf(v);
            ans[i++] = j;
            p.remove(j);
            p.add(0, v);
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        vector<int> p(m);
        iota(p.begin(), p.end(), 1);
        vector<int> ans;
        for (int v : queries) {
            int j = 0;
            for (int i = 0; i < m; ++i) {
                if (p[i] == v) {
                    j = i;
                    break;
                }
            }
            ans.push_back(j);
            p.erase(p.begin() + j);
            p.insert(p.begin(), v);
        }
        return ans;
    }
};

  • class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        p = list(range(1, m + 1))
        ans = []
        for v in queries:
            j = p.index(v)
            ans.append(j)
            p.pop(j)
            p.insert(0, v)
        return ans


  • func processQueries(queries []int, m int) []int {
	p := make([]int, m)
	for i := range p {
		p[i] = i + 1
	}
	ans := []int{}
	for _, v := range queries {
		j := 0
		for i := range p {
			if p[i] == v {
				j = i
				break
			}
		}
		ans = append(ans, j)
		p = append(p[:j], p[j+1:]...)
		p = append([]int{v}, p...)
	}
	return ans
}

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