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1409. Queries on a Permutation With Key

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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:

<li><code>1 &lt;= m &lt;= 10^3</code></li>

<li><code>1 &lt;= queries.length &lt;= m</code></li>

<li><code>1 &lt;= queries[i] &lt;= m</code></li>

Solutions

Python3

class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        nums = list(range(1, m + 1))
        res = []
        for num in queries:
            res.append(nums.index(num))
            nums.remove(num)
            nums.insert(0, num)
        return res

Java

class Solution {
    public int[] processQueries(int[] queries, int m) {
        List<Integer> nums = new LinkedList<>();
        for (int i = 0; i < m; ++i) {
            nums.add(i + 1);
        }
        int[] res = new int[queries.length];
        int i = 0;
        for (int num : queries) {
            res[i++] = nums.indexOf(num);
            nums.remove(Integer.valueOf(num));
            nums.add(0, num);
        }
        return res;
    }
}

C++

class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        vector<int> nums(m);
        iota(nums.begin(), nums.end(), 1);
        vector<int> res;
        for (int num : queries)
        {
            int idx = -1;
            for (int i = 0; i < m; ++i)
            {
                if (nums[i] == num) {
                    idx = i;
                    break;
                }
            }
            res.push_back(idx);
            nums.erase(nums.begin() + idx);
            nums.insert(nums.begin(), num);
        }
        return res;
    }
};

Go

func processQueries(queries []int, m int) []int {
	nums := make([]int, m)
	for i := 0; i < m; i++ {
		nums[i] = i + 1
	}
	var res []int
	for _, num := range queries {
		idx := -1
		for i := 0; i < m; i++ {
			if nums[i] == num {
				idx = i
				break
			}
		}
		res = append(res, idx)
		nums = append(nums[:idx], nums[idx+1:]...)
		nums = append([]int{num}, nums...)
	}
	return res
}

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