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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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