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

Solution 1: Simulation

The problem's data scale is not large, so we can directly simulate it.

Python3JavaC++Go

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

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;
    }
};

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
}

Solution 2: Binary Indexed Tree

The Binary Indexed Tree (BIT), also known as the Fenwick Tree, efficiently supports the following two operations:

1. Point Update update(x, delta): Adds a value delta to the element at position x in the sequence.
2. Prefix Sum Query query(x): Queries the sum of the sequence over the interval [1,...,x], i.e., the prefix sum at position x.

Both operations have a time complexity of $O(\log n)$.

The fundamental functionality of the Binary Indexed Tree is to count the number of elements smaller than a given element x. This comparison is abstract and can refer to size, coordinate, mass, etc.

For example, given the array a[5] = {2, 5, 3, 4, 1}, the task is to compute b[i] = the number of elements to the left of position i that are less than or equal to a[i]. For this example, b[5] = {0, 1, 1, 2, 0}.

The solution is to traverse the array, first calculating query(a[i]) for each position, and then updating the Binary Indexed Tree with update(a[i], 1). When the range of numbers is large, discretization is necessary, which involves removing duplicates, sorting, and then assigning an index to each number.

Python3JavaC++Go

class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [