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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

Solution 1: Simulation

Thinking

\(m\le 10^3\). Locating a value in a length-\(m\) permutation and moving it to the front is \(O(m)\) per query, so \(O(m^2)\) overall is acceptable.

Keep the permutation in a list, record index, then pop and insert at the front.

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

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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

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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;
    }
}

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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;
    }
};

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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

Thinking

Method 1 scans the list each time. If we can ask “how many elements lie to the left of this value” in \(O(\log(m+n))\), the same process is faster.

Place the initial permutation on indices \([n+1,n+m]\) and move each queried value to a free slot on the left. A Fenwick tree stores occupancy; a prefix sum is the current index.

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.

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class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x > 0:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def processQueries(self, queries: List[int], m: int) -> List[int]:
        n = len(queries)
        pos = [0] * (m + 1)
        tree = BinaryIndexedTree(m + n)
        for i in range(1, m + 1):
            pos[i] = n + i
            tree.update(n + i, 1)

        ans = []
        for i, v in enumerate(queries):
            j = pos[v]
            tree.update(j, -1)
            ans.append(tree.query(j))
            pos[v] = n - i
            tree.update(n - i, 1)
        return ans

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class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

class Solution {
    public int[] processQueries(int[] queries, int m) {
        int n = queries.length;
        BinaryIndexedTree tree = new BinaryIndexedTree(m + n);
        int[] pos = new int[m + 1];
        for (int i = 1; i <= m; ++i) {
            pos[i] = n + i;
            tree.update(n + i, 1);
        }
        int[] ans = new int[n];
        int k = 0;
        for (int i = 0; i < n; ++i) {
            int v = queries[i];
            int j = pos[v];
            tree.update(j, -1);
            ans[k++] = tree.query(j);
            pos[v] = n - i;
            tree.update(n - i, 1);
        }
        return ans;
    }
}

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class BinaryIndexedTree {
public:
    int n;
    vector<int> c;

    BinaryIndexedTree(int _n)
        : n(_n)
        , c(_n + 1) {}

    void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    int lowbit(int x) {
        return x & -x;
    }
};

class Solution {
public:
    vector<int> processQueries(vector<int>& queries, int m) {
        int n = queries.size();
        vector<int> pos(m + 1);
        BinaryIndexedTree* tree = new BinaryIndexedTree(m + n);
        for (int i = 1; i <= m; ++i) {
            pos[i] = n + i;
            tree->update(n + i, 1);
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            int v = queries[i];
            int j = pos[v];
            tree->update(j, -1);
            ans.push_back(tree->query(j));
            pos[v] = n - i;
            tree->update(n - i, 1);
        }
        return ans;
    }
};

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type BinaryIndexedTree struct {
    n int
    c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
    c := make([]int, n+1)
    return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
    return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
    for x <= this.n {
        this.c[x] += delta
        x += this.lowbit(x)
    }
}

func (this *BinaryIndexedTree) query(x int) int {
    s := 0
    for x > 0 {
        s += this.c[x]
        x -= this.lowbit(x)
    }
    return s
}

func processQueries(queries []int, m int) []int {
    n := len(queries)
    pos := make([]int, m+1)
    tree := newBinaryIndexedTree(m + n)
    for i := 1; i <= m; i++ {
        pos[i] = n + i
        tree.update(n+i, 1)
    }
    ans := []int{}
    for i, v := range queries {
        j := pos[v]
        tree.update(j, -1)
        ans = append(ans, tree.query(j))
        pos[v] = n - i
        tree.update(n-i, 1)
    }
    return ans
}

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