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 <= 10^31 <= queries.length <= m1 <= queries[i] <= m
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
| 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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| 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;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
| 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#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
| 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
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
| 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
| 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;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
| 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
}
|
