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Nov 22, 2024
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feat: add solutions to lc problem: No.2261 #3801

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83 changes: 51 additions & 32 deletions solution/2200-2299/2261.K Divisible Elements Subarrays/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -82,11 +82,11 @@ nums 中的所有元素都可以被 p = 1 整除。

<!-- solution:start -->

### 方法一：哈希表 + 枚举
### 方法一：枚举 + 字符串哈希

我们可以枚举子数组的左右端点 $i$ 和 $j$，其中 $0 \leq i \leq j < n$。对于每个子数组 $nums[i,..j]$，我们可以统计其中可以被 $p$ 整除的元素的个数 $cnt$，如果 $cnt \leq k$，则该子数组满足条件。我们将所有满足条件的子数组的元素序列作为字符串存入哈希表中，最后哈希表中的元素个数即为答案。
我们可以枚举子数组的左端点 $i$，再在 $[i, n)$ 的范围内枚举子数组的右端点 $j$，在枚举右端点的过程中，我们通过双哈希的方式，将子数组的哈希值存入集合中，最后返回集合的大小即可。

时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 为数组 $nums$ 的长度。
时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为数组的长度。

<!-- tabs:start -->

Expand All @@ -95,15 +95,19 @@ nums 中的所有元素都可以被 p = 1 整除。
```python
class Solution:
def countDistinct(self, nums: List[int], k: int, p: int) -> int:
n = len(nums)
s = set()
n = len(nums)
base1, base2 = 131, 13331
mod1, mod2 = 10**9 + 7, 10**9 + 9
for i in range(n):
cnt = 0
h1 = h2 = cnt = 0
for j in range(i, n):
cnt += nums[j] % p == 0
if cnt > k:
break
s.add(tuple(nums[i : j + 1]))
h1 = (h1 * base1 + nums[j]) % mod1
h2 = (h2 * base2 + nums[j]) % mod2
s.add(h1 << 32 | h2)
return len(s)
```

Expand All @@ -112,17 +116,21 @@ class Solution:
```java
class Solution {
public int countDistinct(int[] nums, int k, int p) {
Set<Long> s = new HashSet<>();
int n = nums.length;
Set<String> s = new HashSet<>();
int base1 = 131, base2 = 13331;
int mod1 = (int) 1e9 + 7, mod2 = (int) 1e9 + 9;
for (int i = 0; i < n; ++i) {
long h1 = 0, h2 = 0;
int cnt = 0;
String t = "";
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0 ? 1 : 0;
if (cnt > k) {
break;
}
t += nums[j] + ",";
s.add(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.add(h1 << 32 | h2);
}
}
return s.size();
Expand All @@ -136,17 +144,21 @@ class Solution {
class Solution {
public:
int countDistinct(vector<int>& nums, int k, int p) {
unordered_set<string> s;
unordered_set<long long> s;
int n = nums.size();
int base1 = 131, base2 = 13331;
int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
for (int i = 0; i < n; ++i) {
long long h1 = 0, h2 = 0;
int cnt = 0;
string t;
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0;
if (cnt > k) {
break;
}
t += to_string(nums[j]) + ",";
s.insert(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.insert(h1 << 32 | h2);
}
}
return s.size();
Expand All @@ -158,18 +170,21 @@ public:

```go
func countDistinct(nums []int, k int, p int) int {
s := map[string]struct{}{}
s := map[int]bool{}
base1, base2 := 131, 13331
mod1, mod2 := 1000000007, 1000000009
for i := range nums {
cnt, t := 0, ""
for _, x := range nums[i:] {
if x%p == 0 {
h1, h2, cnt := 0, 0, 0
for j := i; j < len(nums); j++ {
if nums[j]%p == 0 {
cnt++
if cnt > k {
break
}
}
t += string(x) + ","
s[t] = struct{}{}
h1 = (h1*base1 + nums[j]) % mod1
h2 = (h2*base2 + nums[j]) % mod2
s[h1<<32|h2] = true
}
}
return len(s)
Expand All @@ -180,17 +195,21 @@ func countDistinct(nums []int, k int, p int) int {

```ts
function countDistinct(nums: number[], k: number, p: number): number {
const n = nums.length;
const s = new Set();
for (let i = 0; i < n; ++i) {
let cnt = 0;
let t = '';
for (let j = i; j < n; ++j) {
if (nums[j] % p === 0 && ++cnt > k) {
break;
const s = new Set<bigint>();
const [base1, base2] = [131, 13331];
const [mod1, mod2] = [1000000007, 1000000009];
for (let i = 0; i < nums.length; i++) {
let [h1, h2, cnt] = [0, 0, 0];
for (let j = i; j < nums.length; j++) {
if (nums[j] % p === 0) {
cnt++;
if (cnt > k) {
break;
}
}
t += nums[j].toString() + ',';
s.add(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.add((BigInt(h1) << 32n) | BigInt(h2));
}
}
return s.size;
Expand Down
83 changes: 53 additions & 30 deletions solution/2200-2299/2261.K Divisible Elements Subarrays/README_EN.md
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Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -79,7 +79,11 @@ Since all subarrays are distinct, the total number of subarrays satisfying all t

<!-- solution:start -->

### Solution 1
### Solution 1: Enumeration + String Hashing

We can enumerate the left endpoint $i$ of the subarray, and then enumerate the right endpoint $j$ in the range $[i, n)$. During the enumeration of the right endpoint, we use double hashing to store the hash value of the subarray into a set. Finally, we return the size of the set.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.

<!-- tabs:start -->

Expand All @@ -88,15 +92,19 @@ Since all subarrays are distinct, the total number of subarrays satisfying all t
```python
class Solution:
def countDistinct(self, nums: List[int], k: int, p: int) -> int:
n = len(nums)
s = set()
n = len(nums)
base1, base2 = 131, 13331
mod1, mod2 = 10**9 + 7, 10**9 + 9
for i in range(n):
cnt = 0
h1 = h2 = cnt = 0
for j in range(i, n):
cnt += nums[j] % p == 0
if cnt > k:
break
s.add(tuple(nums[i : j + 1]))
h1 = (h1 * base1 + nums[j]) % mod1
h2 = (h2 * base2 + nums[j]) % mod2
s.add(h1 << 32 | h2)
return len(s)
```

Expand All @@ -105,17 +113,21 @@ class Solution:
```java
class Solution {
public int countDistinct(int[] nums, int k, int p) {
Set<Long> s = new HashSet<>();
int n = nums.length;
Set<String> s = new HashSet<>();
int base1 = 131, base2 = 13331;
int mod1 = (int) 1e9 + 7, mod2 = (int) 1e9 + 9;
for (int i = 0; i < n; ++i) {
long h1 = 0, h2 = 0;
int cnt = 0;
String t = "";
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0 ? 1 : 0;
if (cnt > k) {
break;
}
t += nums[j] + ",";
s.add(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.add(h1 << 32 | h2);
}
}
return s.size();
Expand All @@ -129,17 +141,21 @@ class Solution {
class Solution {
public:
int countDistinct(vector<int>& nums, int k, int p) {
unordered_set<string> s;
unordered_set<long long> s;
int n = nums.size();
int base1 = 131, base2 = 13331;
int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
for (int i = 0; i < n; ++i) {
long long h1 = 0, h2 = 0;
int cnt = 0;
string t;
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0;
if (cnt > k) {
break;
}
t += to_string(nums[j]) + ",";
s.insert(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.insert(h1 << 32 | h2);
}
}
return s.size();
Expand All @@ -151,18 +167,21 @@ public:

```go
func countDistinct(nums []int, k int, p int) int {
s := map[string]struct{}{}
s := map[int]bool{}
base1, base2 := 131, 13331
mod1, mod2 := 1000000007, 1000000009
for i := range nums {
cnt, t := 0, ""
for _, x := range nums[i:] {
if x%p == 0 {
h1, h2, cnt := 0, 0, 0
for j := i; j < len(nums); j++ {
if nums[j]%p == 0 {
cnt++
if cnt > k {
break
}
}
t += string(x) + ","
s[t] = struct{}{}
h1 = (h1*base1 + nums[j]) % mod1
h2 = (h2*base2 + nums[j]) % mod2
s[h1<<32|h2] = true
}
}
return len(s)
Expand All @@ -173,17 +192,21 @@ func countDistinct(nums []int, k int, p int) int {

```ts
function countDistinct(nums: number[], k: number, p: number): number {
const n = nums.length;
const s = new Set();
for (let i = 0; i < n; ++i) {
let cnt = 0;
let t = '';
for (let j = i; j < n; ++j) {
if (nums[j] % p === 0 && ++cnt > k) {
break;
const s = new Set<bigint>();
const [base1, base2] = [131, 13331];
const [mod1, mod2] = [1000000007, 1000000009];
for (let i = 0; i < nums.length; i++) {
let [h1, h2, cnt] = [0, 0, 0];
for (let j = i; j < nums.length; j++) {
if (nums[j] % p === 0) {
cnt++;
if (cnt > k) {
break;
}
}
t += nums[j].toString() + ',';
s.add(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.add((BigInt(h1) << 32n) | BigInt(h2));
}
}
return s.size;
Expand Down
16 changes: 10 additions & 6 deletions solution/2200-2299/2261.K Divisible Elements Subarrays/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,19 +1,23 @@
class Solution {
public:
int countDistinct(vector<int>& nums, int k, int p) {
unordered_set<string> s;
unordered_set<long long> s;
int n = nums.size();
int base1 = 131, base2 = 13331;
int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
for (int i = 0; i < n; ++i) {
long long h1 = 0, h2 = 0;
int cnt = 0;
string t;
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0;
if (cnt > k) {
break;
}
t += to_string(nums[j]) + ",";
s.insert(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.insert(h1 << 32 | h2);
}
}
return s.size();
}
};
};
17 changes: 10 additions & 7 deletions solution/2200-2299/2261.K Divisible Elements Subarrays/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,17 +1,20 @@
func countDistinct(nums []int, k int, p int) int {
s := map[string]struct{}{}
s := map[int]bool{}
base1, base2 := 131, 13331
mod1, mod2 := 1000000007, 1000000009
for i := range nums {
cnt, t := 0, ""
for _, x := range nums[i:] {
if x%p == 0 {
h1, h2, cnt := 0, 0, 0
for j := i; j < len(nums); j++ {
if nums[j]%p == 0 {
cnt++
if cnt > k {
break
}
}
t += string(x) + ","
s[t] = struct{}{}
h1 = (h1*base1 + nums[j]) % mod1
h2 = (h2*base2 + nums[j]) % mod2
s[h1<<32|h2] = true
}
}
return len(s)
}
}
16 changes: 10 additions & 6 deletions solution/2200-2299/2261.K Divisible Elements Subarrays/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,22 @@
class Solution {
public int countDistinct(int[] nums, int k, int p) {
Set<Long> s = new HashSet<>();
int n = nums.length;
Set<String> s = new HashSet<>();
int base1 = 131, base2 = 13331;
int mod1 = (int) 1e9 + 7, mod2 = (int) 1e9 + 9;
for (int i = 0; i < n; ++i) {
long h1 = 0, h2 = 0;
int cnt = 0;
String t = "";
for (int j = i; j < n; ++j) {
if (nums[j] % p == 0 && ++cnt > k) {
cnt += nums[j] % p == 0 ? 1 : 0;
if (cnt > k) {
break;
}
t += nums[j] + ",";
s.add(t);
h1 = (h1 * base1 + nums[j]) % mod1;
h2 = (h2 * base2 + nums[j]) % mod2;
s.add(h1 << 32 | h2);
}
}
return s.size();
}
}
}
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