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feat: update solutions to lc problem: No.1566 #3562

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Sep 25, 2024
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feat: update solutions to lc problem: No.1566 #3562

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104 changes: 56 additions & 48 deletions ...on/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -79,11 +79,15 @@ tags:

<!-- solution:start -->

### 方法一：枚举
### 方法一：一次遍历

枚举数组的左端点 `i`，判断是否存在一个 `i`，满足对于任意 `j∈[0, m * k)`，`arr[i + j] == arr[i + (j % m)]`。存在则返回 `true`，否则返回 `false`。
首先，如果数组的长度小于 $m \times k$，那么肯定不存在长度为 $m$ 且至少重复 $k$ 次的模式，直接返回 $\textit{false}$。

时间复杂度 $O((n-m\times k)\times m \times k)$。
接下来，我们定义一个变量 $\textit{cnt}$ 来记录当前连续重复的次数，如果数组存在连续的 $(k - 1) \times m$ 个元素 $a_i$，使得 $a_i = a_{i - m}$，那么我们就找到了一个长度为 $m$ 且至少重复 $k$ 次的模式，返回 $\textit{true}$。否则，我们将 $\textit{cnt}$ 置为 $0$，继续遍历数组。

最后，如果遍历完数组都没有找到符合条件的模式，返回 $\textit{false}$。

时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -92,15 +96,16 @@ tags:
```python
class Solution:
def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
n = len(arr)
for i in range(n - m * k + 1):
j = 0
while j < m * k:
if arr[i + j] != arr[i + (j % m)]:
break
j += 1
if j == m * k:
return True
if len(arr) < m * k:
return False
cnt, target = 0, (k - 1) * m
for i in range(m, len(arr)):
if arr[i] == arr[i - m]:
cnt += 1
if cnt == target:
return True
else:
cnt = 0
return False
```

Expand All @@ -109,16 +114,17 @@ class Solution:
```java
class Solution {
public boolean containsPattern(int[] arr, int m, int k) {
int n = arr.length;
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.length < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.length; ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand All @@ -132,16 +138,17 @@ class Solution {
class Solution {
public:
bool containsPattern(vector<int>& arr, int m, int k) {
int n = arr.size();
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.size() < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.size(); ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand All @@ -153,16 +160,15 @@ public:

```go
func containsPattern(arr []int, m int, k int) bool {
n := len(arr)
for i := 0; i <= n-m*k; i++ {
j := 0
for ; j < m*k; j++ {
if arr[i+j] != arr[i+(j%m)] {
break
cnt, target := 0, (k-1)*m
for i := m; i < len(arr); i++ {
if arr[i] == arr[i-m] {
cnt++
if cnt == target {
return true
}
}
if j == m*k {
return true
} else {
cnt = 0
}
}
return false
Expand All @@ -173,16 +179,18 @@ func containsPattern(arr []int, m int, k int) bool {

```ts
function containsPattern(arr: number[], m: number, k: number): boolean {
const n = arr.length;
for (let i = 0; i <= n - m * k; ++i) {
let j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.length < m * k) {
return false;
}
const target = (k - 1) * m;
let cnt = 0;
for (let i = m; i < arr.length; ++i) {
if (arr[i] === arr[i - m]) {
if (++cnt === target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand Down
104 changes: 58 additions & 46 deletions ...1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -66,7 +66,15 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Single Traversal

First, if the length of the array is less than $m \times k$, then there is definitely no pattern of length $m$ that repeats at least $k$ times, so we directly return $\textit{false}$.

Next, we define a variable $\textit{cnt}$ to record the current count of consecutive repetitions. If there are $(k - 1) \times m$ consecutive elements $a_i$ in the array such that $a_i = a_{i - m}$, then we have found a pattern of length $m$ that repeats at least $k$ times, and we return $\textit{true}$. Otherwise, we reset $\textit{cnt}$ to $0$ and continue traversing the array.

Finally, if we finish traversing the array without finding a pattern that meets the conditions, we return $\textit{false}$.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -75,15 +83,16 @@ tags:
```python
class Solution:
def containsPattern(self, arr: List[int], m: int, k: int) -> bool:
n = len(arr)
for i in range(n - m * k + 1):
j = 0
while j < m * k:
if arr[i + j] != arr[i + (j % m)]:
break
j += 1
if j == m * k:
return True
if len(arr) < m * k:
return False
cnt, target = 0, (k - 1) * m
for i in range(m, len(arr)):
if arr[i] == arr[i - m]:
cnt += 1
if cnt == target:
return True
else:
cnt = 0
return False
```

Expand All @@ -92,16 +101,17 @@ class Solution:
```java
class Solution {
public boolean containsPattern(int[] arr, int m, int k) {
int n = arr.length;
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.length < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.length; ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand All @@ -115,16 +125,17 @@ class Solution {
class Solution {
public:
bool containsPattern(vector<int>& arr, int m, int k) {
int n = arr.size();
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.size() < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.size(); ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand All @@ -136,16 +147,15 @@ public:

```go
func containsPattern(arr []int, m int, k int) bool {
n := len(arr)
for i := 0; i <= n-m*k; i++ {
j := 0
for ; j < m*k; j++ {
if arr[i+j] != arr[i+(j%m)] {
break
cnt, target := 0, (k-1)*m
for i := m; i < len(arr); i++ {
if arr[i] == arr[i-m] {
cnt++
if cnt == target {
return true
}
}
if j == m*k {
return true
} else {
cnt = 0
}
}
return false
Expand All @@ -156,16 +166,18 @@ func containsPattern(arr []int, m int, k int) bool {

```ts
function containsPattern(arr: number[], m: number, k: number): boolean {
const n = arr.length;
for (let i = 0; i <= n - m * k; ++i) {
let j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.length < m * k) {
return false;
}
const target = (k - 1) * m;
let cnt = 0;
for (let i = m; i < arr.length; ++i) {
if (arr[i] === arr[i - m]) {
if (++cnt === target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
Expand Down
21 changes: 11 additions & 10 deletions solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,18 +1,19 @@
class Solution {
public:
bool containsPattern(vector<int>& arr, int m, int k) {
int n = arr.size();
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.size() < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.size(); ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
}
};
};
19 changes: 9 additions & 10 deletions solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,15 +1,14 @@
func containsPattern(arr []int, m int, k int) bool {
n := len(arr)
for i := 0; i <= n-m*k; i++ {
j := 0
for ; j < m*k; j++ {
if arr[i+j] != arr[i+(j%m)] {
break
cnt, target := 0, (k-1)*m
for i := m; i < len(arr); i++ {
if arr[i] == arr[i-m] {
cnt++
if cnt == target {
return true
}
}
if j == m*k {
return true
} else {
cnt = 0
}
}
return false
}
}
21 changes: 11 additions & 10 deletions solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,17 +1,18 @@
class Solution {
public boolean containsPattern(int[] arr, int m, int k) {
int n = arr.length;
for (int i = 0; i <= n - m * k; ++i) {
int j = 0;
for (; j < m * k; ++j) {
if (arr[i + j] != arr[i + (j % m)]) {
break;
if (arr.length < m * k) {
return false;
}
int cnt = 0, target = (k - 1) * m;
for (int i = m; i < arr.length; ++i) {
if (arr[i] == arr[i - m]) {
if (++cnt == target) {
return true;
}
}
if (j == m * k) {
return true;
} else {
cnt = 0;
}
}
return false;
}
}
}
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