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

No.1566.Detect Pattern of Length M Repeated K or More Times

Author: yanglbme

solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/README.md

@@ -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 -->
 
@@ -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
 ```
 
@@ -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;
@@ -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;
@@ -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
@@ -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;

solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/README_EN.md

@@ -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 -->
 
@@ -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
 ```
 
@@ -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;
@@ -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;
@@ -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
@@ -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;

solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.cpp

@@ -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;
     }
-};
+};

solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.go

@@ -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
-}
+}

solution/1500-1599/1566.Detect Pattern of Length M Repeated K or More Times/Solution.java

@@ -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;
     }
-}
+}