feat: add solutions to lc problem: No.2831 (doocs#2870)

yanglbme · web-flow · commit 3c5c2d62dc57 · 2024-05-21T17:51:04.000+08:00
No.2831.Find the Longest Equal Subarray
diff --git a/.github/workflows/deploy.yml b/.github/workflows/deploy.yml
@@ -4,10 +4,7 @@ on:
   push:
     branches:
       - main
-      - docs
     paths:
-      - package.json
-      - requirements.txt
       - solution/**
       - lcs/**
       - lcp/**
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/README.md b/solution/2800-2899/2831.Find the Longest Equal Subarray/README.md
@@ -179,4 +179,134 @@ function longestEqualSubarray(nums: number[], k: number): number {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### 方法二：哈希表 + 双指针（写法二）
+
+我们可以用一个哈希表 $g$ 维护每个元素的下标列表。
+
+接下来，我们枚举每个元素作为等值元素，我们从哈希表 $g$ 中取出这个元素的下标列表 $ids$，然后我们定义两个指针 $l$ 和 $r$，用于维护一个窗口，使得窗口内的元素个数减去等值元素的个数，结果不超过 $k$。那么我们只需要求出最大的满足条件的窗口即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def longestEqualSubarray(self, nums: List[int], k: int) -> int:
+        g = defaultdict(list)
+        for i, x in enumerate(nums):
+            g[x].append(i)
+        ans = 0
+        for ids in g.values():
+            l = 0
+            for r in range(len(ids)):
+                while ids[r] - ids[l] - (r - l) > k:
+                    l += 1
+                ans = max(ans, r - l + 1)
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int longestEqualSubarray(List<Integer> nums, int k) {
+        int n = nums.size();
+        List<Integer>[] g = new List[n + 1];
+        Arrays.setAll(g, i -> new ArrayList<>());
+        for (int i = 0; i < n; ++i) {
+            g[nums.get(i)].add(i);
+        }
+        int ans = 0;
+        for (List<Integer> ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids.get(r) - ids.get(l) - (r - l) > k) {
+                    ++l;
+                }
+                ans = Math.max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int longestEqualSubarray(vector<int>& nums, int k) {
+        int n = nums.size();
+        vector<int> g[n + 1];
+        for (int i = 0; i < n; ++i) {
+            g[nums[i]].push_back(i);
+        }
+        int ans = 0;
+        for (const auto& ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids[r] - ids[l] - (r - l) > k) {
+                    ++l;
+                }
+                ans = max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func longestEqualSubarray(nums []int, k int) (ans int) {
+	g := make([][]int, len(nums)+1)
+	for i, x := range nums {
+		g[x] = append(g[x], i)
+	}
+	for _, ids := range g {
+		l := 0
+		for r := range ids {
+			for ids[r]-ids[l]-(r-l) > k {
+				l++
+			}
+			ans = max(ans, r-l+1)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function longestEqualSubarray(nums: number[], k: number): number {
+    const n = nums.length;
+    const g: number[][] = Array.from({ length: n + 1 }, () => []);
+    for (let i = 0; i < n; ++i) {
+        g[nums[i]].push(i);
+    }
+    let ans = 0;
+    for (const ids of g) {
+        let l = 0;
+        for (let r = 0; r < ids.length; ++r) {
+            while (ids[r] - ids[l] - (r - l) > k) {
+                ++l;
+            }
+            ans = Math.max(ans, r - l + 1);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/README_EN.md b/solution/2800-2899/2831.Find the Longest Equal Subarray/README_EN.md
@@ -177,4 +177,134 @@ function longestEqualSubarray(nums: number[], k: number): number {
 
 <!-- solution:end -->
 
+<!-- source:start -->
+
+### Solution 2: Hash Table + Two Pointers (Method 2)
+
+We can use a hash table $g$ to maintain the index list of each element.
+
+Next, we enumerate each element as the equal value element. We take out the index list $ids$ of this element from the hash table $g$. Then we define two pointers $l$ and $r$ to maintain a window, so that the number of elements in the window minus the number of equal value elements does not exceed $k$. Therefore, we only need to find the largest window that meets the condition.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def longestEqualSubarray(self, nums: List[int], k: int) -> int:
+        g = defaultdict(list)
+        for i, x in enumerate(nums):
+            g[x].append(i)
+        ans = 0
+        for ids in g.values():
+            l = 0
+            for r in range(len(ids)):
+                while ids[r] - ids[l] - (r - l) > k:
+                    l += 1
+                ans = max(ans, r - l + 1)
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int longestEqualSubarray(List<Integer> nums, int k) {
+        int n = nums.size();
+        List<Integer>[] g = new List[n + 1];
+        Arrays.setAll(g, i -> new ArrayList<>());
+        for (int i = 0; i < n; ++i) {
+            g[nums.get(i)].add(i);
+        }
+        int ans = 0;
+        for (List<Integer> ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids.get(r) - ids.get(l) - (r - l) > k) {
+                    ++l;
+                }
+                ans = Math.max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int longestEqualSubarray(vector<int>& nums, int k) {
+        int n = nums.size();
+        vector<int> g[n + 1];
+        for (int i = 0; i < n; ++i) {
+            g[nums[i]].push_back(i);
+        }
+        int ans = 0;
+        for (const auto& ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids[r] - ids[l] - (r - l) > k) {
+                    ++l;
+                }
+                ans = max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func longestEqualSubarray(nums []int, k int) (ans int) {
+	g := make([][]int, len(nums)+1)
+	for i, x := range nums {
+		g[x] = append(g[x], i)
+	}
+	for _, ids := range g {
+		l := 0
+		for r := range ids {
+			for ids[r]-ids[l]-(r-l) > k {
+				l++
+			}
+			ans = max(ans, r-l+1)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function longestEqualSubarray(nums: number[], k: number): number {
+    const n = nums.length;
+    const g: number[][] = Array.from({ length: n + 1 }, () => []);
+    for (let i = 0; i < n; ++i) {
+        g[nums[i]].push(i);
+    }
+    let ans = 0;
+    for (const ids of g) {
+        let l = 0;
+        for (let r = 0; r < ids.length; ++r) {
+            while (ids[r] - ids[l] - (r - l) > k) {
+                ++l;
+            }
+            ans = Math.max(ans, r - l + 1);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.cpp b/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.cpp
@@ -0,0 +1,21 @@
+class Solution {
+public:
+    int longestEqualSubarray(vector<int>& nums, int k) {
+        int n = nums.size();
+        vector<int> g[n + 1];
+        for (int i = 0; i < n; ++i) {
+            g[nums[i]].push_back(i);
+        }
+        int ans = 0;
+        for (const auto& ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids[r] - ids[l] - (r - l) > k) {
+                    ++l;
+                }
+                ans = max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+};
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.go b/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.go
@@ -0,0 +1,16 @@
+func longestEqualSubarray(nums []int, k int) (ans int) {
+	g := make([][]int, len(nums)+1)
+	for i, x := range nums {
+		g[x] = append(g[x], i)
+	}
+	for _, ids := range g {
+		l := 0
+		for r := range ids {
+			for ids[r]-ids[l]-(r-l) > k {
+				l++
+			}
+			ans = max(ans, r-l+1)
+		}
+	}
+	return
+}
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.java b/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.java
@@ -0,0 +1,21 @@
+class Solution {
+    public int longestEqualSubarray(List<Integer> nums, int k) {
+        int n = nums.size();
+        List<Integer>[] g = new List[n + 1];
+        Arrays.setAll(g, i -> new ArrayList<>());
+        for (int i = 0; i < n; ++i) {
+            g[nums.get(i)].add(i);
+        }
+        int ans = 0;
+        for (List<Integer> ids : g) {
+            int l = 0;
+            for (int r = 0; r < ids.size(); ++r) {
+                while (ids.get(r) - ids.get(l) - (r - l) > k) {
+                    ++l;
+                }
+                ans = Math.max(ans, r - l + 1);
+            }
+        }
+        return ans;
+    }
+}
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.py b/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.py
@@ -0,0 +1,13 @@
+class Solution:
+    def longestEqualSubarray(self, nums: List[int], k: int) -> int:
+        g = defaultdict(list)
+        for i, x in enumerate(nums):
+            g[x].append(i)
+        ans = 0
+        for ids in g.values():
+            l = 0
+            for r in range(len(ids)):
+                while ids[r] - ids[l] - (r - l) > k:
+                    l += 1
+                ans = max(ans, r - l + 1)
+        return ans
diff --git a/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.ts b/solution/2800-2899/2831.Find the Longest Equal Subarray/Solution2.ts
@@ -0,0 +1,18 @@
+function longestEqualSubarray(nums: number[], k: number): number {
+    const n = nums.length;
+    const g: number[][] = Array.from({ length: n + 1 }, () => []);
+    for (let i = 0; i < n; ++i) {
+        g[nums[i]].push(i);
+    }
+    let ans = 0;
+    for (const ids of g) {
+        let l = 0;
+        for (let r = 0; r < ids.length; ++r) {
+            while (ids[r] - ids[l] - (r - l) > k) {
+                ++l;
+            }
+            ans = Math.max(ans, r - l + 1);
+        }
+    }
+    return ans;
+}
