feat: add solutions to lc problem: No.2200 (doocs#2090)

No.2200.Find All K-Distant Indices in an Array

Author: yanglbme

solution/2200-2299/2200.Find All K-Distant Indices in an Array/README.md

@@ -52,6 +52,26 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：枚举**
+
+我们在 $[0, n)$ 的范围内枚举下标 $i$，对于每个下标 $i$，我们在 $[0, n)$ 的范围内枚举下标 $j$，如果 $|i - j| \leq k$ 且 $nums[j] == key$，那么 $i$ 就是一个 K 近邻下标，我们将 $i$ 加入答案数组中，然后跳出内层循环，枚举下一个下标 $i$。
+
+时间复杂度 $O(n^2)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
+
+**方法二：预处理 + 二分查找**
+
+我们可以预处理得到所有等于 $key$ 的元素的下标，记录在数组 $idx$ 中。数组 $idx$ 中的所有下标元素是按照升序排列的，
+
+接下来，我们枚举下标 $i$，对于每个下标 $i$，我们可以使用二分查找的方法在数组 $idx$ 中查找 $[i - k, i + k]$ 范围内的元素，如果存在元素，那么 $i$ 就是一个 K 近邻下标，我们将 $i$ 加入答案数组中。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
+
+**方法三：双指针**
+
+我们枚举下标 $i$，用一个指针 $j$ 指向满足 $j \geq i - k$ 且 $nums[j] = key$ 的最小下标，如果 $j$ 存在且 $j \leq i + k$，那么 $i$ 就是一个 K 近邻下标，我们将 $i$ 加入答案数组中。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -64,10 +84,34 @@ class Solution:
         ans = []
         n = len(nums)
         for i in range(n):
-            for j in range(n):
-                if abs(i - j) <= k and nums[j] == key:
-                    ans.append(i)
-                    break
+            if any(abs(i - j) <= k and nums[j] == key for j in range(n)):
+                ans.append(i)
+        return ans
+```
+
+```python
+class Solution:
+    def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
+        idx = [i for i, x in enumerate(nums) if x == key]
+        ans = []
+        for i in range(len(nums)):
+            l = bisect_left(idx, i - k)
+            r = bisect_right(idx, i + k) - 1
+            if l <= r:
+                ans.append(i)
+        return ans
+```
+
+```python
+class Solution:
+    def findKDistantIndices(self, nums: List[int], key: int, k: int) -> List[int]:
+        ans = []
+        j, n = 0, len(nums)
+        for i in range(n):
+            while j < i - k or (j < n and nums[j] != key):
+                j += 1
+            if j < n and j <= (i + k):
+                ans.append(i)
         return ans
 ```
 
@@ -93,22 +137,45 @@ class Solution {
 }
 ```
 
-### **TypeScript**
+```java
+class Solution {
+    public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
+        List<Integer> idx = new ArrayList<>();
+        for (int i = 0; i < nums.length; i++) {
+            if (nums[i] == key) {
+                idx.add(i);
+            }
+        }
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 0; i < nums.length; ++i) {
+            int l = Collections.binarySearch(idx, i - k);
+            int r = Collections.binarySearch(idx, i + k + 1);
+            l = l < 0 ? -l - 1 : l;
+            r = r < 0 ? -r - 2 : r - 1;
+            if (l <= r) {
+                ans.add(i);
+            }
+        }
+        return ans;
+    }
+}
+```
 
-```ts
-function findKDistantIndices(nums: number[], key: number, k: number): number[] {
-    const n = nums.length;
-    let ans = [];
-    for (let j = 0; j < n; j++) {
-        if (nums[j] == key) {
-            for (let i = j - k; i <= j + k; i++) {
-                if (i >= 0 && i < n && !ans.includes(i)) {
-                    ans.push(i);
-                }
+```java
+class Solution {
+    public List<Integer> findKDistantIndices(int[] nums, int key, int k) {
+        int n = nums.length;
+        List<Integer> ans = new ArrayList<>();
+        for (int i = 0, j = 0; i < n; ++i) {
+            while (j < i - k || (j < n && nums[j] != key)) {
+                ++j;
+            }
+            if (j < n && j <= i + k) {
+                ans.add(i);
             }
         }
+        return ans;
     }
-    return ans;
 }
 ```
 
@@ -133,15 +200,56 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
+        vector<int> idx;
+        int n = nums.size();
+        for (int i = 0; i < n; ++i) {
+            if (nums[i] == key) {
+                idx.push_back(i);
+            }
+        }
+        vector<int> ans;
+        for (int i = 0; i < n; ++i) {
+            auto it1 = lower_bound(idx.begin(), idx.end(), i - k);
+            auto it2 = upper_bound(idx.begin(), idx.end(), i + k) - 1;
+            if (it1 <= it2) {
+                ans.push_back(i);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+```cpp
+class Solution {
+public:
+    vector<int> findKDistantIndices(vector<int>& nums, int key, int k) {
+        int n = nums.size();
+        vector<int> ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            while (j < i - k || (j < n && nums[j] != key)) {
+                ++j;
+            }
+            if (j < n && j <= i + k) {
+                ans.push_back(i);
+            }
+        }
+        return ans;
+    }
+};
+```
+
 ### **Go**
 
 ```go
-func findKDistantIndices(nums []int, key int, k int) []int {
-	n := len(nums)
-	var ans []int
-	for i := 0; i < n; i++ {
-		for j, v := range nums {
-			if abs(i-j) <= k && v == key {
+func findKDistantIndices(nums []int, key int, k int) (ans []int) {
+	for i := range nums {
+		for j, x := range nums {
+			if abs(i-j) <= k && x == key {
 				ans = append(ans, i)
 				break
 			}
@@ -158,6 +266,107 @@ func abs(x int) int {
 }
 ```
 
+```go
+func findKDistantIndices(nums []int, key int, k int) (ans []int) {
+	idx := []int{}
+	for i, x := range nums {
+		if x == key {
+			idx = append(idx, i)
+		}
+	}
+	for i := range nums {
+		l := sort.SearchInts(idx, i-k)
+		r := sort.SearchInts(idx, i+k+1) - 1
+		if l <= r {
+			ans = append(ans, i)
+		}
+	}
+	return
+}
+```
+
+```go
+func findKDistantIndices(nums []int, key int, k int) (ans []int) {
+	n := len(nums)
+	for i, j := 0, 0; i < n; i++ {
+		for j < i-k || (j < n && nums[j] != key) {
+			j++
+		}
+		if j < n && j <= i+k {
+			ans = append(ans, i)
+		}
+	}
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function findKDistantIndices(nums: number[], key: number, k: number): number[] {
+    const n = nums.length;
+    const ans: number[] = [];
+    for (let i = 0; i < n; ++i) {
+        for (let j = 0; j < n; ++j) {
+            if (Math.abs(i - j) <= k && nums[j] === key) {
+                ans.push(i);
+                break;
+            }
+        }
+    }
+    return ans;
+}
+```
+
+```ts
+function findKDistantIndices(nums: number[], key: number, k: number): number[] {
+    const n = nums.length;
+    const idx: number[] = [];
+    for (let i = 0; i < n; i++) {
+        if (nums[i] === key) {
+            idx.push(i);
+        }
+    }
+    const search = (x: number): number => {
+        let [l, r] = [0, idx.length];
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (idx[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    const ans: number[] = [];
+    for (let i = 0; i < n; ++i) {
+        const l = search(i - k);
+        const r = search(i + k + 1) - 1;
+        if (l <= r) {
+            ans.push(i);
+        }
+    }
+    return ans;
+}
+```
+
+```ts
+function findKDistantIndices(nums: number[], key: number, k: number): number[] {
+    const n = nums.length;
+    const ans: number[] = [];
+    for (let i = 0, j = 0; i < n; ++i) {
+        while (j < i - k || (j < n && nums[j] !== key)) {
+            ++j;
+        }
+        if (j < n && j <= i + k) {
+            ans.push(i);
+        }
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```