feat: add python and java solutions to leetcode problem: No.0219

Author: yanglbme

solution/0200-0299/0219.Contains Duplicate II/README.md

@@ -33,15 +33,36 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
+        helper = {}
+        for i, v in enumerate(nums):
+            if v in helper and i - helper[v] <= k:
+                return True
+            helper[v] = i
+        return False
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean containsNearbyDuplicate(int[] nums, int k) {
+        Map<Integer, Integer> helper = new HashMap<>();
+        for (int i = 0, n = nums.length; i < n; ++i) {
+            if (helper.containsKey(nums[i])) {
+                int j = helper.get(nums[i]);
+                if (i - j <= k) {
+                    return true;
+                }
+            }
+            helper.put(nums[i], i);
+        }
+        return false;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0219.Contains Duplicate II/README_EN.md

@@ -55,13 +55,34 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
+        helper = {}
+        for i, v in enumerate(nums):
+            if v in helper and i - helper[v] <= k:
+                return True
+            helper[v] = i
+        return False
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public boolean containsNearbyDuplicate(int[] nums, int k) {
+        Map<Integer, Integer> helper = new HashMap<>();
+        for (int i = 0, n = nums.length; i < n; ++i) {
+            if (helper.containsKey(nums[i])) {
+                int j = helper.get(nums[i]);
+                if (i - j <= k) {
+                    return true;
+                }
+            }
+            helper.put(nums[i], i);
+        }
+        return false;
+    }
+}
 ```
 
 ### **...**

solution/0200-0299/0219.Contains Duplicate II/Solution.java

@@ -1,16 +1,15 @@
 class Solution {
     public boolean containsNearbyDuplicate(int[] nums, int k) {
-        if(nums == null || nums.length == 0) return false;
-        Map<Integer, Integer> map = new HashMap<>();
-        
-        for(int i = 0; i < nums.length; i++) {
-            if(map.containsKey(nums[i]) && i - map.get(nums[i]) <= k) {
-                return true;
-            } else {
-                map.put(nums[i], i);
+        Map<Integer, Integer> helper = new HashMap<>();
+        for (int i = 0, n = nums.length; i < n; ++i) {
+            if (helper.containsKey(nums[i])) {
+                int j = helper.get(nums[i]);
+                if (i - j <= k) {
+                    return true;
+                }
             }
+            helper.put(nums[i], i);
         }
-        
         return false;
     }
 }

solution/0200-0299/0219.Contains Duplicate II/Solution.py

@@ -1,32 +1,8 @@
-'''
-https://leetcode.com/problems/contains-duplicate-ii/
-
-Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.
-'''
-# Performance
-'''
-Runtime: 96 ms, faster than 93.53% of Python3 online submissions for Contains Duplicate II.
-Memory Usage: 20.5 MB, less than 62.50% of Python3 online submissions for Contains Duplicate II.
-'''
-
-# Algorithm Explained
-'''
-Create a hashmap to remember the most recent position of unique values
-If we found duplicate and the range is less than k, then return true
-Else remember that index
-
-Space: O(n) with n is the number of original array
-Complexity: O(n)
-'''
 class Solution:
-  def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
-    pos = {}
-    for idx, element in enumerate(nums):
-      if element in pos and idx - pos[element] <= k:
-        return True
-      pos[element] = idx
-    return False
-            
-            
-                
-        
+    def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
+        helper = {}
+        for i, v in enumerate(nums):
+            if v in helper and i - helper[v] <= k:
+                return True
+            helper[v] = i
+        return False