Merge pull request #6 from xchux/feature/15-3Sum

✨ (solution): Problem 15. 3Sum solution

Author: xchux

solutions/15. 3Sum/README.md

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+---
+comments: true
+difficulty: easy
+# Follow `Topics` tags
+tags:
+    - Array
+    - Two Pointers
+    - Sorting
+---
+
+# [15. 3Sum](https://leetcode.com/problems/3sum/description/)
+
+## Description
+You're given an array of integers, `nums`. Your task is to find all unique sets of three numbers \([nums[i], nums[j], nums[k]]\) such that:  
+
+1. The indices \(i\), \(j\), and \(k\) are distinct (\(i \neq j \neq k\)).  
+2. The sum of the three numbers is zero: \(nums[i] + nums[j] + nums[k] = 0\).  
+3. No duplicate triplets should appear in the result set.  
+
+Return a list of all such triplets.
+
+
+**Example 1:**
+```
+Input: nums = [[-2, -1, 0, 1, 3]]
+Output: [[-1, 0, 1], [-2, -1, 3]]
+```
+
+**Example 2:**
+```
+Input: nums = [[-2, -1, 1, -1, 2, 1]]
+Output: [[-1, -1, 2], [-2, 1, 1]]
+```
+
+
+**Constraints:**
+`3 <= nums.length <= 3000`
+`-10^5 <= nums[i] <= 10^5`
+
+## Solution
+
+Fix a value and use two pointers from bengin and end to loop all combinations sum equal 0.
+
+```java
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> res = new ArrayList<>();
+        Arrays.sort(nums);
+        for (int k = 0; k < nums.length - 2; k++) {
+            if (k > 0 && nums[k] == nums[k - 1])
+                continue;
+            int i = k + 1, j = nums.length - 1;
+            while (i < j) {
+                int sum = nums[k] + nums[i] + nums[j];
+                if (sum < 0) {
+                    i++;
+                } else if (sum > 0) {
+                    j--;
+                } else {
+                    res.add(Arrays.asList(nums[k], nums[i], nums[j]));
+                    while (i < j && nums[i] == nums[i + 1])
+                        i++;
+                    while (i < j && nums[j] == nums[j - 1])
+                        j--;
+                    i++;
+                    j--;
+                }
+            }
+        }
+        return res;
+    }
+}
+```
+
+```python
+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        nums.sort()
+        n = len(nums)
+        returnList = []
+        for i, i_num in enumerate(nums):
+            if i and i_num == nums[i-1]:
+                continue
+            l = i + 1
+            r = n - 1
+            while l < r:
+                sum = i_num + nums[l] + nums[r]
+                if sum == 0:
+                    returnList.append([i_num, nums[l], nums[r]])
+                    while l < r and nums[l] == nums[l+1]:
+                        l += 1
+                    while l < r and nums[r] == nums[r-1]:
+                        r -= 1
+                    l += 1
+                    r -= 1
+                elif sum < 0:
+                    l += 1
+                else:
+                    r -= 1
+        return returnList
+```
+
+
+## Complexity
+
+- Time complexity: $$O(n^2)$$
+<!-- Add time complexity here, e.g. $$O(n)$$ -->
+
+- Space complexity: $$O(n)$$
+<!-- Add space complexity here, e.g. $$O(n)$$ -->
+

solutions/15. 3Sum/Solution.java

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+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        List<List<Integer>> res = new ArrayList<>();
+        Arrays.sort(nums);
+        for (int k = 0; k < nums.length - 2; k++) {
+            if (k > 0 && nums[k] == nums[k - 1])
+                continue;
+            int i = k + 1, j = nums.length - 1;
+            while (i < j) {
+                int sum = nums[k] + nums[i] + nums[j];
+                if (sum < 0) {
+                    i++;
+                } else if (sum > 0) {
+                    j--;
+                } else {
+                    res.add(Arrays.asList(nums[k], nums[i], nums[j]));
+                    while (i < j && nums[i] == nums[i + 1])
+                        i++;
+                    while (i < j && nums[j] == nums[j - 1])
+                        j--;
+                    i++;
+                    j--;
+                }
+            }
+        }
+        return res;
+    }
+}

solutions/15. 3Sum/Solution.py

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+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        nums.sort()
+        n = len(nums)
+        returnList = []
+        for i, i_num in enumerate(nums):
+            if i and i_num == nums[i-1]:
+                continue
+            l = i + 1
+            r = n - 1
+            while l < r:
+                sum = i_num + nums[l] + nums[r]
+                if sum == 0:
+                    returnList.append([i_num, nums[l], nums[r]])
+                    while l < r and nums[l] == nums[l+1]:
+                        l += 1
+                    while l < r and nums[r] == nums[r-1]:
+                        r -= 1
+                    l += 1
+                    r -= 1
+                elif sum < 0:
+                    l += 1
+                else:
+                    r -= 1
+        return returnList