Merge pull request #4091 from SadafKausar2025/madeChanges

Points Not added in the leaderboard

Author: ajay-dhangar

dsa-solutions/lc-solutions/0000-0099/0015-3sum.md

@@ -6,6 +6,10 @@ tags:
   - Hash Table
   - Math
   - String
+  - python
+  - javascript
+  - java
+  - c++
 description: find the number of triplets with given sum .
 sidebar_position: 15
 ---
@@ -16,7 +20,7 @@ Given an integer array `nums`, return all the triplets `[nums[i], nums[j], nums[
 
 Notice that the solution set must not contain duplicate triplets.
 
-### Example 1
+### Example 1:
 
 - **Input:** `nums = [-1,0,1,2,-1,-4]`
 - **Output:** `[[-1,-1,2],[-1,0,1]]`
@@ -27,13 +31,13 @@ Notice that the solution set must not contain duplicate triplets.
   - The distinct triplets are `[-1,0,1]` and `[-1,-1,2]`.
   - Notice that the order of the output and the order of the triplets does not matter.
 
-### Example 2
+### Example 2:
 
 - **Input:** `nums = [0,1,1]`
 - **Output:** `[]`
 - **Explanation:** The only possible triplet does not sum up to 0.
 
-### Example 3
+### Example 3:
 
 - **Input:** `nums = [0,0,0]`
 - **Output:** `[[0,0,0]]`
@@ -237,23 +241,23 @@ public:
    res = []
    ```
 
-2. **Sort the Input Array:**
+2. **Sort the Input Array :**
 
    - Sort the input array `nums` in non-decreasing order.
 
    ```python
    nums.sort()
    ```
 
-3. **Iterate Through the Array:**
+3. **Iterate Through the Array :**
 
    - Iterate through each element in the sorted array `nums`.
 
    ```python
    for i in range(len(nums)):
    ```
 
-4. **Skip Duplicate Elements:**
+4. **Skip Duplicate Elements :**
 
    - Check if the current element is a duplicate of the previous element and skip it if it is.
 
@@ -262,7 +266,7 @@ public:
        continue
    ```
 
-5. **Initialize Pointers:**
+5. **Initialize Pointers :**
 
    - Initialize two pointers `j` and `k` to point to the elements next to the current element `i` and at the end of the array, respectively.
 
@@ -271,23 +275,23 @@ public:
    k = len(nums) - 1
    ```
 
-6. **Two-Pointer Approach:**
+6. **Two-Pointer Approach :**
 
    - Use a two-pointer approach with pointers `j` and `k` to find triplets whose sum equals zero.
 
    ```python
    while j < k:
    ```
 
-7. **Calculate Total:**
+7. **Calculate Total :**
 
    - Calculate the total sum of the current triplet.
 
    ```python
    total = nums[i] + nums[j] + nums[k]
    ```
 
-8. **Adjust Pointers Based on Total:**
+8. **Adjust Pointers Based on Total :**
 
    - If the total sum is greater than zero, decrement the `k` pointer to decrease the total sum.
 
@@ -311,7 +315,7 @@ public:
        j += 1
    ```
 
-9. **Handle Duplicate Triplets:**
+9. **Handle Duplicate Triplets :**
 
    - Increment the `j` pointer to skip any duplicate elements.
 
@@ -320,12 +324,12 @@ public:
        j += 1
    ```
 
-10. **Return Result:**
+10. **Return Result :**
     - Return the list `res` containing all the unique triplets whose sum is zero.
     ```python
     return res
     ```
 
 This algorithm efficiently finds all unique trip
 
-lets in the given array `nums` whose sum equals zero using a two-pointer approach. It avoids duplicate triplets by skipping duplicate elements during traversal.
+lets in the given array `nums` whose sum equals zero using a two-pointer approach. It avoids duplicate triplets by skipping duplicate elements during traversal.