Create ShellSortAditya.java

[aditya-chouksey]

Shell Sort is an optimized version of Insertion Sort that allows the exchange of far-apart elements.

The array is sorted using gaps that reduce by half each iteration.

When the gap becomes 1, it effectively becomes a normal insertion sort — but the array is mostly sorted by then.

Complexity:
Best case: O(n log n)

Average case: O(n^(3/2))

Worst case: O(n²)

Space complexity: O(1)

[github-actions]

💎 Code Quality Check Results

❌ Missing Complexity Analysis

These files don't include time/space complexity:

• Java/algorithms/sorting/ShellSortAditya.java

Required: Add comments explaining time and space complexity (e.g., Time: O(n log n), Space: O(n))

❌ Missing Algorithm Description

These files don't explain what the algorithm does:

• Java/algorithms/sorting/ShellSortAditya.java

Required: Add a description explaining the algorithm, its purpose, and how it works

📚 Quality Standards

To maintain high quality, every contribution should include:

1. ✍️ Algorithm Description

   • Explain what the algorithm does
   • Describe the approach and methodology
   • Include use cases or applications

2. 📊 Complexity Analysis

   • Time complexity (e.g., O(n log n))
   • Space complexity (e.g., O(n))
   • Brief explanation of why

3. 💬 Meaningful Comments

   • Explain complex logic
   • Document function parameters
   • Add inline comments for clarity

4. ✅ Test Cases/Examples

   • Demonstrate the code works
   • Show different input scenarios
   • Include edge cases

💡 Example Template

"""
Binary Search Algorithm

Description: Searches for a target value in a sorted array using divide-and-conquer

Time Complexity: O(log n) - halves search space each iteration
Space Complexity: O(1) - only uses constant extra space
"""

def binary_search(arr, target):
    # Initialize pointers
    left, right = 0, len(arr) - 1

    while left <= right:
        mid = (left + right) // 2

        # Check if target found
        if arr[mid] == target:
            return mid
        # Search right half
        elif arr[mid] < target:
            left = mid + 1
        # Search left half
        else:
            right = mid - 1

    return -1  # Not found

# Test cases
if __name__ == "__main__":
    test_arr = [1, 3, 5, 7, 9]
    print(binary_search(test_arr, 5))  # Output: 2
    print(binary_search(test_arr, 6))  # Output: -1

🔧 How to Fix

1. Review each file mentioned above
2. Add the missing documentation
3. Push your changes
4. The workflow will re-run automatically

💪 You've Got This!

These checks help maintain quality and make your contribution more valuable to learners. Thank you for taking the time to improve! 🙏

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Quality over quantity - let's build something amazing together! 🌟

[github-actions]

🎉 Welcome to Hacktoberfest 2025, @aditya-chouksey! 🎃

Thank you for your first contribution to our DSA repository! Here's what happens next:

🔍 Automatic Checks

• ✅ Code Validation: Passed
• 🧪 Compilation Tests: Failed

📋 Next Steps

⚠️ Action needed: Please fix the compilation errors and push your changes.

🎁 What You Get

• 🏆 Hacktoberfest Credit: This PR counts toward your 6 PR goal for exclusive T-shirt + Tree!
• 🌟 Hall of Fame: You'll be featured in our contributors list
• 📚 Learning: Code review feedback from experienced developers

💡 Tips for Success

• Follow our Contributing Guidelines
• Add comments explaining your algorithm
• Include time/space complexity analysis
• Test your code before submitting

Welcome to the community! 🚀

[github-actions]

🤖 Automated PR Status

🔍 Code Validation

✅ Passed - File naming and structure look good!

🧪 Compilation Tests

❌ Failed - Please fix compilation errors and try again.

📋 Overall Status

⚠️ Needs Work - Please address the issues above.
💡 Push new commits to automatically re-run these checks.

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This comment was generated automatically. Checks will re-run when you push new commits.