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1. Understanding the Problem:
   • Can I assume that arr1 and arr2 are always given and are both non-empty?
   • Will arr2 always contain distinct integers, and will all elements of arr2 be present in arr1?
   • How should elements in arr1 that do not appear in arr2 be handled in the final sorted array?
2. Constraints and Edge Cases:
   • What are the constraints on the size of arr1 and arr2?
   • How should we handle duplicate elements in arr1?
   • Is there any constraint on the range of integer values in arr1 and arr2?

With these questions, we can comprehend the requirements and constraints needed to tailor an efficient solution for this problem.

Let’s assume the following until further details are provided:

• arr2 will only contain unique values, and each of these values will definitely be present in arr1.
• Any elements in arr1 not in arr2 should be sorted in ascending order at the end of the output list.

Strategy:

1. Create a Mapping for Order in arr2: Use a dictionary to map each element in arr2 to its index, which will help in ordering elements of arr1 that are present in arr2.
2. Segregate and Sort: Traverse arr1:
   • Collect elements that are in arr2 and sort them using the mapping from the dictionary.
   • Collect elements that are not in arr2 and sort them in ascending order.
3. Combine Results: Concatenate the sorted elements from arr2 and the sorted elements not in arr2.

Time Complexity:

• Mapping: Creating the dictionary from arr2 will take (O(n2)) time, where (n2) is the length of arr2.
• Segregating: Traversing arr1 and categorizing elements will take (O(n1)) time, where (n1) is the length of arr1.
• Sorting: Sorting the elements in arr1 that are not in arr2 will take (O(m \log m)) time, where (m) is the number of elements in arr1 not found in arr2.

Hence, the overall complexity will be (O(n1 + m \log m + n2)), which simplifies to (O(n \log n)) in the worst case where n is the length of arr1.

Code:

def relativeSortArray(arr1, arr2):
    # Create a mapping from arr2 to its indices
    order_map = {val: idx for idx, val in enumerate(arr2)}

    # Function to get sort key
    def sort_key(val):
        if val in order_map:
            return (0, order_map[val])  # Elements in arr2 are prioritized
        else:
            return (1, val)  # Elements not in arr2 are sorted naturally

    # Sort arr1 using the custom key
    arr1.sort(key=sort_key)

    return arr1

# Example usage
arr1 = [2,3,1,3,2,4,6,7,9,2,19]
arr2 = [2,1,4,3,9,6]
print(relativeSortArray(arr1, arr2))  # Output: [2,2,2,1,4,3,3,9,6,7,19]

In the given code:

• order_map holds the indices of elements in arr2.
• The sort_key function ensures that elements present in arr2 are prioritized and sorted according to their indices.
• Elements not in arr2 are sorted naturally in ascending order.

Feel free to run this code with different inputs to verify its correctness!

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