algoadvance

Given an integer array nums, separate the numbers at even indices from those at odd indices. The indices must follow the following conditions:

• Sort the even-indexed elements in non-decreasing order.
• Sort the odd-indexed elements in non-increasing order.

Reconstruct the array by maintaining the original index structure but with the sorted even and odd index elements.

For example:

• Input: nums = [4,1,2,3]
• Output: nums = [2,3,4,1]

Clarifying Questions

1. Can nums contain negative numbers?
   • Yes, all inputs are valid integers.
2. What if the input array is empty?
   • Returning an empty array in that case is acceptable.
3. Is the input array guaranteed to have an even number of elements?
   • No, the number of elements can be either even or odd.

Strategy

1. Separate the elements that are at even indices and the elements that are at odd indices.
2. Sort the elements at even indices in ascending (non-decreasing) order.
3. Sort the elements at odd indices in descending (non-increasing) order.
4. Place the sorted even elements back at their original even positions and odd elements back at their original odd positions.

Code

Here is the implementation in Python:

def sortEvenOdd(nums):
    # Separate even and odd indexed elements
    even_nums = [nums[i] for i in range(0, len(nums), 2)]
    odd_nums = [nums[i] for i in range(1, len(nums), 2)]
    
    # Sort even indexed elements in non-decreasing order
    even_nums.sort()
    
    # Sort odd indexed elements in non-increasing order
    odd_nums.sort(reverse=True)
    
    # Reconstruct the array
    result = []
    
    even_idx, odd_idx = 0, 0
    for i in range(len(nums)):
        if i % 2 == 0:
            result.append(even_nums[even_idx])
            even_idx += 1
        else:
            result.append(odd_nums[odd_idx])
            odd_idx += 1
    
    return result

# Example usage:
nums = [4, 1, 2, 3]
sorted_nums = sortEvenOdd(nums)
print(sorted_nums)  # Output should be [2, 3, 4, 1]

Explanation

1. We first use list comprehensions to separate the elements at even and odd indices.
2. We then sort the even-indexed elements in non-decreasing order.
3. We sort the odd-indexed elements in non-increasing order.
4. Finally, we reconstruct the array by iterating through the original indices and placing the sorted elements back in the respective positions.

Time Complexity

• Separating the elements into even and odd indexed lists takes O(n) time.
• Sorting the even indexed list takes O(k log k) time and the odd indexed list takes O(m log m) time, where k and m are the lengths of these lists respectively. Since k + m = n, the sorting operations are approximately O((n/2) log (n/2)) = O(n log n).
• Reconstructing the array takes O(n) time.

Thus, the overall time complexity is O(n log n).

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