Naive Approach

The most straightforward approach is to iterate through the arrays and, for each index, perform the necessary increment or decrement operations on subarrays starting from that index until the nums element matches the corresponding target element. This involves repeated subarray modifications, making it inefficient.

Code Example (Python)

def min_operations_naive(nums, target):
    n = len(nums)
    operations = 0
    for i in range(n):
        diff = target[i] - nums[i]
        if diff > 0:
            for j in range(i, n):
                nums[j] += diff
            operations += diff
        elif diff < 0:
            for j in range(i, n):
                nums[j] += diff
            operations += abs(diff)
    return operations

Time Complexity

O(n^2) in the worst case, where n is the length of the arrays. This is due to the nested loops for subarray modifications.

Space Complexity

O(1), as it modifies the input array in place.

Optimal Approach

A more efficient approach is to focus on the differences between consecutive elements. The key insight is that the number of operations needed is the sum of the absolute differences between adjacent elements in the difference array.

Algorithm

1. Calculate the difference array diff[i] = target[i] - nums[i].
2. Initialize operations = abs(diff[0]).
3. Iterate from i = 1 to n-1 and add abs(diff[i] - diff[i-1]) to operations.
4. Return operations.

Code Example (Python)

def min_operations_optimal(nums, target):
    n = len(nums)
    operations = abs(target[0] - nums[0])
    for i in range(1, n):
        operations += abs(target[i] - nums[i] - (target[i-1] - nums[i-1]))
    return operations

Time Complexity

O(n), where n is the length of the arrays. We iterate through the arrays once.

Space Complexity

O(1), as it only uses a constant amount of extra space.

Edge Cases

• Empty Arrays: If the input arrays are empty, return 0.
• Single Element Arrays: The number of operations is the absolute difference between the single elements in nums and target.
• Large Integer Values: The absolute differences could potentially cause integer overflow, so it may be necessary to use larger integer types (e.g., long in Java/C++ or handle it via modulo if the constraints allow).

Summary

The optimal solution provides a significant improvement in time complexity compared to the naive approach. By focusing on the differences between consecutive elements, it avoids the need for repeated subarray modifications.