Problem Description

Given an array of integers nums, you can perform any number of operations on this array. In each operation, you can choose a prefix of the array and choose an integer k (which can be negative) and add k to each element in the chosen prefix. A prefix of an array is a subarray that starts from the beginning of the array and extends to any point within it. You need to return the minimum number of operations required to make all elements in nums equal.

Intuition

To solve this problem, the key observation is that the minimum number of operations depends on the number of distinct contiguous segments with different values. As you can add or subtract any value to a prefix, the best approach is to minimize the operation count by only adjusting segments where consecutive elements differ.

By iterating over the array, we count how many times two consecutive elements are different. Each time they differ, an operation is needed to make them the same. As a result, you only need to track changes in value between consecutive elements and use this count to determine the number of operations required.

Solution Approach

The solution uses a simple, single-pass technique to determine the number of operations needed. The main idea is to traverse the nums array and compare each element with its previous element to identify segments where value changes occur.

Here’s how the solution unfolds:

1. Initialize a Counter: Begin by initializing a counter to keep track of the number of operations required.

2. Traverse the Array: Iterate through the array using a loop, starting from the second element. For each element, compare it with the previous one.

3. Count Changes: Whenever you find that the current element is different from the previous one, increment the counter. This is because a change in value indicates a distinct segment that requires an operation to make it consistent with adjacent segments.

4. Return the Counter: Once the loop completes, the counter represents the minimum number of operations needed to make all elements equal. Return this counter as the result.

The solution leverages Python's pairwise function from the itertools module to compare consecutive elements efficiently:

class Solution:
    def minOperations(self, nums: List[int]) -> int:
        return sum(x != y for x, y in pairwise(nums))

This approach efficiently counts the number of operations needed with a time complexity of O(n), where n is the length of the array.

Example Walkthrough

Let's walk through a simple example to illustrate the solution approach:

Consider the array nums = [1, 2, 2, 3, 2].

1. Initialization:

   • Start with a counter operations = 0.

2. Traverse the Array:

   • Compare elements pair-by-pair using a loop, beginning at the second element.

3. Count Changes:

   • Compare nums[0] and nums[1]: (1) and (2) are different, so increment operations to (1).
   • Compare nums[1] and nums[2]: (2) and (2) are the same, no change to operations.
   • Compare nums[2] and nums[3]: (2) and (3) are different, increment operations to (2).
   • Compare nums[3] and nums[4]: (3) and (2) are different, increment operations to (3).

4. Return the Counter:

   • After the loop completes, operations equals (3), which is the minimum number of operations to make all elements equal by adjusting the segments.

The result is (3), showing that three operations are required to make the entire array homogeneous.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the list nums. This is because the pairwise function effectively compares each consecutive pair of elements in the list, resulting in a linear scan through the list.

The space complexity of the code is O(1). This is due to the fact that we are only using a constant amount of extra space to perform the pairwise comparisons and sum the count of differing pairs.

Learn more about how to find time and space complexity quickly.