How do you solve Minimum Moves to Equal Array Elements II on LeetCode?

Minimum Moves to Equal Array Elements II (LeetCode #462) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: The median minimizes the total distance when moving numbers in a one-dimensional space.

What is the time complexity of Minimum Moves to Equal Array Elements II?

The optimal solution for Minimum Moves to Equal Array Elements II runs in O(n log n) time and uses O(1) space. The time complexity is O(n log n) due to the sorting step. The space complexity is O(1) because we are using a constant amount of extra space.

What algorithm or data structure is used to solve Minimum Moves to Equal Array Elements II?

Minimum Moves to Equal Array Elements II uses the following concepts: Array, Math, Sorting. The recommended patterns to study are: Sorting, Median, Array.

What are the interview tips for Minimum Moves to Equal Array Elements II?

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Be prepared to discuss the implications of choosing the median over other potential targets. Practice explaining the time and space complexities of your solutions.

What are common mistakes when solving Minimum Moves to Equal Array Elements II?

Not considering the median and trying to move all numbers to the mean instead. Overlooking the need to sort the array before finding the median.

Minimum Moves to Equal Array Elements II — LeetCode #462 (Medium)

Tags: Array, Math, Sorting

Related patterns: Sorting, Median, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

In the brute force approach, we calculate the cost to make all elements equal to each possible target value in the array. This means we check every number in the array as a potential target and sum the moves needed to reach that target from all other numbers.

The time complexity is O(n²) because for each of the n elements, we calculate the moves for all n elements, resulting in n * n operations. The space complexity is O(1) as we are using only a few extra variables.

Step 1: Initialize a variable to store the minimum moves as infinity.
Step 2: For each element in the array, consider it as a target value.
Step 3: For each target, calculate the total moves required to make all elements equal to this target.
Step 4: Update the minimum moves if the current target's moves are less than the previously recorded minimum.
Step 5: Return the minimum moves found.

For nums = [1, 2, 3]: 1. Target = 1: Moves = |1-1| + |2-1| + |3-1| = 0 + 1 + 2 = 3 2. Target = 2: Moves = |1-2| + |2-2| + |3-2| = 1 + 0 + 1 = 2 3. Target = 3: Moves = |1-3| + |2-3| + |3-3| = 2 + 1 + 0 = 3 Final minimum moves = 2.

Optimal Solution approach

Time complexity: O(n log n). Space complexity: O(1).

The optimal solution leverages the median of the array. By moving all elements to the median, we minimize the total distance moved. This is because the median minimizes the sum of absolute deviations.

The time complexity is O(n log n) due to the sorting step. The space complexity is O(1) because we are using a constant amount of extra space.

Step 1: Sort the array.
Step 2: Find the median of the array. If n is odd, it's the middle element; if even, either of the two middle elements will work.
Step 3: Calculate the total moves required to make all elements equal to the median.
Step 4: Return the total moves.

For nums = [1, 10, 2, 9]: 1. Sort: [1, 2, 9, 10] 2. Median = 9 (at index 2) 3. Moves = |1-9| + |2-9| + |9-9| + |10-9| = 8 + 7 + 0 + 1 = 16.

Key Insights

The median minimizes the total distance when moving numbers in a one-dimensional space.
Sorting the array is crucial for efficiently finding the median.

Common Mistakes

Not considering the median and trying to move all numbers to the mean instead.
Overlooking the need to sort the array before finding the median.

Interview Tips

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions.
Be prepared to discuss the implications of choosing the median over other potential targets.
Practice explaining the time and space complexities of your solutions.

#462

Minimum Moves to Equal Array Elements II

Medium

Array↗Math↗Sorting↗Sorting↗Median↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

⚙️

Algorithm

4 steps

1Step 1: Sort the array.
2Step 2: Find the median of the array. If n is odd, it's the middle element; if even, either of the two middle elements will work.
3Step 3: Calculate the total moves required to make all elements equal to the median.
4Step 4: Return the total moves.

solution.py6 lines

1# Full working Python code
2
3def minMoves2(nums):
4    nums.sort()
5    median = nums[len(nums) // 2]
6    return sum(abs(num - median) for num in nums)

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step. The space complexity is O(1) because we are using a constant amount of extra space.

1The median minimizes the total distance when moving numbers in a one-dimensional space.
2Sorting the array is crucial for efficiently finding the median.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.