LeetCode 462: Minimum Moves to Equal Array Elements II Solution in Python – A Step-by-Step Guide

What Is LeetCode 462: Minimum Moves to Equal Array Elements II?

In LeetCode 462: Minimum Moves to Equal Array Elements II, you’re given an integer array nums, and your task is to find the minimum number of moves to make all elements equal, where each move increments or decrements any element by 1. Unlike its predecessor (453), you can adjust any element, not just n-1, making the median the key target. For example, in [1, 2, 3], it takes 2 moves to reach 2 (1→2, 3→2). It’s like nudging numbers toward a central meeting point with the least effort.

Problem Statement

• Input: nums (List[int])—array of integers.
• Output: int—minimum moves to equalize all elements.
• Rules:
• Each move: +1 or -1 to any element.
• All elements must end up equal.

Constraints

• 1 <= nums.length <= 10^5.
• -10^9 <= nums[i] <= 10^9.

Examples to Get Us Started

• Input: nums = [1,2,3]
• Output: 2 (Move 1→2, 3→2).
• Input: nums = [1,10,2,9]
• Output: 16 (All to median 5 or 6).
• Input: nums = [1]
• Output: 0 (Already equal).

Understanding the Problem: Finding the Sweet Spot

To solve LeetCode 462: Minimum Moves to Equal Array Elements II in Python, we need to minimize moves to align all elements to a single value. A naive approach—testing every possible target—could be O(n²) or worse. The key insight? The optimal target is the median, as it minimizes the total absolute distance (moves) in a 1D array. We’ll explore:

• Best Solution (Median-Based with Sorting): O(n log n) time, O(1) space—fast and optimal.
• Alternative Solution (Brute Force with Mean Testing): O(n²) time, O(1) space—simple but slow.

Let’s dive into the median solution—it’s the coach’s perfect playbook we need.

Best Solution: Median-Based Approach with Sorting

Why This Is the Best Solution

The median-based approach with sorting is the top pick because it’s O(n log n) time and O(1) space, leveraging the mathematical property that the median minimizes the sum of absolute differences in a list. It sorts the array and calculates moves by pairing elements around the median, ensuring the least effort to equalize. It’s like finding the middle ground where the team naturally balances—smart and efficient!

How It Works

Here’s the strategy:

• Step 1: Sort the array in ascending order.
• Step 2: For n elements:
• If n is odd, median is the middle element.
• If n is even, median is between two middle elements (any value between works).
• Step 3: Sum absolute differences from median (or use two-pointer trick).
• Step 4: Return total moves.
• Why It Works:
• Median minimizes total distance in 1D (proof via calculus or geometry).
• Sorting lets us compute moves efficiently.

Step-by-Step Example

Example: nums = [1,2,3]

• Sort: [1,2,3].
• Median: n=3 (odd), middle = 2 (index 1).
• Moves:
• 1→2: 1 move.
• 2→2: 0 moves.
• 3→2: 1 move.
• Total: 1 + 0 + 1 = 2.
• Result: 2.

Example: nums = [1,10,2,9]

• Sort: [1,2,9,10].
• Median: n=4 (even), middle = (2+9)/2 = 5.5, use 2 and 9.
• Two-Pointer:
• Left=1, Right=10: 10-1 = 9.
• Left=2, Right=9: 9-2 = 7.
• Total = 9 + 7 = 16.
• Result: 16.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def minMoves2(self, nums: List[int]) -> int:
        # Step 1: Sort the array
        nums.sort()

        # Step 2: Use two pointers from ends
        left = 0
        right = len(nums) - 1
        moves = 0

        # Step 3: Sum differences
        while left < right:
            moves += nums[right] - nums[left]
            left += 1
            right -= 1

        return moves

• Line 4: Sort nums (e.g., [1,2,9,10]).
• Line 7-8: Set pointers to ends.
• Line 11-14: Pairwise differences:
• Add right - left (e.g., 10-1, 9-2).
• Move pointers inward.
• Line 16: Return total moves.
• Time Complexity: O(n log n)—sorting.
• Space Complexity: O(1)—in-place.

It’s like a team-balancing zip-line!

Why Median Works

• Sum of absolute differences |x - a| is minimized at median a.
• For even n, any value between two middle elements works, but two-pointer sums diffs directly.

Alternative Solution: Brute Force with Mean Testing

Why an Alternative Approach?

The brute force method tests possible target values—O(n²) time, O(1) space—computing moves for each and taking the minimum. It’s slow but intuitive, like trying every score until you find the best fit. Good for small arrays or understanding!

How It Works

• Step 1: Define range of targets (min to max of nums).
• Step 2: For each target, sum absolute differences.
• Step 3: Return minimum moves found.

Step-by-Step Example

Example: nums = [1,2,3]

• Targets: 1 to 3.
• Target 1: |1-1| + |2-1| + |3-1| = 0 + 1 + 2 = 3.
• Target 2: |1-2| + |2-2| + |3-2| = 1 + 0 + 1 = 2.
• Target 3: |1-3| + |2-3| + |3-3| = 2 + 1 + 0 = 3.
• Result: Min = 2 (target 2).

Code for Brute Force

Copy codeclass Solution:
    def minMoves2(self, nums: List[int]) -> int:
        if len(nums) <= 1:
            return 0

        min_val = min(nums)
        max_val = max(nums)
        min_moves = float('inf')

        # Test each target value
        for target in range(min_val, max_val + 1):
            moves = sum(abs(num - target) for num in nums)
            min_moves = min(min_moves, moves)

        return min_moves

• Line 3-4: Early exit.
• Line 6-7: Find range.
• Line 10-12: Test targets, compute moves, track min.
• Time Complexity: O(n²)—n targets, n diffs each.
• Space Complexity: O(1)—minimal space.

It’s a slow score tester!

Comparing the Two Solutions

• Median-Based (Best):
• Pros: O(n log n), optimal, fast.
• Cons: Requires sorting insight.
• Brute Force (Alternative):
• Pros: O(n²), intuitive.
• Cons: Slow for large n.

Median wins for efficiency.

Edge Cases and More Examples

• Input: [1] → 0.
• Input: [1,1] → 0.
• Input: [1,1000000000] → 999999999.

Median handles all perfectly.

Complexity Recap

• Median: Time O(n log n), Space O(1).
• Brute Force: Time O(n²), Space O(1).

Median’s the champ.

Key Takeaways

• Median: Middle minimizes moves.
• Brute Force: Test all targets.
• Python Tip: Sorting solves—see [Python Basics](/python/basics).

Final Thoughts: Level That Team

LeetCode 462: Minimum Moves to Equal Array Elements II in Python is a strategic balancing act. The median solution is your fast playbook, while brute force is a slow drill. Want more array challenges? Try LeetCode 453: Minimum Moves to Equal Array Elements or LeetCode 215: Kth Largest Element in an Array. Ready to adjust some scores? Head to Solve LeetCode 462 on LeetCode and equalize today—your coding skills are perfectly aligned!