LeetCode 453: Minimum Moves to Equal Array Elements Solution in Python – A Step-by-Step Guide

What Is LeetCode 453: Minimum Moves to Equal Array Elements?

In LeetCode 453: Minimum Moves to Equal Array Elements, you’re given an integer array nums of length n, and your task is to find the minimum number of moves to make all elements equal. A move increments exactly n-1 elements by 1. For example, with [1, 2, 3], you need 2 moves to reach [3, 3, 3]. It’s like raising everyone’s score but one player per round until the team’s all on the same level.

Problem Statement

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

Constraints

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

Examples to Get Us Started

• Input: nums = [1,2,3]
• Output: 2 ([1,2,3] → [2,3,3] → [3,3,3]).
• Input: nums = [1,1,1]
• Output: 0 (Already equal).
• Input: nums = [1,10]
• Output: 9 (9 moves to [10,10]).

Understanding the Problem: Leveling the Field

To solve LeetCode 453: Minimum Moves to Equal Array Elements in Python, we need to determine how many moves—where we increment n-1 elements by 1—get all numbers to the same value. A naive approach—simulating each move—works but could be O(n * m) where m is moves, potentially huge. The key insight? Incrementing n-1 elements by 1 is equivalent to decreasing 1 element by 1 relative to the others, leading to a brilliant math shortcut. We’ll explore:

• Best Solution (Mathematical Formula): O(n) time, O(1) space—fast and brilliant.
• Alternative Solution (Simulation with Sorting): O(n log n + n * m) time, O(1) space—intuitive but slow.

Let’s dive into the mathematical solution—it’s the equalizer we need.

Best Solution: Mathematical Approach Using Minimum Element

Why This Is the Best Solution

The mathematical approach is the top pick because it’s O(n) time and O(1) space, solving the problem with a single pass and a clever formula. It realizes that each move effectively reduces the gap between elements and the minimum value, and the total moves equal the sum of differences from the minimum to reach a common target. It’s like calculating how many boosts the lowest player needs to catch up, then applying that across the team—genius and efficient!

How It Works

Here’s the strategy:

• Step 1: Find the minimum element (min_val).
• Step 2: Compute the total moves as the sum of (nums[i] - min_val) for all i.
• Step 3: Return the result.
• Why It Works:
• Each move on n-1 elements = -1 on one element relative to others.
• To equalize, all elements must drop to min_val in this relative sense.
• Total moves = sum of excesses above min_val.

Step-by-Step Example

Example: nums = [1,2,3]

• Min: 1.
• Differences:
• 1 - 1 = 0.
• 2 - 1 = 1.
• 3 - 1 = 2.
• Sum: 0 + 1 + 2 = 3 - 1 = 2 moves.
• Verify:
• [1,2,3] → [2,3,3] (increment 1,2).
• [2,3,3] → [3,3,3] (increment 2).
• Result: 2.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

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

        # Step 1: Find minimum
        min_val = min(nums)

        # Step 2: Sum differences from minimum
        moves = 0
        for num in nums:
            moves += num - min_val

        return moves

• Line 3-4: Handle trivial cases.
• Line 7: Find min_val in one pass.
• Line 10-11: Sum (num - min_val) for each element.
• Line 13: Return total moves.
• Time Complexity: O(n)—one pass for min, one for sum.
• Space Complexity: O(1)—just a few variables.

It’s like a math shortcut to team harmony!

Mathematical Insight

• Let final value = f.
• Each move adds 1 to n-1 elements, total sum increases by n-1.
• Initial sum = s, moves = m, final sum = f * n.
• s + m * (n-1) = f * n.
• But simpler: m = sum(nums[i] - min_val), as all must reach min_val relative to moves.

Alternative Solution: Simulation with Sorting

Why an Alternative Approach?

The simulation with sorting method mimics the process—O(n log n + n * m) time, O(1) space—by repeatedly incrementing n-1 smallest elements until all are equal. It’s intuitive, like coaching the team step-by-step, but slow for large gaps. Good for understanding or small n!

How It Works

• Step 1: Sort array initially.
• Step 2: While not all equal:
• Increment n-1 smallest by 1.
• Resort if needed (or track manually).
• Step 3: Return move count.

Step-by-Step Example

Example: nums = [1,2,3]

• Sort: [1,2,3].
• Move 1: Increment [1,2] → [2,2,3], moves = 1.
• Move 2: Increment [2,2] → [3,2,3] → [3,3,3], moves = 2.
• Result: 2.

Code for Simulation Approach

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

        moves = 0
        nums.sort()  # Initial sort

        while nums[0] != nums[-1]:  # Until all equal
            # Increment n-1 smallest
            for i in range(len(nums) - 1):
                nums[i] += 1
            nums.sort()  # Resort after increment
            moves += 1

        return moves

• Line 3-4: Handle trivial cases.
• Line 7: Sort initially.
• Line 9-12: Loop until equal, increment n-1, resort.
• Time Complexity: O(n log n + n * m)—sorting plus moves.
• Space Complexity: O(1)—in-place.

It’s a hands-on team leveler!

Comparing the Two Solutions

• Mathematical (Best):
• Pros: O(n), fast, no simulation.
• Cons: Requires insight.
• Simulation (Alternative):
• Pros: O(n log n + n * m), intuitive.
• Cons: Slow for large gaps.

Math wins for speed.

Edge Cases and More Examples

• Input: [] → 0.
• Input: [1] → 0.
• Input: [1,1,100] → 99.

Math handles all efficiently.

Complexity Recap

• Mathematical: Time O(n), Space O(1).
• Simulation: Time O(n log n + n * m), Space O(1).

Math’s the champ.

Key Takeaways

• Mathematical: Formula beats brute force.
• Simulation: Step-by-step clarity.
• Python Tip: Math simplifies—see [Python Basics](/python/basics).

Final Thoughts: Equalize That Team

LeetCode 453: Minimum Moves to Equal Array Elements in Python is a math-driven coaching challenge. The mathematical solution is your instant equalizer, while simulation builds intuition. Want more array fun? Try LeetCode 462: Minimum Moves to Equal Array Elements II or LeetCode 561: Array Partition I. Ready to level some arrays? Head to Solve LeetCode 453 on LeetCode and balance that team today—your coding skills are in sync!