LeetCode 517: Super Washing Machines Solution in Python – A Step-by-Step Guide

What Is LeetCode 517: Super Washing Machines?

In LeetCode 517: Super Washing Machines, you’re given an array machines where each element represents the number of dresses in a washing machine. You can move one dress per step from a machine to an adjacent one (left or right), and your task is to find the minimum number of moves to make all machines have the same number of dresses—or return -1 if impossible. For example, with [1, 0, 5], it takes 3 moves to balance to [2, 2, 2]. This problem tests optimization, related to LeetCode 122: Best Time to Buy and Sell Stock II for greedy concepts.

Problem Statement

• Input: List of integers machines—dresses per machine.
• Output: Integer—minimum moves to balance, or -1 if impossible.
• Rules: Move one dress per step to adjacent machine; all must end equal.

Constraints

• 1 <= machines.length <= 10⁶
• 0 <= machines[i] <= 10⁵

Examples

• Input: [1,0,5]
• Output: 3
• Steps: [1,0,5] → [1,1,4] → [2,1,3] → [2,2,2].
• Input: [0,3,0]
• Output: 2
• Steps: [0,3,0] → [1,2,0] → [1,1,1].
• Input: [0,2,0]
• Output: -1
• Sum = 2, n = 3, not divisible.

Understanding the Problem: Balancing the Load

To solve LeetCode 517: Super Washing Machines in Python, we need a method to minimize moves to equalize dresses across machines, where each move shifts one dress to a neighbor. The total dresses must be divisible by the number of machines, and we must consider bottlenecks where dresses pass through. A naive approach might simulate every move, but with up to 10⁶ machines, we need efficiency. We’ll explore:

• Best Solution (Greedy with Cumulative Difference): O(n) time, O(1) space—smart and fast.
• Alternative Solution (Simulation): O(n * moves) time, O(n) space—intuitive but slow.

Let’s dive into the greedy solution with a friendly breakdown!

Best Solution: Greedy with Cumulative Difference Tracking

Why Greedy Wins

The greedy approach with cumulative difference tracking is the best for LeetCode 517 because it computes the minimum moves in a single pass by analyzing dress flow and bottlenecks. It runs in O(n) time and O(1) space, making it highly efficient. It’s like figuring out the busiest traffic point in a dress-moving highway!

How It Works

Think of this as a dress-balancing roadmap:

• Step 1: Check Feasibility:
• Total dresses must be divisible by number of machines (sum % n == 0).
• Target = sum / n.
• Step 2: Track Flow:
• For each machine, calculate dresses to move left (cum_diff).
• Max moves = max of:
• Current machine’s excess/deficit (abs(cum_diff)).
• Dresses passing through current position (max(0, cum_diff)).
• Step 3: Update Max Moves:
• Track the maximum bottleneck across all positions.
• Why It Works:
• Cumulative difference shows net flow.
• Bottleneck determines min moves.

It’s like a dress-flow optimizer!

Step-by-Step Example

Example: [1,0,5]

• Init:
• n = 3, sum = 6, target = 2.
• max_moves = 0, cum_diff = 0.
• Step 1: 6 % 3 = 0, feasible.
• Step 2: Iterate:
• i=0: 1 < 2, need 1 more.
• cum_diff = 1 - 2 = -1.
• max_moves = max(0, abs(-1), 0) = 1.
• i=1: 0 < 2, need 2 more.
• cum_diff = -1 + (0 - 2) = -3.
• max_moves = max(1, abs(-3), 0) = 3.
• i=2: 5 > 2, give 3.
• cum_diff = -3 + (5 - 2) = 0.
• max_moves = max(3, abs(0), 0) = 3.
• Result: 3.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def findMinMoves(self, machines: List[int]) -> int:
        n = len(machines)
        total = sum(machines)

        # Step 1: Check if balancing is possible
        if total % n != 0:
            return -1
        target = total // n

        # Step 2: Track cumulative difference and max moves
        cum_diff = 0
        max_moves = 0

        # Step 3: Process each machine
        for dresses in machines:
            # Difference from target
            diff = dresses - target
            # Update cumulative difference
            cum_diff += diff
            # Max moves: bottleneck or current excess/deficit
            max_moves = max(max_moves, abs(cum_diff), max(0, diff))

        # Step 4: Return minimum moves
        return max_moves

• Lines 3-4: Get length and total dresses.
• Lines 7-8: If sum not divisible, impossible.
• Line 9: Target dresses per machine.
• Lines 12-13: Track flow and max moves.
• Lines 16-20: For each machine:
• diff: Excess or deficit.
• cum_diff: Net dresses moved left.
• max_moves: Max of bottleneck, deficit, or excess.
• Line 23: Return result.
• Time Complexity: O(n)—single pass.
• Space Complexity: O(1)—few variables.

It’s like a dress-balancing genius!

Alternative Solution: Simulation-Based Approach

Why an Alternative Approach?

The simulation-based solution mimics the actual process of moving dresses step-by-step until balanced. It’s O(n * moves) time and O(n) space—intuitive but impractical for large inputs due to potentially millions of moves. Great for understanding the problem physically!

How It Works

Picture this as a dress-moving game:

• Step 1: Check feasibility (sum % n == 0).
• Step 2: Simulate moves:
• While unbalanced, move dresses from high to low neighbors.
• Step 3: Count total moves.

It’s like playing dress distributor!

Step-by-Step Example

Example: [1,0,5]

• Init: target = 2, moves = 0.
• Step 1: 6 % 3 = 0, feasible.
• Step 2: Simulate:
• [1,0,5] → [1,1,4] (5→4, 0→1), moves = 1.
• [1,1,4] → [2,1,3] (4→3, 1→2), moves = 2.
• [2,1,3] → [2,2,2] (3→2, 1→2), moves = 3.
• Result: 3.

Code for Simulation Approach

Copy codeclass Solution:
    def findMinMoves(self, machines: List[int]) -> int:
        n = len(machines)
        total = sum(machines)

        if total % n != 0:
            return -1
        target = total // n

        moves = 0
        while True:
            # Check if balanced
            if all(x == target for x in machines):
                return moves

            # Make a move
            changed = False
            new_machines = machines.copy()
            for i in range(n):
                if machines[i] > target:
                    if i > 0 and machines[i-1] < target:
                        new_machines[i] -= 1
                        new_machines[i-1] += 1
                        changed = True
                    elif i < n-1 and machines[i+1] < target:
                        new_machines[i] -= 1
                        new_machines[i+1] += 1
                        changed = True
            if not changed:  # No progress possible
                break
            machines = new_machines
            moves += 1

        return moves

• Line 10: Loop until balanced.
• Lines 17-26: Move dresses to neighbors if possible.
• Time Complexity: O(n * moves)—moves can be large.
• Space Complexity: O(n)—array copy.

It’s a slow dress balancer!

Comparing the Two Solutions

• Greedy (Best):
• Pros: O(n), O(1), fast and smart.
• Cons: Abstract logic.
• Simulation (Alternative):
• Pros: O(n * moves), intuitive.
• Cons: Impractical for large inputs.

Greedy wins hands down!

Additional Examples and Edge Cases

• ** [1]: 0.
• ** [4,0,0,4]: 2.
• ** [0,1]: -1.

Greedy handles them all!

Complexity Recap

• Greedy: Time O(n), Space O(1).
• Simulation: Time O(n * moves), Space O(n).

Greedy’s the efficiency king!

Key Takeaways

• Greedy: Flow bottlenecks—learn at Python Basics!
• Simulation: Step-by-step clarity.
• Optimization: Balance is key.
• Python: Simple loops shine!

Final Thoughts: Balance Those Machines!

LeetCode 517: Super Washing Machines in Python is a thrilling greedy challenge. The cumulative difference approach is your fast track, while simulation shows the process. Want more? Try LeetCode 122: Best Time to Buy and Sell Stock II or LeetCode 135: Candy. Ready to balance? Head to Solve LeetCode 517 on LeetCode and even out those dresses today!