You have n super washing machines on a line. Initially, each washing machine has some dresses or is empty.
For each move, you could choose any m (1 <= m <= n) washing machines, and pass one dress of each washing machine to one of its adjacent washing machines at the same time.
Given an integer array machines representing the number of dresses in each washing machine from left to right on the line, return the minimum number of moves to make all the washing machines have the same number of dresses. If it is not possible to do it, return -1.
Example 1:
Input: machines = [1,0,5] Output: 3 Explanation: 1st move: 1 0 <-- 5 => 1 1 4 2nd move: 1 <-- 1 <-- 4 => 2 1 3 3rd move: 2 1 <-- 3 => 2 2 2
Example 2:
Input: machines = [0,3,0] Output: 2 Explanation: 1st move: 0 <-- 3 0 => 1 2 0 2nd move: 1 2 --> 0 => 1 1 1
Example 3:
Input: machines = [0,2,0] Output: -1 Explanation: It's impossible to make all three washing machines have the same number of dresses.
Constraints:
n == machines.length1 <= n <= 1040 <= machines[i] <= 105The #517 Super Washing Machines problem asks for the minimum number of moves required so that every washing machine holds the same number of dresses. The key observation is that dresses can move only between adjacent machines in each step, making this a load balancing problem.
First compute the total number of dresses and determine whether equal distribution is possible. If total % n != 0, the task cannot be completed. Otherwise, calculate the target = total / n. Traverse the array and track the running imbalance (prefix surplus or deficit). For each machine, compute how many dresses it must send or receive relative to the target.
The answer is determined by the maximum of two values: the absolute running imbalance and the local surplus at a machine. This greedy insight ensures that we capture both cumulative flow across machines and the maximum load moved in a single step. The algorithm processes the array once, achieving O(n) time and O(1) extra space.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Greedy Load Balancing with Prefix Imbalance | O(n) | O(1) |
Greg Hogg
The greedy approach involves determining the imbalance of dresses between adjacent machines and making moves to balance it. We find the maximum number of dresses that need to be moved either into or out of any machine to balance the system.
If it's impossible to balance the dresses equally among all machines (i.e., the total number of dresses is not divisible by the number of machines), the solution is -1.
Time Complexity: O(n), where n is the number of machines.
Space Complexity: O(1), since we use a constant amount of extra space.
1#include <stdio.h>
2#include <stdlib.h>
3
4int findMinMoves(int* machines, int machinesSize) {
5 int totalDresses =
This implementation starts by calculating the total number of dresses. If it isn't divisible by the number of machines, we return -1. Next, we calculate the difference between each machine's load and the average load, keeping a running balance, and determine the maximum number of moves by considering both the load difference and the accumulative balance, which is necessary due to passing dresses through intermediate machines.
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The difficulty lies in identifying the correct greedy insight rather than implementing complex code. Candidates must recognize that both local surplus and cumulative imbalance affect the number of moves required.
Yes, this problem is considered a classic greedy and load-balancing interview question. Variants of it appear in technical interviews at large tech companies because it tests reasoning about flows, constraints, and optimal distribution.
The problem mainly relies on array traversal and arithmetic calculations. No advanced data structures are required; a few variables are used to track totals, target distribution, and running imbalance.
The optimal approach uses a greedy strategy with prefix imbalance tracking. By computing the target number of dresses per machine and monitoring cumulative surplus or deficit, we determine the maximum number of moves required. This approach runs in linear time with constant extra space.
Prefix imbalance represents the net number of dresses that must flow across a boundary between machines. Tracking it helps capture the cumulative transfer needed across multiple machines. The maximum imbalance often determines the minimum number of required moves.