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 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.
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.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the number of machines.
Space Complexity: O(1), since we use a constant amount of extra space.
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