LeetCode Super Washing Machines

Trick:
Need to consider that, a machine can only put 1 clothes at each time, but not for both directions.
But it can accept clothes from both sides.

 

 

 

517. Super Washing Machines

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 representing the number of dresses in each washing machine from left to right on the line, you should find 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.

Example1

Input: [1,0,5]Output: 3Explanation: 1st move:    1     0 <-- 5    =>    1     1     42nd move:    1 <-- 1 <-- 4    =>    2     1     3    3rd move:    2     1 <-- 3    =>    2     2     2   

Example2

Input: [0,3,0]Output: 2Explanation: 1st move:    0 <-- 3     0    =>    1     2     0    2nd move:    1     2 --> 0    =>    1     1     1     

Example3

Input: [0,2,0]Output: -1Explanation: It's impossible to make all the three washing machines have the same number of dresses. 

 

Note:

1. The range of n is [1, 10000].
2. The range of dresses number in a super washing machine is [0, 1e5].

 

 

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My solution:

class Solution(object):
    def findMinMoves(self, machines):
        """
        :type machines: List[int]
        :rtype: int
        """
        max = 0
        sum = 0
        steps = 0
        l = len(machines)
        if (l == 0):  return 0
        i = 0
        while (i < l):
            sum = sum + machines[i]
            i = i +1
        if (sum % l != 0):
            return -1
        each = sum / l
        sum=0
        i = 0
        max2 = 0
        step = 0
        while (i < l):
            sum = sum + machines[i]
            steps=each * ( i + 1 ) - sum 
            step = machines[i] - each
            if (step > max2): max2 = step
            if (steps <0): steps = 0 - steps
            if (steps > max): max = steps
            i = i +1
        if (max2 > max): max = max2
        return max