https://leetcode.com/problems/super-washing-machines/#/description

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 theminimum 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:
 
 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   

Example2

Input: [0,3,0]

Output: 2

Explanation: 
1st move:    0 <-- 3     0    =>    1     2     0    
2nd move:    1     2 --> 0    =>    1     1     1     

Example3

Input: [0,2,0]

Output: -1

Explanation: 
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].

class Solution {
public:
    int findMinMoves(vector<int>& machines) {
        int size = machines.size();
        int count = 0;
        for(int i = 0; i < size; i++){
            count += machines[i];
        }
        if(count%size != 0)return -1;
        else
        return MinMoves(machines, count / size);
    }
    
    int MinMoves(vector<int>& mac, int key){
        int move = 0;
        int size = mac.size();
        if(size == 1)return 0;
        if(mac[size - 1] == key);
        else if(mac[size - 1] > key){
            move = mac[size - 1] - key;
            mac[size - 2] += move;
        }
        else if(mac[size - 1] < key){
            int nums = key - mac[size - 1];
            for(int i = size - 2; i >= 0; i --){
                if(mac[i] > 0 && mac[i] < nums){
                    move += mac[i] * (size - 1 - i);
                    nums -= mac[i];
                    mac[i] = 0;
                }
                else if(mac[i] > 0 && mac[i] >= nums){
                    move += nums * (size - 1 - i);
                    mac[i] -= nums;
                    break;
                }
            }
        }
        mac.pop_back();
        return move + MinMoves(mac, key);
    }

};

int findMinMoves(vector<int>& machines) {
    int totalDresses = 0, size = machines.size();
    for (auto i = 0; i < size; ++i) totalDresses += machines[i];
    if (totalDresses % size != 0) return -1;
    
    auto targetDresses = totalDresses / size, totalMoves = 0, ballance = 0;
    for (auto i = 0; i < size; ++i) {
        ballance += machines[i] - targetDresses;
        totalMoves = max(totalMoves, max(machines[i] - targetDresses, abs(ballance)));
    }
    return totalMoves;
}