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: 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].

 

Approach #1: Math. [C++]

class Solution {
    public int findMinMoves(int[] machines) {
        int total = 0;
        
        for (int i: machines) total += i;
        if (total % machines.length != 0) return -1;
        int avg = total / machines.length, cnt = 0, max = 0;
        for (int load: machines) {
            cnt += load - avg;
            max = Math.max(Math.max(max, Math.abs(cnt)), load - avg);
        }
        
        return max;
    }
}

　　

Approach #2: Math. [C++]

class Solution {
public:
    int findMinMoves(vector<int>& machines) {
        int n = machines.size();
        vector<int> sum(n+1, 0);
        for (int i = 0; i < n; ++i) 
            sum[i+1] = sum[i] + machines[i];
        if (sum[n] % n) return -1;
        int avg = sum[n] / n;
        int res = 0;
        for (int i = 0; i < n; ++i) {
            int l_sum = i * avg - sum[i];
            int r_sum = (n - i - 1) * avg - (sum[n] - sum[i] - machines[i]);
            
            if (l_sum > 0 && r_sum > 0)
                res = std::max(res, std::abs(l_sum) + std::abs(r_sum));
            else 
                res = std::max(res, std::max(std::abs(l_sum), std::abs(r_sum)));
        }
        
        return res;
    }
};

Analysis:

First we check the sum of dresses in all machines. is that number cannot be divided by count of machines, there is no solution.

Otherwise, we can always transfer a dress from one machine to another, one at a time untill every machines reach the same number, so there must be a solution. In this way, the total actions is sum of operations on every machine.

Since we can operate several machines at the same time, the minium number of moves is the maximum number of necessary operations on every machine.

For a single machine, necessary operations is to transfer dresses from one side to another untill sum of both sides and itself reaches the average number. We can calculate (required dresses) - (contained dresses) of each side as L and R:

L > 0 && R > 0: both sides lacks dresses, and we can only export one dress from current machines at a time, so result is abs(L) + abs(R)

L < 0 && R < 0: both sides contains too many dresses, and we can import dresses from both sides ata the same time, so result is max(abs(L), abs(R))

L < 0 && R > 0 or L > 0 && R < 0: the side with a larger absolute value will import/export its extra dresses from/to current machine or other side, so result is max(abs(L), abs(R))

Reference:

https://leetcode.com/problems/super-washing-machines/discuss/99181/C%2B%2B-16ms-O(n)-solution-(with-trivial-proof)

https://blog.csdn.net/TstsUgeg/article/details/62427718