# Very simple Java O(n) solution with explaination in the comment

• '''

``````public int findMinMoves(int[] machines) {
if ((machines == null) || (machines.length <= 1)) return 0;
long sum = 0;
for (final int machine : machines) sum += machine;
if ((sum % machines.length) != 0) return -1;
final int avg = (int)(sum / machines.length);
if (machines.length == 2) return Math.max(machines[0], machines[1]) - avg;

/*
* The main idea is that we only need to find out the machine which needs the maximum moves.
* This maximum moves is the min moves to make all the machine having same dresses.
*/
int moves = 0, giveToNext = 0, tmp = 0;
for (final int machine : machines) {
tmp = machine + giveToNext -avg;
moves = Math.max(moves, getMoves(giveToNext, tmp));
giveToNext = tmp;
}

return moves;
}

private static int getMoves(final int gotFromPrev, final int giveToNext) {
if (gotFromPrev >= 0) {
/*
* In this case, this machine has got some dresses from prev machine.
* Then
* (1) if giveToNext >= 0, it means this machine also needs to move some dresses to next machine,
* so the giveToNext = the necessary moves happening on this machine.
* (2) if giveToNext < 0, it means the next machine needs to move some dresses to this machine,
* so the necessary moves happening on this machine itself = 0.
*/
return giveToNext >= 0 ? giveToNext : 0;
} else {
/*
* In this case, this machine has to move some dresses to prev machine.
* So at least, the necessary moves on this machine is already (-gotFromPrev).
* Then
* (1) if giveToNext >= 0, it means this machine also needs to move some dresses to next machine,
* so the necessary moves happening on this machine = giveToNext + (-gotFromPrev)
* (2) if giveToNext < 0, it means the next machine needs to move some dresses to this machine,
* so the necessary moves happening on this machine itself is just (-gotFromPrev).
*/
return giveToNext >= 0 ? giveToNext-gotFromPrev : -gotFromPrev;
}
}
``````

'''

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