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I think it's not quite easy!

Suppose the array is nums, and n is the length, sum is the sum of the array, and minn is the minimal item of the array.

Suppose x is the minimum moves, we should reach that:

(sum + (n - 1) * x) % n === 0		// should be true

Simplify it, we can get:

(sum - x) % n === 0   // should be true

and we can get the smallest x by a loop:

for (let i = 0; ; i++) {
  if ((sum - i) % n)
    continue;

  return i; 
}

And we can get the final value of each item, it's (sum + (n - 1) * x) / n, define it as finalValue, and we should reach finalValue - minn <= x. Think for a while, finalValue - minn is the time the minimal item of the array should be added, and in fact we will add for x times, so finalValue should be smaller than x (or equal to).