Distant Barcodes LeetCode solution in Java

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class Solution {
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  public int[] rearrangeBarcodes(int[] barcodes) {
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    if (barcodes.length <= 2) {
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      return barcodes; // Problem says solution always exist.
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    }
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    Map<Integer, Integer> count = new HashMap<>();
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    Integer maxKey = null; // Character having max frequency
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    for (int i : barcodes) {
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      count.put(i, count.getOrDefault(i, 0) + 1);
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      if (maxKey == null || count.get(i) > count.get(maxKey)) {
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        maxKey = i;
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      }
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    }
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    int pos = 0;
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    // Fill maxChar
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    int curr = count.get(maxKey);
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    while (curr-- > 0) {
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      barcodes[pos] = maxKey;
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      pos += 2;
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      if (pos >= barcodes.length) {
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        pos = 1;
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      }
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    }
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    count.remove(maxKey); // Since that character is done, we don't need to fill it again
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    // Fill the remaining Characters.
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    for (int key : count.keySet()) {
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      curr = count.get(key);
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      while (curr-- > 0) {
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        barcodes[pos] = key;
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        pos += 2;
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        if (pos >= barcodes.length) {
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          pos = 1;
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        }
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      }
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    }
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    return barcodes;
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  }
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}