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class Solution {
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  public boolean stoneGameIX(int[] stones) {
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    Map<Integer, Integer> div3 = new HashMap<>();
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    div3.put(0, 0);
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    div3.put(1, 0);
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    div3.put(2, 0);
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    for (int stone : stones) {
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      div3.put(stone % 3, div3.get(stone % 3) + 1);
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    }
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    // the count of 3's don't matter, only whether it is even or odd
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    div3.put(0, div3.get(0) % 2);
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    if (div3.get(1) == 0 && div3.get(2) == 0) {
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      return false;
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    }
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    int smaller = Math.min(div3.get(1), div3.get(2));
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    int larger = Math.max(div3.get(2), div3.get(1));
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    // the combinations of 1's and 2's will work with each other in a complementary way.
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    // A pair of 1 and 2 makes modulo 3 to be 0
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    // Three counts of 1 or 2 makes modulo 3 to be 0
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    // so, we need only relative counts
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    // if there are even 3's, then bob can't reverse alice's win
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    // so, if all three digits chosen are the same then bob wins, but if there is another option
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    // then alice wins
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    // [1,2,2,2] -> alice picks 1 and wins
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    // [1,3,3,2] -> alice picks 1 or two and wins
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    // [2,2,2] -> alice has to pick the third 2 and loses
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    if (div3.get(0) == 0) {
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      return smaller != 0;
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    }
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    // all cases now have odd number of 3's, so result can be reversed
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    // [1,1,1,1,3] -> 1,1,3,1 picked or 1,3,1,1 picked means alice wins
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    // similar for 2 because the other number doesn't exist to make a %3 pair
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    // if the difference of number counts is more than 2 then alice can always force bob
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    // [3,1,2,2,2] ->
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    // [3,1,2,2,2,2] ->
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    if (larger > smaller + 2) {
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      return true;
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    }
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    return false;
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  }
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}