Equal Distinct

We need to decide if we can split array A into two arrays B and C such that:

• Every element of A goes to exactly one of B or C.

• F(B) = F(C), where F(S) = number of distinct elements in S.

Key Observations

Let’s define:

• d = number of distinct values in A.

• multi = number of distinct values that appear at least twice in A.

Important points:

1. Every distinct value in A must appear in at least one of B or C.

2. To have F(B) = F(C), the total number of distinct values we can distribute between them must be even, and we must be able to give each side the same count.

3. A value that appears at least twice can be placed in both B and C simultaneously by putting one occurrence in each.

• This increases both F(B) and F(C) by 1 if the value wasn’t present there before.

4. A value that appears exactly once can be placed only in one of B or C, not both.

So:

• Values with frequency ≥ 2: “flexible” – can contribute to both sides.

• Values with frequency = 1: “rigid” – can contribute only to one side.

Core Logic

1. Count:

• d = total distinct elements.

• multi = number of elements with frequency ≥2.

2. If no value appears twice (multi = 0):

• Every distinct value can only be on one side.

• So all distinct values are strictly partitioned between B and C, and the size of the distinct sets on each side will sum to d.

• For equality, we need to split them as d2 and d2.

• This is only possible if d is even.

• So:

• If multi = 0, answer is YES iff d is even.

3. If at least one value appears twice (multi ≥ 1):

• We can always use those multi-occurring values to balance the distinct counts between B and C.

• Any difference caused by assigning single-occurrence values unevenly can be compensated by assigning a multi-occurrence value to the side with smaller distinct count (giving it a new distinct element) and also to the other side if needed.

• It turns out that this always allows a valid partition when multi≥1.

• So:

• If multi ≥ 1, answer is always YES.

Final Rule:

For each test case:

• Compute frequency of each value.

• Let:

• d = number of distinct values.

• multi= count of values with frequency ≥2.

• Then:

• If multi ≥ 1: print YES.

• Else (all frequencies are 1):

• If d is even: print YES.

• Else: print NO.

Check Against Samples:

1. A = [1, 2]

• Frequencies: 1,1

• d = 2,multi = 1, d even → YES.

2. A = [1, 2, 3]

• Frequencies: 1,1,1

• d = 3, multi = 0, d odd → NO.

3. A = [3, 1, 1, 2]

• Frequencies: {1:2, 2:1, 3:1}

• d = 3, multi = 1 (value 1) → YES.

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