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• RE: Bricks Falling When Hit

@ibmtp380 Queries with repeated values are not allowed. This was not clearly explained in the problem, so I updated the explanation.

The solution can be amended easily to work with repeated queries: all subsequent repeated queries are ignored and given an answer of zero.

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• RE: Minimum Swaps To Make Sequences Increasing

@beyourself @shriram2112 @proenca @Bug_Exceeded I will update the article shortly, but here is the idea:

First, let's work through an example.

Say we have
A = [1, 3, 8]
and
B = [2, 7, 4]

At the beginning, just looking at one column, the cost to end in column (1, 2) is natural = 0, and the cost to end in (2, 1) is swapped = 1. (Cost can be infinity if it's not possible.)

Now, when i = 1, if we ended in (1, 2), we could end in (3, 7) for no cost; we could end in (7, 3) for 1 cost [the cost of ending in (1,2), plus swapping the second column]; if we ended in (2, 1), we could end in (3, 7) for no additional cost [total cost 1], or end in (7, 3) for 1 additional cost [total cost 2]. In total, ending in (3, 7) has lowest total cost 0, and ending in (7, 3) has lowest total cost 1.

Now, when i = 2, if we ended in (3, 7), we can't end in (8, 4) because the sequences won't be increasing [specifically, B won't be]. We could end in (4, 8) for an additional cost of 1 [total cost 1]. If we ended in (7, 3), we can end in (8, 4) for an additional cost of 0 [total cost 1], and we can't end in (4, 8) because A won't be increasing. In total, ending in (4, 8) has lowest total cost 1, and ending in (8, 4) also has lowest total cost 1.

I know these above paragraphs are kind of confusing, but here is a recap:
Cost to end the first i columns with the i-th column not flipped: 0, 0, 1
Cost to end the first i columns with the i-th column flipped: 1, 1, 1

Now to answer a question about what the second if condition means. It means "if you swap column i, and both A and B will become increasing from the (i-1)th column to the i-th column". In the worked example, when i = 2, we had n1 = 0 and s1 = 1 from before (from the end of the i=1 for-block). Then, the first if condition fails, but because of the second if condition, we were able to deduce that n2 = s1 : that the cost n2 of having a legal sequence up to column i that ends with column i not flipped, is going to be the cost s1 of having a legal sequence up to column i-1 that ends in column i-1 flipped, and a similar explanation for s2 = n2 + 1.

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• Minimum Swaps To Make Sequences Increasing

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• Find Eventual Safe States

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• Bricks Falling When Hit

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• Similar RGB Color

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