The starting position can be anywhere with a footprint. The footprints can be categorized into 3 types.
- only L s
- only R s
- R s followed by L s
In case 1, we end in the left of all footprints. In case 2, we end in the right of all footprints. In case 3, we either end in the rightmost R or the leftmost L
We can simply move by greedy method — only moves when it takes the boat closer to the destination.
Fact 0: If a has odd parity, we can apply operation 1 to increase its number of 1 by 1.
Fact 1: If a has even parity, its number of 1 cannot increase anymore
If a has parity 0, unless we pop a 1, otherwise we cannot write a new 1 into a.
Fact 2: If the number of 1 in a is not less than the one in b, we can always turn a to b
The idea is to make a copy of b at the right of a. Lets assume a has even parity now. If we need a 0, simply apply operation 1. If we need a 1, keep remove from head until we remove a 1. Notice that we never remove digits from 'new part' of a. Now the parity of a will be odd and we can apply operation 1. After that, the parity of a becomes even again, the number of 1 in the 'old part' of a decrease by 1 and we handle a 1 in b.
Finally, remove the remaining 'old part' of a. Now we get b.
Combine all those facts, we can conclude that we can turn a into b if and only if
If n > m, set every weight to 1 and we are done. Otherwise, sort a and b in non-increasing order, and trim the last part of b such that its length equals a.
Claim: answer is YES if and only if exists i such that ai > bi
If for every i, ai ≤ bi, that means for every Alice's fish, there is a corresponding Bob's fish which is as heavy as Alice's.
Let i be the smallest indices such that ai > bi. We can amplify the gap between wai and wbi large enough to make Alice wins.
An equivalent definition for almost unique, is an array with at least ⌊ 2n / 3⌋ different elements. The idea is to split s into three parts: In the first part, we give uniqueness to a. In the second part, we give uniqueness to b. In the third part, we give uniqueness to both.
Lets assume s is sorted. Since s is an unique array, si ≥ i for all i (0-based). The image below will give some intuition on how we will split it. a is red, b is blue, the length of the bar represent the magnitude of the number. In the first and second part, we do not care about the array that we are not giving uniqueness to.
We will make an example with n = 30.
i = 0... 9: assign ai = i (do not care values of b)
i = 10... 19: assign bi = i (do not care values of a)
i = 20... 29: assign bi = 29 - i. a takes the remains. From i = 20, a will have strictly increasing values starting from at least 11.
For k = 1 there is only one coloring so we just need to check the number of "E" constraints. When k ≥ 2, it turns out that using only 2 colors is sufficient to satisfy the constraints.
We will call the constraints that involves cells in different row "vertical constraints", and similar for "horizontal constraints".
First of all, we color the first row such that all horizontal constraints in row 1 are satisfied. We will color the remaining rows one by one.
To color row i, first we color it such that all horizontal constraints in row i are satisfied. Then consider the vertical constraints between row i and row i - 1. Count the number of satisfied and unsatisfied vertical constraints. If there are more unsatisfied constraints than satisfied constraints, flip the coloring of row i. Flipping a row means turning 2211212 → 1122121, for example.
If we flip the coloring of row i, all horizontal constraints in row i are still satisfied, but for the vertical constraints between row i and row i - 1, satisfied will turn to unsatisfied, unsatisfied will turn to satisfied. Therefore, we can always satisfy at least half the vertical constraints between row i and row i - 1.
Now for each row, all horizontal constraints are satisfied (m - 1 each row), at least half vertical constraints with the upper row are satisfied (⌈ m / 2⌉). In total we satisfied n(m - 1) + (n - 1)⌈ m / 2⌉ constraints. Finally notice that when m ≥ n, we satisfied at least 3 / 4 fraction of the constraints — if it is not the case that m ≥ n, simply rotate the table.