| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3993 |
| 2 | jiangly | 3743 |
| 3 | orzdevinwang | 3707 |
| 4 | Radewoosh | 3627 |
| 5 | jqdai0815 | 3620 |
| 6 | Benq | 3564 |
| 7 | Kevin114514 | 3443 |
| 8 | ksun48 | 3434 |
| 9 | Rewinding | 3397 |
| 10 | Um_nik | 3396 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 155 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 10 | djm03178 | 152 |
Mode of an array is the most frequent element, mode frequency is frequency of most frequent element.
For example, $$$[1,3,5,5]$$$ has mode $$$5$$$ and it's frequency is $$$2$$$ so mode frequency of this array is $$$2$$$.
For an array of length $$$N$$$ we need to find sum of mode frequency of all subarrays.
Is there any way to do this better than $$$O(N^2)$$$?
see codechef last contest question 6
A very similar problem was asked in Codechef Starters 136. A slight difference was that the string provided was binary, so there were only two elements you needed to keep a track of, but the overall process of implementation will be the same.
Not slight difference, binary is simple with prefix sums, I don't see any way to extend this to even ternary, let alone general array.
Since there are at most $$$\sqrt N$$$ different frequencies, so first we collect them, then for each frequency $$$f$$$ scan array with two pointers counting start of $$$f$$$-segment and end, but I can't immediately see is clean implementation of this because there are many cases where segments start and end.
What will we exactly check for each possible frequency
Number of pairs $$$(l,r)$$$ where it's the mode frequency
Then what will be the worst complexity?
The issue I see with this approach is array like
5 5 5 5has single frequency, but subarrays have 4 of them, if you find a way handle this (maybe scan for values with given freq instead of frequencies itself) it should be $$$O(n\sqrt n)$$$how to be expert in 8 months
I still don't get how would you implement counting the pairs when considering a particular frequency.