Help me please.
I have problem.
Given array.
You have Q queries.
You need to find most frequent number between l and r. n, Q <= 500 000.
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3408 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 154 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 146 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 133 |
| 10 | cry | 130 |
I have problem.
Given array.
You have Q queries.
You need to find most frequent number between l and r. n, Q <= 500 000.
| Name |
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Maybe, it can be solved by MO's algorithm with big constant(~5).
Let's maintain:
cntx -> the number of occurance of x,
cntBlockx -> the number of elements, that occurrence lie in this block(if K is size of blocks, occurrence must be between [K * Block: (K + 1) * Block)).
valsx -> the vector with elements, that occurrence is equal to x.
When adding/removing elements, we must update each of them. Deleting can be done in 1 operation. Swap with last, delete last(maintain extra array posx, the position of x in vector).
Finding answer will be ez. Find maximum block mx that, cntBlockmx > 0, find maximum number x in this block, that vals[x].size() > 0.
Sorry for my poor English.
Do you know smth about segment tree?