Find number of unique longest increasing subsequence in array a.
Constraints : 1 <= Lenght of a <= 5000, 0 < a[i] < 2^31
Example input :
4
1 2 2 3
Example output :
1
Help me :v.
| # | User | Rating |
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| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
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| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
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| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
Find number of unique longest increasing subsequence in array a.
Constraints : 1 <= Lenght of a <= 5000, 0 < a[i] < 2^31
Example input :
4
1 2 2 3
Example output :
1
Help me :v.
| Name |
|---|



Calculate :
dp[i] = longest increasing subsequence that ends at position i.
dp[i] = max(dp[j] + 1) with a[j] < a[i], j < i
cnt[i] = number of unique longest increasing subsequences that end at position i.
If there was no restriction for the subsequences to be unique,
cnt[i] would be equal to sum of (cnt[j] with j < i, a[j] < a[i], dp[j] is maximum).
Notice that when calculating cnt[i], if we have 2 positions k < j < i, with A[j] = A[k] and A[j] < A[i], and both dp[j] and dp[k] are maximum, we only need to add cnt[j] to cnt[i]. (i.e every unique longest subsequence that ends at k, can also end at j.)
So cnt[i] = sum(cnt[j] | j < i, a[j] < a[i], dp[j] is maximum and j = last appearance of A[j] in [1, j]).
Number of total unique longest increasing subsequences will be :
sum of (cnt[pos] such that dp[pos] in maximum, pos = last appearance of A[pos] in [1, N] where N = number of elements).
can you share the question link if possible ?