Given an integer array nums, return the number of longest increasing subsequences.
Notice that the sequence has to be strictly increasing.
Example 1:
Input: nums = [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: nums = [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.
Solution: (Using LIS and Count)
class Solution
{
public:
int findNumberOfLIS(vector<int> &nums)
{
int n = nums.size();
vector<int> dp(n, 1);
vector<int> count(n, 1);
int res = 1;
for (int i = 1; i < n; i++)
{
for (int j = 0; j < i; j++)
{
if (nums[j] < nums[i] && dp[j] >= dp[i])
{
dp[i] = max(dp[i], dp[j] + 1);
count[i] = count[j];
}
else if (nums[j] < nums[i] && dp[j] + 1 == dp[i])
{
dp[i] = max(dp[i], dp[j] + 1);
count[i] += count[j];
}
}
res = max(res, dp[i]);
}
int ct = 0;
for (int i = 0; i < n; i++)
{
if (dp[i] == res)
{
ct += count[i];
}
}
return ct;
}
};