Medium
Number of Longest Increasing Subsequence — C++
Full explanation · Time O(n^2) · Space O(n)
// Time: O(n^2)
// Space: O(n)
class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
auto result = 0, max_len = 0;
vector<pair<int, int>> dp(nums.size(), {1, 1}); // {length, number} pair
for (int i = 0; i < nums.size(); ++i) {
for (int j = 0; j < i; ++j) {
if (nums[i] > nums[j]) {
if (dp[i].first == dp[j].first + 1) {
dp[i].second += dp[j].second;
} else if (dp[i].first < dp[j].first + 1) {
dp[i] = {dp[j].first + 1, dp[j].second};
}
}
}
if (max_len == dp[i].first) {
result += dp[i].second;
} else if (max_len < dp[i].first) {
max_len = dp[i].first;
result = dp[i].second;
}
}
return result;
}
};