Medium
Find the Maximum Length of a Good Subsequence I — C++
Full explanation · Time O(n * k) · Space O(n * k)
// Time: O(n * k)
// Space: O(n * k)
// dp
class Solution {
public:
int maximumLength(vector<int>& nums, int k) {
unordered_map<int, int> lookup;
for (const auto& x : unordered_set<int>(cbegin(nums), cend(nums))) {
lookup[x] = size(lookup);
}
vector<vector<int>> dp(k + 1, vector<int>(size(lookup)));
vector<int> result(k + 1);
for (auto x : nums) {
x = lookup[x];
for (int i = k; i >= 0; --i) {
dp[i][x] = max(dp[i][x], i - 1 >= 0 ? result[i - 1] : 0) + 1;
result[i] = max(result[i], dp[i][x]);
}
}
return result[k];
}
};
// Time: O(n * k)
// Space: O(n * k)
// dp
class Solution2 {
public:
int maximumLength(vector<int>& nums, int k) {
vector<unordered_map<int, int>> dp(k + 1);
vector<int> result(k + 1);
for (const auto& x : nums) {
for (int i = k; i >= 0; --i) {
dp[i][x] = max(dp[i][x], i - 1 >= 0 ? result[i - 1] : 0) + 1;
result[i] = max(result[i], dp[i][x]);
}
}
return result[k];
}
};
// Time: O(n^2 * k)
// Space: O(n * k)
// dp
class Solution3 {
public:
int maximumLength(vector<int>& nums, int k) {
vector<vector<int>> dp(size(nums), vector<int>(k + 1));
int result = 0;
for (int i = 0; i < size(nums); ++i) {
dp[i][0] = 1;
for (int l = 0; l <= k; ++l) {
for (int j = 0; j < i; ++j) {
dp[i][l] = max({dp[i][l], nums[j] == nums[i] ? dp[j][l]+ 1 : 1, l - 1 >= 0 ? dp[j][l - 1] + 1 : 1});
}
result = max(result, dp[i][l]);
}
}
return result;
}
};