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[Flynn] WEEK 01 Solutions #302
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/** | ||
* for given size of input nums N, | ||
* | ||
* Time complexity: O(N) | ||
* - iteration: O(N) | ||
* - unorderd_set find method: O(1) on average | ||
* - unorderd_set insert method: O(1) on average | ||
* | ||
* Space complexity: O(N) | ||
*/ | ||
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class Solution { | ||
public: | ||
bool containsDuplicate(vector<int>& nums) { | ||
unordered_set<int> us; | ||
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auto it = nums.begin(); | ||
while (it != nums.end()) { | ||
if (us.find(*it) == us.end()) us.insert(*(it++)); | ||
else return true; | ||
} | ||
return false; | ||
} | ||
}; |
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/** | ||
* For the height H of the given BST, | ||
* | ||
* Time complexity: O(H) at worst | ||
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* | ||
* Space complexity: O(H + K) at worst | ||
* - call stack + additional vector to save nums | ||
*/ | ||
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/** | ||
* Definition for a binary tree node. | ||
* struct TreeNode { | ||
* int val; | ||
* TreeNode *left; | ||
* TreeNode *right; | ||
* TreeNode() : val(0), left(nullptr), right(nullptr) {} | ||
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} | ||
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} | ||
* }; | ||
*/ | ||
class Solution { | ||
public: | ||
void inorder(TreeNode* node, vector<int>& v, int max_size) { | ||
if (!node || v.size() == max_size) return; | ||
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if (node->left) inorder(node->left, v, max_size); | ||
v.push_back(node->val); | ||
if (node->right) inorder(node->right, v, max_size); | ||
} | ||
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int kthSmallest(TreeNode* root, int k) { | ||
vector<int> nums; | ||
inorder(root, nums, k); | ||
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return nums[k - 1]; | ||
} | ||
}; |
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/** | ||
* For given integer N, | ||
* | ||
* Time complexity: O(logN) | ||
* | ||
* Space complexity: O(1) | ||
*/ | ||
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class Solution { | ||
public: | ||
int hammingWeight(int n) { | ||
int res = 0; | ||
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while (n) { | ||
if (n & 1) res++; | ||
n >>= 1; | ||
} | ||
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return res; | ||
} | ||
}; |
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/** | ||
* For the length N of given string s, | ||
* | ||
* Time complexity: O(N^3) | ||
* | ||
* Space complexity: O(1) | ||
*/ | ||
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class Solution { | ||
public: | ||
int countSubstrings(string s) { | ||
int res = 0; | ||
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for (int i = 0; i < s.size(); i++) { | ||
for (int j = i; j < s.size(); j++) { | ||
int start = i, end = j, flag = 0; | ||
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while (start <= end) { | ||
if (s[start] != s[end]) { | ||
flag = 1; | ||
break; | ||
} | ||
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start++; | ||
end--; | ||
} | ||
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if (!flag) res++; | ||
} | ||
} | ||
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return res; | ||
} | ||
}; |
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/** | ||
* For given size of nums N and K, | ||
* | ||
* Time complexity: O(N + MlogM + KlogM) <= O(NlogN) | ||
* - the first iteration: O(N) | ||
* - N times of | ||
* - unordered_map find: O(1) on average | ||
* - unordered_map insert: O(1) on average | ||
* - the second iteration: O(MlogM) such that M is the number of unique numbers | ||
* - M times of | ||
* - priority_queue push: O(logM) | ||
* - the last iteration of making result: O(KlogM) | ||
* - K times of | ||
* - priority_queue pop: O(logM) | ||
* | ||
* Space complexity: O(N) at worst | ||
*/ | ||
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class Solution { | ||
public: | ||
vector<int> topKFrequent(vector<int>& nums, int k) { | ||
vector<int> res; | ||
priority_queue<pair<int, int>> pq; | ||
unorderd_map<int, int> m; | ||
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auto nums_it = nums.begin(); | ||
while (nums_it != nums.end()) { | ||
if (m.find(*nums_it) == m.end()) m.insert({*nums_it, 1}); | ||
else m[*nums_it]++; | ||
nums_it++; | ||
} | ||
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auto m_it = m.begin(); | ||
while (m_it != m.end()) { | ||
pq.push({(*m_it).second, (*m_it).first}); | ||
m_it++; | ||
} | ||
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while (k--) { | ||
res.push_back(pq.top().second); | ||
pq.pop(); | ||
} | ||
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return res; | ||
} | ||
}; |
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C++ 기억이 잘 안났는데 이렇게 적어주시니 편하네요 👍
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c++로 푸시는 분들이 많이 안 계신 것 같아서, python이나 ts를 사용해야 하나 고민도 되네요 ㅎㅎ 감사합니다