Huahua’s Tech Road

花花酱 LeetCode 173. Binary Search Tree Iterator

By zxi on November 28, 2021

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output

[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); bSTIterator.next(); // return 3 bSTIterator.next(); // return 7 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 9 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 15 bSTIterator.hasNext(); // return True bSTIterator.next(); // return 20 bSTIterator.hasNext(); // return False

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
0 <= Node.val <= 10⁶
At most 10⁵ calls will be made to hasNext, and next.

Follow up:

Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

Solution: In-order traversal using a stack

Use a stack to simulate in-order traversal.

Each next, we walk to the left most (smallest) node and push all the nodes along the path to the stack.

Then pop the top one t as return val, our next node to explore is the right child of t.

Time complexity: amortized O(1) for next() call.
Space complexity: O(n)

C++

// Author: Huahua

/**

* 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 BSTIterator {

public:

BSTIterator(TreeNode* root) : root_(root) {}

int next() {

while (root_) {

s_.push(root_);

root_ = root_->left;

}

TreeNode* t = s_.top(); s_.pop();

root_ = t->right;

return t->val;

}

bool hasNext() {

return (root_ || !s_.empty());

}

private:

TreeNode* root_;

stack<TreeNode*> s_;

};

/**

* Your BSTIterator object will be instantiated and called as such:

* BSTIterator* obj = new BSTIterator(root);

* int param_1 = obj->next();

* bool param_2 = obj->hasNext();

花花酱 LeetCode 187. Repeated DNA Sequences

By zxi on November 28, 2021

The DNA sequence is composed of a series of nucleotides abbreviated as 'A', 'C', 'G', and 'T'.

For example, "ACGAATTCCG" is a DNA sequence.

When studying DNA, it is useful to identify repeated sequences within the DNA.

Given a string s that represents a DNA sequence, return all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule. You may return the answer in any order.

Example 1:

Input: s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT"
Output: ["AAAAACCCCC","CCCCCAAAAA"]

Example 2:

Input: s = "AAAAAAAAAAAAA"
Output: ["AAAAAAAAAA"]

Constraints:

1 <= s.length <= 10⁵
s[i] is either 'A', 'C', 'G', or 'T'.

Solution: Hashtable

Store each subsequence into the hashtable, add it into the answer array when it appears for the second time.

Time complexity: O(n*l)
Space complexity: O(n*l) -> O(n) / string_view

C++

// Author: Huahua

class Solution {

public:

vector<string> findRepeatedDnaSequences(string_view s) {

constexpr int kLen = 10;

const int n = s.length();

unordered_map<string_view, int> m;

vector<string> ans;

for (int i = 0; i + kLen <= n; ++i)

if (++m[s.substr(i, kLen)] == 2)

ans.emplace_back(s.substr(i, kLen));

return ans;

}

};

Optimization

There are 4 type of letters, each can be encoded into 2 bits. We can represent the 10-letter-long string using 20 lowest bit of a int32. We can use int as key for the hashtable.

A -> 00
C -> 01
G -> 10
T -> 11

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

vector<string> findRepeatedDnaSequences(string s) {

constexpr int kLen = 10;

constexpr int mask = (1 << (2 * kLen)) -1;

const int n = s.length();

unordered_map<int, int> m;

array<int, 128> km;

km['A'] = 0;

km['C'] = 1;

km['G'] = 2;

km['T'] = 3;

vector<string> ans;

for(int i = 0, key = 0; i < n; ++i) {

key = ((key << 2) & mask) | km[s[i]];

if (i < kLen - 1) continue;

if (++m[key] == 2)

ans.push_back(s.substr(i - kLen + 1, kLen));

}

return ans;

}

};

花花酱 LeetCode 188. Best Time to Buy and Sell Stock IV

By zxi on November 28, 2021

You are given an integer array prices where prices[i] is the price of a given stock on the i^th day, and an integer k.

Find the maximum profit you can achieve. You may complete at most k transactions.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: k = 2, prices = [2,4,1]
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4. Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

Constraints:

0 <= k <= 100
0 <= prices.length <= 1000
0 <= prices[i] <= 1000

Solution: DP

profit[i][j] := max profit by making up to j sells in first i days. (Do not hold any share)
balance[i][j] := max balance by making at to up j buys in first i days. (Most hold a share)

balance[i][j] = max(balance[i-1][j], profit[i-1][j-1] – prices[i]) // do nothing or buy at i-th day.
profit[i][j] = max(profit[i-1][j], balance[i-1][j] + prices[i]) // do nothing or sell at i-th day.

ans = profit[n-1][k]

Time complexity: O(n*k)
Space complexity: O(k)

C++

// Author: Huahua

class Solution {

public:

int maxProfit(int k, vector<int>& prices) {

const int n = prices.size();

vector<int> balance(k + 1, INT_MIN);

vector<int> profit(k + 1, 0);

for (int i = 0; i < n; ++i)

for (int j = 1; j <= k; ++j) {

balance[j] = max(balance[j], profit[j - 1] - prices[i]);

profit[j] = max(profit[j], balance[j] + prices[i]);

}

return profit[k];

}

};

花花酱 LeetCode 199. Binary Tree Right Side View

By zxi on November 28, 2021

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example 1:

Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

Input: root = [1,null,3]
Output: [1,3]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Solution: Pre-order traversal

By using pre-order traversal, the right most node will be the last one to visit in each level.

Time complexity: O(n)
Space complexity: O(|height|)

C++

// Author: Huahua

class Solution {

public:

vector<int> rightSideView(TreeNode* root) {

vector<int> ans;

function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int d) {

if (!root) return;

if (ans.size() < d + 1) ans.push_back(0);

ans[d] = root->val;

dfs(root->left, d + 1);

dfs(root->right, d + 1);

};

dfs(root, 0);

return ans;

}

};

花花酱 LeetCode 191. Number of 1 Bits

By zxi on November 28, 2021

Write a function that takes an unsigned integer and returns the number of ‘1’ bits it has (also known as the Hamming weight).

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

Example 1:

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Constraints:

The input must be a binary string of length 32.

Follow up: If this function is called many times, how would you optimize it?

Solution: Bit Operation

Use n & 1 to get the lowest bit of n.
Use n >>= 1 to right shift n for 1 bit, e.g. removing the last bit.

Time complexity: O(logn)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int hammingWeight(uint32_t n) {

int ans = 0;

while (n) {

ans += n & 1;

n >>= 1;

}

return ans;

}

};

Huahua's Tech Road

花花酱 LeetCode 173. Binary Search Tree Iterator

Solution: In-order traversal using a stack

C++

花花酱 LeetCode 187. Repeated DNA Sequences

Solution: Hashtable

C++

Optimization

C++

花花酱 LeetCode 188. Best Time to Buy and Sell Stock IV

Solution: DP

C++

Related Problems

花花酱 LeetCode 199. Binary Tree Right Side View

Solution: Pre-order traversal

C++

花花酱 LeetCode 191. Number of 1 Bits

Solution: Bit Operation

C++