Huahua’s Tech Road

花花酱 LeetCode 1305. All Elements in Two Binary Search Trees

By zxi on December 29, 2019

Given two binary search trees root1 and root2.

Return a list containing all the integers from both trees sorted in ascending order.

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]

Example 2:

Input: root1 = [0,-10,10], root2 = [5,1,7,0,2]
Output: [-10,0,0,1,2,5,7,10]

Example 3:

Input: root1 = [], root2 = [5,1,7,0,2]
Output: [0,1,2,5,7]

Example 4:

Input: root1 = [0,-10,10], root2 = []
Output: [-10,0,10]

Example 5:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]

Constraints:

Each tree has at most 5000 nodes.
Each node’s value is between [-10^5, 10^5].

Solution: Inorder traversal + Merge Sort

Time complexity: O(t1 + t2)
Space complexity: O(t1 + t2)

C++

// Author: Huahua

class Solution {

public:

vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {

function<void(TreeNode*, vector<int>&)> inorder = [&](TreeNode* root, vector<int>& t) {

if (!root) return;

inorder(root->left, t);

t.push_back(root->val);

inorder(root->right, t);

};

vector<int> t1;

vector<int> t2;

inorder(root1, t1);

inorder(root2, t2);

vector<int> m;

int i = 0;

int j = 0;

while (m.size() != t1.size() + t2.size()) {

if (j == t2.size()) m.push_back(t1[i++]);

else if (i == t1.size()) m.push_back(t2[j++]);

else m.push_back(t1[i] < t2[j] ? t1[i++] : t2[j++]);

}

return m;

}

};

C++/STL

// Author: Huahua

class Solution {

public:

vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {

function<void(TreeNode*, vector<int>&)> inorder = [&](TreeNode* root, vector<int>& t) {

if (!root) return;

inorder(root->left, t);

t.push_back(root->val);

inorder(root->right, t);

};

vector<int> t1;

vector<int> t2;

inorder(root1, t1);

inorder(root2, t2);

vector<int> m;

std::merge(begin(t1), end(t1), begin(t2), end(t2), back_inserter(m));

return m;

}

};

花花酱 LeetCode 1304. Find N Unique Integers Sum up to Zero

By zxi on December 29, 2019

Given an integer n, return any array containing n unique integers such that they add up to 0.

Example 1:

Input: n = 5
Output: [-7,-1,1,3,4]
Explanation: These arrays also are accepted [-5,-1,1,2,3] , [-3,-1,2,-2,4].

Example 2:

Input: n = 3
Output: [-1,0,1]

Example 3:

Input: n = 1
Output: [0]

Constraints:

1 <= n <= 1000

Solution: Generation

Use numbers from {-n/2, … n/2} + {0 if n is odd}

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

C++

// Author: Huahua

class Solution {

public:

vector<int> sumZero(int n) {

vector<int> ans;

for (int i = 1; i <= n / 2; ++i) {

ans.push_back(i);

ans.push_back(-i);

}

if (ans.size() != n) ans.push_back(0);

return ans;

}

};

Binary search template 二分搜索模板

By zxi on December 29, 2019

Python

# Author: Huahua

# Returns the smallest m in [l, r),

# s.t. cond(m) == True

# If not found returns r.

def binarySearch(l, r)

while l < r:

m = l + (r - l) // 2

if cond(m):

r = m

else

l = m + 1

return l

C++

// Author: Huahua

// Returns the smallest m in [l, r),

// s.t. cond(m) == True

// If not found returns r.

int binarySearch(int l, int r) {

while (l < r) {

int m = l + (r - l) / 2;

if (cond(m)) r = m;

else l = m + 1;

}

return l;

}

花花酱 LeetCode 1301. Number of Paths with Max Score

By zxi on December 28, 2019

You are given a square board of characters. You can move on the board starting at the bottom right square marked with the character 'S'.

You need to reach the top left square marked with the character 'E'. The rest of the squares are labeled either with a numeric character 1, 2, ..., 9 or with an obstacle 'X'. In one move you can go up, left or up-left (diagonally) only if there is no obstacle there.

Return a list of two integers: the first integer is the maximum sum of numeric characters you can collect, and the second is the number of such paths that you can take to get that maximum sum, taken modulo 10^9 + 7.

In case there is no path, return [0, 0].

Example 1:

Input: board = ["E23","2X2","12S"]
Output: [7,1]

Example 2:

Input: board = ["E12","1X1","21S"]
Output: [4,2]

Example 3:

Input: board = ["E11","XXX","11S"]
Output: [0,0]

Constraints:

2 <= board.length == board[i].length <= 100

Solution: DP

dp[i][j] := max score when reach (j, i)
count[i][j] := path to reach (j, i) with max score

m = max(dp[i + 1][j], dp[i][j+1], dp[i+1][j+1])
dp[i][j] = board[i][j] + m
count[i][j] += count[i+1][j] if dp[i+1][j] == m
count[i][j] += count[i][j+1] if dp[i][j+1] == m
count[i][j] += count[i+1][j+1] if dp[i+1][j+1] == m

Time complexity: O(n^2)
Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

vector<int> pathsWithMaxScore(vector<string>& board) {

constexpr int kMod = 1e9 + 7;

const int n = board.size();

vector<vector<int>> dp(n + 1, vector<int>(n + 1));

vector<vector<int>> cc(n + 1, vector<int>(n + 1, 0));

board[n - 1][n - 1] = board[0][0] = '0';

cc[n - 1][n - 1] = 1;

for (int i = n - 1; i >= 0; --i)

for (int j = n - 1; j >= 0; --j) {

if (board[i][j] != 'X') {

int m = max({dp[i + 1][j], dp[i][j + 1], dp[i + 1][j + 1]});

dp[i][j] = (board[i][j] - '0') + m;

if (dp[i + 1][j] == m)

cc[i][j] = (cc[i][j] + cc[i + 1][j]) % kMod;

if (dp[i][j + 1] == m)

cc[i][j] = (cc[i][j] + cc[i][j + 1]) % kMod;

if (dp[i + 1][j + 1] == m)

cc[i][j] = (cc[i][j] + cc[i + 1][j + 1]) % kMod;

}

return {cc[0][0] ? dp[0][0] : 0, cc[0][0]};

}

};

花花酱 LeetCode 1299. Replace Elements with Greatest Element on Right Side

By zxi on December 28, 2019

Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1.

After doing so, return the array.

Example 1:

Input: arr = [17,18,5,4,6,1]
Output: [18,6,6,6,1,-1]

Constraints:

1 <= arr.length <= 10^4
1 <= arr[i] <= 10^5

Solution: Scan from right to left

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

C++

// Author: Huahua

class Solution {

public:

vector<int> replaceElements(vector<int>& arr) {

int g = -1;

for (int i = arr.size() - 1; i >= 0; --i)

g = max(g, exchange(arr[i], g));

return arr;

}

};

Huahua's Tech Road

花花酱 LeetCode 1305. All Elements in Two Binary Search Trees

Solution: Inorder traversal + Merge Sort

C++

C++/STL

花花酱 LeetCode 1304. Find N Unique Integers Sum up to Zero

Solution: Generation

C++

Binary search template 二分搜索模板

Python

C++

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花花酱 LeetCode 1299. Replace Elements with Greatest Element on Right Side

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