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花花酱 LeetCode 1443. Minimum Time to Collect All Apples in a Tree

By zxi on May 10, 2020

Given an undirected tree consisting of n vertices numbered from 0 to n-1, which has some apples in their vertices. You spend 1 second to walk over one edge of the tree. Return the minimum time in seconds you have to spend in order to collect all apples in the tree starting at vertex 0 and coming back to this vertex.

The edges of the undirected tree are given in the array edges, where edges[i] = [from_i, to_i] means that exists an edge connecting the vertices from_i and to_i. Additionally, there is a boolean array hasApple, where hasApple[i] = true means that vertex i has an apple, otherwise, it does not have any apple.

Example 1:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,true,true,false]
Output: 8 
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,true,false,false,true,false]
Output: 6
Explanation: The figure above represents the given tree where red vertices have an apple. One optimal path to collect all apples is shown by the green arrows.

Example 3:

Input: n = 7, edges = [[0,1],[0,2],[1,4],[1,5],[2,3],[2,6]], hasApple = [false,false,false,false,false,false,false]
Output: 0

Constraints:

1 <= n <= 10^5
edges.length == n-1
edges[i].length == 2
0 <= from_i, to_i <= n-1
from_i < to_i
hasApple.length == n

Solution: DFS

Build the graph (tree) and DFS.

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

C++

// Author: Huahua

class Solution {

public:

int minTime(int n, vector<vector<int>>& edges, vector<bool>& hasApple) {

vector<vector<int>> g(n);

for (const auto& e : edges)

g[e[0]].push_back(e[1]);

function<int(int)> dfs = [&] (int cur) {

int total = 0;

for (int nxt : g[cur]) {

int cost = dfs(nxt);

if (cost > 0 || hasApple[nxt]) total += 2 + cost;

}

return total;

};

return dfs(0);

}

};

if edge is not ordered

C++

// Author: Huahua

class Solution {

public:

int minTime(int n, vector<vector<int>>& edges,

vector<bool>& hasApple) {

vector<vector<int>> g(n);

for (const auto& e : edges) {

g[e[0]].push_back(e[1]);

g[e[1]].push_back(e[0]);

}

vector<int> seen(n);

function<int(int)> dfs = [&] (int cur) {

seen[cur] = 1;

int total = 0;

for (int child : g[cur]) {

if (seen[child]) continue;

int cost = dfs(child);

if (cost > 0 || hasApple[child])

total += 2 + cost;

}

return total;

};

return dfs(0);

}

};

花花酱 LeetCode 1281. Subtract the Product and Sum of Digits of an Integer

By zxi on December 8, 2019

Given an integer number n, return the difference between the product of its digits and the sum of its digits.

Example 1:

Input: n = 234

Output: 15

Explanation:

Product of digits = 2 * 3 * 4 = 24

Sum of digits = 2 + 3 + 4 = 9

Result = 24 - 9 = 15

Example 2:

Input: n = 4421

Output: 21

Explanation:

Product of digits = 4 * 4 * 2 * 1 = 32

Sum of digits = 4 + 4 + 2 + 1 = 11

Result = 32 - 11 = 21

Constraints:

1 <= n <= 10^5

Solution: Simulation

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

C++

// Author: Huahua

class Solution {

public:

int subtractProductAndSum(int n) {

int p = 1;

int s = 0;

while (n) {

const int d = n % 10;

p *= d;

s += d;

n /= 10;

}

return p - s;

}

};

花花酱 LeetCode 59. Spiral Matrix II

By zxi on September 29, 2019

Given a positive integer n, generate a square matrix filled with elements from 1 to n² in spiral order.

Example:

Input: 3
Output:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

Solution: Simulation

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

C++

// Author: Huahua

class Solution {

public:

vector<vector<int> > generateMatrix(int n) {

vector<vector<int>> ans(n, vector<int>(n));

int dir = 2, x = 0, y = 0;

int max_x = n - 1, max_y = n - 1;

int min_x = 0, min_y = 1;

int i = 0;

while (++i <= n*n) {

ans[y][x] = i;

switch (dir) {

// up

case 1: if (--y == min_y) { dir = 2; ++min_y; } break;

// right

case 2: if (++x == max_x) { dir = 3; --max_x; } break;

// down

case 3: if (++y == max_y) { dir = 4; --max_y; } break;

// left

case 4: if (--x == min_x) { dir = 1; ++min_x; } break;

}

return ans;

}

};

花花酱 LeetCode 226. Invert Binary Tree

By zxi on July 22, 2019

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

Solution: Recursion

Recursive invert the left and right subtrees and swap them.

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

C++

// Author: Huahua

class Solution {

public:

TreeNode* invertTree(TreeNode* root) {

if (!root) return nullptr;

TreeNode* tmp = root->right;

root->right = invertTree(root->left);

root->left = invertTree(tmp);

return root;

}

};

Python3

# Author: Huahua

class Solution:

def invertTree(self, root: TreeNode) -> TreeNode:

if not root: return None

root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)

return root

花花酱 LeetCode 1124. Longest Well-Performing Interval

By zxi on July 14, 2019

We are given hours, a list of the number of hours worked per day for a given employee.

A day is considered to be a tiring day if and only if the number of hours worked is (strictly) greater than 8.

A well-performing interval is an interval of days for which the number of tiring days is strictly larger than the number of non-tiring days.

Return the length of the longest well-performing interval.

Example 1:

Input: hours = [9,9,6,0,6,6,9]
Output: 3
Explanation: The longest well-performing interval is [9,9,6].

Constraints:

1 <= hours.length <= 10000
0 <= hours[i] <= 16

Solution: Target sum == 1

This problem can be reduced to find the longest subarray with sum of 1, or the longest subarray starting from left-most that has a sum greater than 0.

e.g. [6, 6, (6, 9, 9), 6, 6] => sum = 1
e.g. [9, 9, 9] => sum = 3

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

C++

// Author: Huahua

class Solution {

public:

int longestWPI(vector<int>& hours) {

const int n = hours.size();

for (int& v : hours) v = v > 8 ? 1 : -1;

int s = 0;

unordered_map<int, int> idx;

int ans = 0;

for (int i = 0; i < hours.size(); ++i) {

s += hours[i];

if (s > 0)

ans = i + 1;

if (!idx.count(s))

idx[s] = i;

if (idx.count(s - 1))

ans = max(ans, i - idx[s - 1]);

}

return ans;

}

};

Posts published in “Uncategorized”

花花酱 LeetCode 1443. Minimum Time to Collect All Apples in a Tree

Solution: DFS

C++

if edge is not ordered

C++

花花酱 LeetCode 1281. Subtract the Product and Sum of Digits of an Integer

Solution: Simulation

C++

花花酱 LeetCode 59. Spiral Matrix II

Solution: Simulation

C++

花花酱 LeetCode 226. Invert Binary Tree

Solution: Recursion

C++

Python3

花花酱 LeetCode 1124. Longest Well-Performing Interval

Solution: Target sum == 1

C++