Huahua’s Tech Road

花花酱 LeetCode 1936. Add Minimum Number of Rungs

By zxi on December 31, 2021

You are given a strictly increasing integer array rungs that represents the height of rungs on a ladder. You are currently on the floor at height 0, and you want to reach the last rung.

You are also given an integer dist. You can only climb to the next highest rung if the distance between where you are currently at (the floor or on a rung) and the next rung is at most dist. You are able to insert rungs at any positive integer height if a rung is not already there.

Return the minimum number of rungs that must be added to the ladder in order for you to climb to the last rung.

Example 1:

Input: rungs = [1,3,5,10], dist = 2
Output: 2
Explanation:
You currently cannot reach the last rung.
Add rungs at heights 7 and 8 to climb this ladder. 
The ladder will now have rungs at [1,3,5,7,8,10].

Example 2:

Input: rungs = [3,6,8,10], dist = 3
Output: 0
Explanation:
This ladder can be climbed without adding additional rungs.

Example 3:

Input: rungs = [3,4,6,7], dist = 2
Output: 1
Explanation:
You currently cannot reach the first rung from the ground.
Add a rung at height 1 to climb this ladder.
The ladder will now have rungs at [1,3,4,6,7].

Constraints:

1 <= rungs.length <= 10⁵
1 <= rungs[i] <= 10⁹
1 <= dist <= 10⁹
rungs is strictly increasing.

Solution: Math

Check two consecutive rungs, if their diff is > dist, we need insert (diff – 1) / dist rungs in between.
ex1 5 -> 11, diff = 6, dist = 2, (diff – 1) / dist = (6 – 1) / 2 = 2. => 5, 7, 9, 11.
ex2 0 -> 3, diff = 3, dist = 1, (diff – 1) / dist = (3 – 1) / 1 = 2 => 0, 1, 2, 3

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

C++

// Author: Huahua

class Solution {

public:

int addRungs(vector<int>& rungs, int dist) {

int ans = 0;

for (size_t i = 0; i < rungs.size(); ++i)

ans += (rungs[i] - (i ? rungs[i - 1] : 0) - 1) / dist;

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def addRungs(self, rungs: List[int], dist: int) -> int:

return sum((r - 1 - (rungs[i - 1] if i else 0)) // dist for i, r in enumerate(rungs))

花花酱 LeetCode 1935. Maximum Number of Words You Can Type

By zxi on December 31, 2021

There is a malfunctioning keyboard where some letter keys do not work. All other keys on the keyboard work properly.

Given a string text of words separated by a single space (no leading or trailing spaces) and a string brokenLetters of all distinct letter keys that are broken, return the number of words in text you can fully type using this keyboard.

Example 1:

Input: text = "hello world", brokenLetters = "ad"
Output: 1
Explanation: We cannot type "world" because the 'd' key is broken.

Example 2:

Input: text = "leet code", brokenLetters = "lt"
Output: 1
Explanation: We cannot type "leet" because the 'l' and 't' keys are broken.

Example 3:

Input: text = "leet code", brokenLetters = "e"
Output: 0
Explanation: We cannot type either word because the 'e' key is broken.

Constraints:

1 <= text.length <= 10⁴
0 <= brokenLetters.length <= 26
text consists of words separated by a single space without any leading or trailing spaces.
Each word only consists of lowercase English letters.
brokenLetters consists of distinct lowercase English letters.

Solution: Hashset / bitset

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

C++

// Author: Huahua

class Solution {

public:

int canBeTypedWords(string text, string brokenLetters) {

const int n = text.size();

bitset<26> b;

for (char c : brokenLetters) b.set(c - 'a');

int ans = 0;

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

if (i == n || text[i] == ' ') {

if (!broken) ++ans;

broken = 0;

} else {

broken |= b[text[i] - 'a'];

}

return ans;

}

};

花花酱 LeetCode 1930. Unique Length-3 Palindromic Subsequences

By zxi on December 30, 2021

Given a string s, return the number of unique palindromes of length three that are a subsequence of s.

Note that even if there are multiple ways to obtain the same subsequence, it is still only counted once.

A palindrome is a string that reads the same forwards and backwards.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

For example, "ace" is a subsequence of "abcde".

Example 1:

Input: s = "aabca"
Output: 3
Explanation: The 3 palindromic subsequences of length 3 are:
- "aba" (subsequence of "aabca")
- "aaa" (subsequence of "aabca")
- "aca" (subsequence of "aabca")

Example 2:

Input: s = "adc"
Output: 0
Explanation: There are no palindromic subsequences of length 3 in "adc".

Example 3:

Input: s = "bbcbaba"
Output: 4
Explanation: The 4 palindromic subsequences of length 3 are:
- "bbb" (subsequence of "bbcbaba")
- "bcb" (subsequence of "bbcbaba")
- "bab" (subsequence of "bbcbaba")
- "aba" (subsequence of "bbcbaba")

Constraints:

3 <= s.length <= 10⁵
s consists of only lowercase English letters.

Solution: Enumerate first character of a palindrome

For a length 3 palindrome, we just need to enumerate the first character c.
We found the first and last occurrence of c in original string and scan the middle part to see how many unique characters there.

e.g. aabca
Enumerate from a to z, looking for a*a, b*b, …, z*z.
For a*a, aabca, we found first and last a, in between is abc, which has 3 unique letters.
We can use a hastable or a bitset to track unique letters.

Time complexity: O(26*n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countPalindromicSubsequence(string s) {

auto count = [&](char c) -> int {

int l = s.find_first_of(c);

int r = s.find_last_of(c);

if (l == string::npos || r == string::npos) return 0;

bitset<26> m;

for (int i = l + 1; i < r; ++i)

m.set(s[i] - 'a');

return m.count();

};

int ans = 0;

for (char c = 'a'; c <= 'z'; ++c)

ans += count(c);

return ans;

}

};

花花酱 LeetCode 1929. Concatenation of Array

By zxi on December 30, 2021

Given an integer array nums of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed).

Specifically, ans is the concatenation of two nums arrays.

Return the array ans.

Example 1:

Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]]
- ans = [1,2,1,1,2,1]

Example 2:

Input: nums = [1,3,2,1]
Output: [1,3,2,1,1,3,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]

Constraints:

n == nums.length
1 <= n <= 1000
1 <= nums[i] <= 1000

Solution: Pre-allocation

Pre-allocate an array of length 2 * n.
ans[i] = nums[i % n]

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

C++

// Author: Huahua

class Solution {

public:

vector<int> getConcatenation(vector<int>& nums) {

const int n = nums.size();

vector<int> ans(2 * n);

for (int i = 0; i < 2 * n; ++i)

ans[i] = nums[i % n];

return ans;

}

};

花花酱 LeetCode 1926. Nearest Exit from Entrance in Maze

By zxi on December 30, 2021

You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+'). You are also given the entrance of the maze, where entrance = [entrance_row, entrance_col] denotes the row and column of the cell you are initially standing at.

In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance. An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

Example 1:

Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
- You can reach [1,0] by moving 2 steps left.
- You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.

Example 2:

Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
- You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.

Example 3:

Input: maze = [[".","+"]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.

Constraints:

maze.length == m
maze[i].length == n
1 <= m, n <= 100
maze[i][j] is either '.' or '+'.
entrance.length == 2
0 <= entrance_row < m
0 <= entrance_col < n
entrance will always be an empty cell.

Solution: BFS

Use BFS to find the shortest path. We can re-use the board for visited array.

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

C++

// Author: Huahua

class Solution {

public:

int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {

const int m = maze.size();

const int n = maze[0].size();

const vector<int> dirs{0, -1, 0, 1, 0};

queue<pair<int, int>> q;

q.emplace(entrance[1], entrance[0]);

for (int steps = 0; !q.empty(); ++steps) {

for (int s = q.size(); s; --s) {

const auto [x, y] = q.front(); q.pop();

if (x == 0 || x == n - 1 || y == 0 || y == m - 1)

if (x != entrance[1] || y != entrance[0])

return steps;

for (int i = 0; i < 4; ++i) {

const int tx = x + dirs[i];

const int ty = y + dirs[i + 1];

if (tx < 0 || tx >= n || ty < 0 || ty >= m || maze[ty][tx] != '.') continue;

maze[ty][tx] = '*';

q.emplace(tx, ty);

}

return -1;

}

};

Huahua's Tech Road

花花酱 LeetCode 1936. Add Minimum Number of Rungs

Solution: Math

C++

Python3

花花酱 LeetCode 1935. Maximum Number of Words You Can Type

Solution: Hashset / bitset

C++

花花酱 LeetCode 1930. Unique Length-3 Palindromic Subsequences

Solution: Enumerate first character of a palindrome

C++

花花酱 LeetCode 1929. Concatenation of Array

Solution: Pre-allocation

C++

花花酱 LeetCode 1926. Nearest Exit from Entrance in Maze

Solution: BFS

C++