Posts tagged as “math”

花花酱 LeetCode 1974. Minimum Time to Type Word Using Special Typewriter

By zxi on December 31, 2021

There is a special typewriter with lowercase English letters 'a' to 'z' arranged in a circle with a pointer. A character can only be typed if the pointer is pointing to that character. The pointer is initially pointing to the character 'a'.

Each second, you may perform one of the following operations:

Move the pointer one character counterclockwise or clockwise.
Type the character the pointer is currently on.

Given a string word, return the minimum number of seconds to type out the characters in word.

Example 1:

Input: word = "abc"
Output: 5
Explanation: 
The characters are printed as follows:
- Type the character 'a' in 1 second since the pointer is initially on 'a'.
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer clockwise to 'c' in 1 second.
- Type the character 'c' in 1 second.

Example 2:

Input: word = "bza"
Output: 7
Explanation:
The characters are printed as follows:
- Move the pointer clockwise to 'b' in 1 second.
- Type the character 'b' in 1 second.
- Move the pointer counterclockwise to 'z' in 2 seconds.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'a' in 1 second.
- Type the character 'a' in 1 second.

Example 3:

Input: word = "zjpc"
Output: 34
Explanation:
The characters are printed as follows:
- Move the pointer counterclockwise to 'z' in 1 second.
- Type the character 'z' in 1 second.
- Move the pointer clockwise to 'j' in 10 seconds.
- Type the character 'j' in 1 second.
- Move the pointer clockwise to 'p' in 6 seconds.
- Type the character 'p' in 1 second.
- Move the pointer counterclockwise to 'c' in 13 seconds.
- Type the character 'c' in 1 second.

Constraints:

1 <= word.length <= 100
word consists of lowercase English letters.

Solution: Clockwise or Counter-clockwise?

For each pair of (prev, curr), choose the shortest distance.
One is abs(p – c), another is 26 – abs(p – c).
Don’t forget to add 1 for typing itself.

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

C++

// Author: Huahua

class Solution {

public:

int minTimeToType(string word) {

int ans = 0;

char p = 'a';

for (char c : word) {

ans += 1 + min(abs(c - p), 26 - abs(c - p));

p = c;

}

return ans;

}

};

花花酱 LeetCode 1953. Maximum Number of Weeks for Which You Can Work

By zxi on December 31, 2021

There are n projects numbered from 0 to n - 1. You are given an integer array milestones where each milestones[i] denotes the number of milestones the i^th project has.

You can work on the projects following these two rules:

Every week, you will finish exactly one milestone of one project. You must work every week.
You cannot work on two milestones from the same project for two consecutive weeks.

Once all the milestones of all the projects are finished, or if the only milestones that you can work on will cause you to violate the above rules, you will stop working. Note that you may not be able to finish every project’s milestones due to these constraints.

Return the maximum number of weeks you would be able to work on the projects without violating the rules mentioned above.

Example 1:

Input: milestones = [1,2,3]
Output: 6
Explanation: One possible scenario is:
- During the 1^st week, you will work on a milestone of project 0.
- During the 2^nd week, you will work on a milestone of project 2.
- During the 3^rd week, you will work on a milestone of project 1.
- During the 4^th week, you will work on a milestone of project 2.
- During the 5^th week, you will work on a milestone of project 1.
- During the 6^th week, you will work on a milestone of project 2.
The total number of weeks is 6.

Example 2:

Input: milestones = [5,2,1]
Output: 7
Explanation: One possible scenario is:
- During the 1^st week, you will work on a milestone of project 0.
- During the 2^nd week, you will work on a milestone of project 1.
- During the 3^rd week, you will work on a milestone of project 0.
- During the 4^th week, you will work on a milestone of project 1.
- During the 5^th week, you will work on a milestone of project 0.
- During the 6^th week, you will work on a milestone of project 2.
- During the 7^th week, you will work on a milestone of project 0.
The total number of weeks is 7.
Note that you cannot work on the last milestone of project 0 on 8^th week because it would violate the rules.
Thus, one milestone in project 0 will remain unfinished.

Constraints:

n == milestones.length
1 <= n <= 10⁵
1 <= milestones[i] <= 10⁹

Solution: Math

Let x be the longest project.

Case 1: x > sum of the rest.

Obviously, we cannot finish it.
The best we can do is : [x, a}, {x, b}, {x, c}, …., {x, z}, x.
Ans = 2 * rest + 1

Case 2: x <= sum of the rest.

We can finish all the projects by alternating them properly.
Ans = sum

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

C++

// Author: Huahua

class Solution {

public:

long long numberOfWeeks(vector<int>& milestones) {

long long sum = accumulate(begin(milestones), end(milestones), 0LL);

long long longest = *max_element(begin(milestones), end(milestones));

return min(sum, 2 * (sum - longest) + 1);

}

};

花花酱 LeetCode 1952. Three Divisors

By zxi on December 31, 2021

Given an integer n, return true if n has exactly three positive divisors. Otherwise, return false.

An integer m is a divisor of n if there exists an integer k such that n = k * m.

Example 1:

Input: n = 2
Output: false
Explantion: 2 has only two divisors: 1 and 2.

Example 2:

Input: n = 4
Output: true
Explantion: 4 has three divisors: 1, 2, and 4.

Constraints:

1 <= n <= 10⁴

Solution: Enumerate divisors.

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

C++

// Author: Huahua

class Solution {

public:

bool isThree(int n) {

int c = 0;

for (int i = 1; i <= n; ++i)

if (n % i == 0) ++c;

return c == 3;

}

};

Optimization

Only need to enumerate divisors up to sqrt(n). Special handle for the d * d == n case.

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

C++

// Author: Huahua

class Solution {

public:

bool isThree(int n) {

int c = 0;

for (int i = 1; i <= sqrt(n); ++i)

if (n % i == 0)

c += 1 + (i * i != n);

return c == 3;

}

};

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 1925. Count Square Sum Triples

By zxi on December 30, 2021

A square triple (a,b,c) is a triple where a, b, and c are integers and a² + b² = c².

Given an integer n, return the number of square triples such that 1 <= a, b, c <= n.

Example 1:

Input: n = 5
Output: 2
Explanation: The square triples are (3,4,5) and (4,3,5).

Example 2:

Input: n = 10
Output: 4
Explanation: The square triples are (3,4,5), (4,3,5), (6,8,10), and (8,6,10).

Constraints:

1 <= n <= 250

Solution: Enumerate a & b

Brute force enumerates a & b & c, which takes O(n³). Just need to enumerate a & b and validate c.

Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countTriples(int n) {

int ans = 0;

for (int a = 1; a <= n; ++a)

for (int b = 1; b <= n; ++b) {

int c = sqrt(a * a + b * b);

if (c <= n && c * c == a * a + b * b) ++ans;

}

return ans;

}

};