Huahua’s Tech Road

花花酱 LeetCode 1961. Check If String Is a Prefix of Array

By zxi on December 31, 2021

Given a string s and an array of strings words, determine whether s is a prefix string of words.

A string s is a prefix string of words if s can be made by concatenating the first k strings in words for some positive k no larger than words.length.

Return true if s is a prefix string of words, or false otherwise.

Example 1:

Input: s = "iloveleetcode", words = ["i","love","leetcode","apples"]
Output: true
Explanation:
s can be made by concatenating "i", "love", and "leetcode" together.

Example 2:

Input: s = "iloveleetcode", words = ["apples","i","love","leetcode"]
Output: false
Explanation:
It is impossible to make s using a prefix of arr.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 20
1 <= s.length <= 1000
words[i] and s consist of only lowercase English letters.

Solution: Concat and test

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

C++

// Author: Huahua

class Solution {

public:

bool isPrefixString(string s, vector<string>& words) {

string t;

for (const string& w : words) {

t += w;

if (t.length() >= s.length()) break;

}

return t == s;

}

};

花花酱 LeetCode 1957. Delete Characters to Make Fancy String

By zxi on December 31, 2021

A fancy string is a string where no three consecutive characters are equal.

Given a string s, delete the minimum possible number of characters from s to make it fancy.

Return the final string after the deletion. It can be shown that the answer will always be unique.

Example 1:

Input: s = "leeetcode"
Output: "leetcode"
Explanation:
Remove an 'e' from the first group of 'e's to create "leetcode".
No three consecutive characters are equal, so return "leetcode".

Example 2:

Input: s = "aaabaaaa"
Output: "aabaa"
Explanation:
Remove an 'a' from the first group of 'a's to create "aabaaaa".
Remove two 'a's from the second group of 'a's to create "aabaa".
No three consecutive characters are equal, so return "aabaa".

Example 3:

Input: s = "aab"
Output: "aab"
Explanation: No three consecutive characters are equal, so return "aab".

Constraints:

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

Solution:

Skip the current letter if there are already two same letters in the output string.

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

C++

// Author: Huahua

class Solution {

public:

string makeFancyString(string s) {

string ans;

int count = 0;

for (char c : s) {

if (ans.empty() || c != ans.back())

count = 1;

else if (++count >= 3) continue;

ans += 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 1946. Largest Number After Mutating Substring

By zxi on December 31, 2021

You are given a string num, which represents a large integer. You are also given a 0-indexed integer array change of length 10 that maps each digit 0-9 to another digit. More formally, digit d maps to digit change[d].

You may choose to mutate a single substring of num. To mutate a substring, replace each digit num[i] with the digit it maps to in change (i.e. replace num[i] with change[num[i]]).

Return a string representing the largest possible integer after mutating (or choosing not to) a single substring of num.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: num = "132", change = [9,8,5,0,3,6,4,2,6,8]
Output: "832"
Explanation: Replace the substring "1":
- 1 maps to change[1] = 8.
Thus, "132" becomes "832".
"832" is the largest number that can be created, so return it.

Example 2:

Input: num = "021", change = [9,4,3,5,7,2,1,9,0,6]
Output: "934"
Explanation: Replace the substring "021":
- 0 maps to change[0] = 9.
- 2 maps to change[2] = 3.
- 1 maps to change[1] = 4.
Thus, "021" becomes "934".
"934" is the largest number that can be created, so return it.

Example 3:

Input: num = "5", change = [1,4,7,5,3,2,5,6,9,4]
Output: "5"
Explanation: "5" is already the largest number that can be created, so return it.

Constraints:

1 <= num.length <= 10⁵
num consists of only digits 0-9.
change.length == 10
0 <= change[d] <= 9

Solution: Greedy

Find the first digit that is less equal to the mutated one as the start of the substring, keep replacing as long as mutated >= current.

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

C++

// Author: Huahua

class Solution {

public:

string maximumNumber(string num, vector<int>& change) {

const int n = num.size();

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

if (num[i] - '0' < change[num[i] - '0']) {

for (int j = i; j < n && num[j] - '0' <= change[num[j] - '0']; ++j)

num[j] = change[num[j] - '0'] + '0';

break;

}

return num;

}

};

Huahua's Tech Road

花花酱 LeetCode 1961. Check If String Is a Prefix of Array

Solution: Concat and test

C++

花花酱 LeetCode 1957. Delete Characters to Make Fancy String

Solution:

C++

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

Solution: Math

C++

花花酱 LeetCode 1952. Three Divisors

Solution: Enumerate divisors.

C++

Optimization

C++

花花酱 LeetCode 1946. Largest Number After Mutating Substring

Solution: Greedy

C++