Posts published in “String”

花花酱 LeetCode 2138. Divide a String Into Groups of Size k

By zxi on January 30, 2022

A string s can be partitioned into groups of size k using the following procedure:

The first group consists of the first k characters of the string, the second group consists of the next k characters of the string, and so on. Each character can be a part of exactly one group.
For the last group, if the string does not have k characters remaining, a character fill is used to complete the group.

Note that the partition is done so that after removing the fill character from the last group (if it exists) and concatenating all the groups in order, the resultant string should be s.

Given the string s, the size of each group k and the character fill, return a string array denoting the composition of every group s has been divided into, using the above procedure.

Example 1:

Input: s = "abcdefghi", k = 3, fill = "x"
Output: ["abc","def","ghi"]
Explanation:
The first 3 characters "abc" form the first group.
The next 3 characters "def" form the second group.
The last 3 characters "ghi" form the third group.
Since all groups can be completely filled by characters from the string, we do not need to use fill.
Thus, the groups formed are "abc", "def", and "ghi".

Example 2:

Input: s = "abcdefghij", k = 3, fill = "x"
Output: ["abc","def","ghi","jxx"]
Explanation:
Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters only.
1 <= k <= 100
fill is a lowercase English letter.

Solution: Pre-fill

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

C++

// Author: Huahua

class Solution {

public:

vector<string> divideString(string s, int k, char fill) {

if (s.length() % k)

s.append(k - s.length() % k, fill);

vector<string> ans;

for (int i = 0; i < s.length(); i += k)

ans.push_back(s.substr(i, k));

return ans;

}

};

花花酱 LeetCode 2129. Capitalize the Title

By zxi on January 8, 2022

You are given a string title consisting of one or more words separated by a single space, where each word consists of English letters. Capitalize the string by changing the capitalization of each word such that:

If the length of the word is 1 or 2 letters, change all letters to lowercase.
Otherwise, change the first letter to uppercase and the remaining letters to lowercase.

Return the capitalized title.

Example 1:

Input: title = "capiTalIze tHe titLe"
Output: "Capitalize The Title"
Explanation:
Since all the words have a length of at least 3, the first letter of each word is uppercase, and the remaining letters are lowercase.

Example 2:

Input: title = "First leTTeR of EACH Word"
Output: "First Letter of Each Word"
Explanation:
The word "of" has length 2, so it is all lowercase.
The remaining words have a length of at least 3, so the first letter of each remaining word is uppercase, and the remaining letters are lowercase.

Example 3:

Input: title = "i lOve leetcode"
Output: "i Love Leetcode"
Explanation:
The word "i" has length 1, so it is lowercase.
The remaining words have a length of at least 3, so the first letter of each remaining word is uppercase, and the remaining letters are lowercase.

Constraints:

1 <= title.length <= 100
title consists of words separated by a single space without any leading or trailing spaces.
Each word consists of uppercase and lowercase English letters and is non-empty.

Solution: Straight forward

Without splitting the sentence into words, we need to take care the word of length one and two.

Tips: use std::tolower, std::toupper to transform letters.

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

C++

// Author: Huahua

class Solution {

public:

string capitalizeTitle(string title) {

const int n = title.size();

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

if ((i == 0 || title[i - 1] == ' ')

&& i + 2 < n && title[i + 1] != ' ' && title[i + 2] != ' ')

title[i] = toupper(title[i]);

else

title[i] = tolower(title[i]);

}

return title;

}

};

Python3

# Author: Huahua

class Solution:

def capitalizeTitle(self, title: str) -> str:

return " ".join(map(lambda w: w.lower() if len(w) <= 2 else w[0].upper() + w[1:].lower(), title.split()))

花花酱 LeetCode 2124. Check if All A’s Appears Before All B’s

By zxi on January 2, 2022

Given a string s consisting of only the characters 'a' and 'b', return true if every 'a' appears before every 'b' in the string. Otherwise, return false.

Example 1:

Input: s = "aaabbb"
Output: true
Explanation:
The 'a's are at indices 0, 1, and 2, while the 'b's are at indices 3, 4, and 5.
Hence, every 'a' appears before every 'b' and we return true.

Example 2:

Input: s = "abab"
Output: false
Explanation:
There is an 'a' at index 2 and a 'b' at index 1.
Hence, not every 'a' appears before every 'b' and we return false.

Example 3:

Input: s = "bbb"
Output: true
Explanation:
There are no 'a's, hence, every 'a' appears before every 'b' and we return true.

Constraints:

`1 <= s.length <= 100`

`s[i]` is either `'a'` or `'b'`.

Solution: Count bs

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

C++

// Author: Huahua

class Solution {

public:

bool checkString(string s) {

int b = 0;

for (char ch : s) {

b += ch == 'b';

if (ch == 'a' && b) return false;

}

return true;

}

};

花花酱 LeetCode 1967. Number of Strings That Appear as Substrings in Word

By zxi on December 31, 2021

Given an array of strings patterns and a string word, return the number of strings in patterns that exist as a substring in word.

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

Example 1:

Input: patterns = ["a","abc","bc","d"], word = "abc"
Output: 3
Explanation:
- "a" appears as a substring in "abc".
- "abc" appears as a substring in "abc".
- "bc" appears as a substring in "abc".
- "d" does not appear as a substring in "abc".
3 of the strings in patterns appear as a substring in word.

Example 2:

Input: patterns = ["a","b","c"], word = "aaaaabbbbb"
Output: 2
Explanation:
- "a" appears as a substring in "aaaaabbbbb".
- "b" appears as a substring in "aaaaabbbbb".
- "c" does not appear as a substring in "aaaaabbbbb".
2 of the strings in patterns appear as a substring in word.

Example 3:

Input: patterns = ["a","a","a"], word = "ab"
Output: 3
Explanation: Each of the patterns appears as a substring in word "ab".

Constraints:

1 <= patterns.length <= 100
1 <= patterns[i].length <= 100
1 <= word.length <= 100
patterns[i] and word consist of lowercase English letters.

Solution: Brute Force

We can use count_if for 1-liner.

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

C++

// Author: Huahua

class Solution {

public:

int numOfStrings(vector<string>& patterns, string word) {

return count_if(begin(patterns), end(patterns), [&](string& p) {

return word.find(p) != string::npos;

});

}

};

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;

}

};

Posts published in “String”

花花酱 LeetCode 2138. Divide a String Into Groups of Size k

Solution: Pre-fill

C++

花花酱 LeetCode 2129. Capitalize the Title

Solution: Straight forward

C++

Python3

花花酱 LeetCode 2124. Check if All A’s Appears Before All B’s

1 <= s.length <= 100

s[i] is either 'a' or 'b'.Solution: Count bs

C++

花花酱 LeetCode 1967. Number of Strings That Appear as Substrings in Word

Solution: Brute Force

C++

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

Solution: Concat and test

C++

`1 <= s.length <= 100`

`s[i]` is either `'a'` or `'b'`.

Solution: Count bs