Posts published in “String”

花花酱 1249. Minimum Remove to Make Valid Parentheses

By zxi on November 9, 2019

Given a string s of '(' , ')' and lowercase English characters.

Your task is to remove the minimum number of parentheses ( '(' or ')', in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

It is the empty string, contains only lowercase characters, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"

Constraints:

1 <= s.length <= 10^5
s[i] is one of '(' , ')' and lowercase English letters.

Solution: Couting

Count how many “(” are open and how many “)” left.

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

C++

// Author: Huahua

class Solution {

public:

string minRemoveToMakeValid(string s) {

int close = count(begin(s), end(s), ')');

int open = 0;

string ans;

for (char c : s) {

if (c == '(') {

if (open == close) continue;

++open;

} else if (c == ')') {

--close;

if (open == 0) continue;

--open;

}

ans += c;

}

return ans;

}

};

花花酱 LeetCode 68. Text Justification

By zxi on October 1, 2019

Given an array of words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly maxWidth characters.

Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

For the last line of text, it should be left justified and no extra space is inserted between words.

Note:

A word is defined as a character sequence consisting of non-space characters only.
Each word’s length is guaranteed to be greater than 0 and not exceed maxWidth.
The input array words contains at least one word.

Example 1:

Input:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]

Example 2:

Input:
words = ["What","must","be","acknowledgment","shall","be"]
maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be",
             because the last line must be left-justified instead of fully-justified.
             Note that the second line is also left-justified becase it contains only one word.

Example 3:

Input:
words = ["Science","is","what","we","understand","well","enough","to","explain",
         "to","a","computer.","Art","is","everything","else","we","do"]
maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]

Solution: Simulation

Time complexity: O(sum(len(s))
Space complexity: O(sum(len(s))

C++

// Author: Huahua

class Solution {

public:

vector<string> fullJustify(vector<string> &words, int L) {

size_t i = 0, n = words.size();

vector<string> ans;

while (i < n) {

bool last_line = false;

int ll = 0;

vector<string_view> tw;

while (ll <= L && i < n) {

size_t wl = words[i].length();

int tll = ll + (tw.size() ? 1 : 0) + wl;

if (tll > L) break;

ll = tll;

tw.push_back(string_view(words[i]));

if (++i == n) last_line = true;

if (ll == L) break;

}

string line;

if (!last_line) {

int tl = 0;

for(const auto& w : tw) tl += w.length();

int avg_space = tw.size()==1 ? 0 : ((L-tl) / (tw.size() - 1));

int extra_space = tw.size()==1 ? 0 : (L - avg_space * (tw.size() - 1) - tl);

for (const auto& w : tw) {

if (line.length() > 0) {

line.append(avg_space, ' ');

if (extra_space > 0) {

line += ' ';

--extra_space;

}

line += w;

}

} else {

for (const auto& w : tw) {

if (line.length() > 0)

line += ' ';

line += w;

}

line.append(L - line.length(), ' ');

ans.push_back(std::move(line));

}

return ans;

}

};

花花酱 LeetCode 58. Length of Last Word

By zxi on September 30, 2019

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

Example:

Input: "Hello World"
Output: 5

Solution:

There can be tailing spaces, e.g. “abc “, we need to trim the string first and then scan the string in reverse order until a space or reach the beginning of the string.

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

C++

// Author: Huahua

class Solution {

public:

int lengthOfLastWord(string s) {

int i = s.length() - 1;

int l = 0;

while (i >= 0 && s[i] == ' ') --i;

while (i >= 0 && s[i] != ' ') {

--i;

++l;

}

return l;

}

};

花花酱 LeetCode 1163. Last Substring in Lexicographical Order

By zxi on August 18, 2019

Given a string s, return the last substring of s in lexicographical order.

Example 1:

Input: "abab"
Output: "bab"
Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab".

Example 2:

Input: "leetcode"
Output: "tcode"

Note:

1 <= s.length <= 10^5
s contains only lowercase English letters.

Key observation: The last substring must be a suffix of the original string, can’t a substring in the middle since we can always extend it.
e.g. leetcode -> tcode, can’t be “t”, “tc”, “tco”, “tcod”

Solution 1: Brute Force

Try all possible suffixes.
Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua, 452 ms, 12.8 MB

class Solution {

public:

string lastSubstring(string_view s) {

string_view ans;

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

if (s.substr(i) > ans) ans = s.substr(i);

return string(ans);

}

};

Solution 2: Keep max and compare with candidates

Find the first largest letter as a starting point, whenever there is a same letter, keep it as a candidate and compare with the current best. If the later is larger, take over the current best.

e.g. “acbacbc”

“c” > “a”, the first “c” becomes the best.
“c” = “c”, the second “c” becomes a candidate
starting compare best and candidate.
“cb” = “cb”
“cba” < “cbc”, cand_i is the new best.

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

C++

// Author: Huahua

class Solution {

public:

string lastSubstring(string s) {

int max_i = 0;

char max_c = 0;

int cand_i = 0;

int l = 0;

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

if (s[i] > max_c) {

// Find a better starting point.

max_c = s[i];

max_i = i;

cand_i = -1;

l = 1;

} else if (cand_i != -1) {

// Has a candidate.

if (s[i] > s[max_i + l]) {

// The candidate is larger.

max_i = cand_i;

cand_i = -1;

} else if (s[i] == s[max_i + l]) {

// The candidate is the same as current best.

++l;

} else {

// The candidate is smaller, no longer needed.

cand_i = -1;

}

} else if (s[i] == max_c) {

// Find a new candidate, starting with length 1.

cand_i = i;

l = 1;

}

return s.substr(max_i);

}

};

花花酱 LeetCode 1154. Day of the Year

By zxi on August 11, 2019

Given a string date representing a Gregorian calendar date formatted as YYYY-MM-DD, return the day number of the year.

Example 1:

Input: date = "2019-01-09"
Output: 9
Explanation: Given date is the 9th day of the year in 2019.

Example 2:

Input: date = "2019-02-10"
Output: 41

Example 3:

Input: date = "2003-03-01"
Output: 60

Example 4:

Input: date = "2004-03-01"
Output: 61

Constraints:

date.length == 10
date[4] == date[7] == '-', and all other date[i]‘s are digits
date represents a calendar date between Jan 1st, 1900 and Dec 31, 2019.

Solution:

Key: checking whether that year is a leap year or not.
is_leap = (year % 4 == 0 and year % 100 !=0) or year % 400 == 0

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

Python

# Author: Huahua

class Solution:

def dayOfYear(self, date: str) -> int:

y, m, d = int(date.split("-")[0]), int(date.split("-")[1]), int(date.split("-")[2])

leap = (y % 4 == 0 and y % 100 != 0) or y % 400 == 0

days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

s = sum(days[0:m-1]) + d

if leap and m > 2: s += 1

return s