String Archives - Page 16 of 33 - Huahua's Tech Road

花花酱 LeetCode 1374. Generate a String With Characters That Have Odd Counts

By zxi on March 8, 2020

Given an integer n, return a string with n characters such that each character in such string occurs an odd number of times.

The returned string must contain only lowercase English letters. If there are multiples valid strings, return any of them.

Example 1:

Input: n = 4
Output: "pppz"
Explanation: "pppz" is a valid string since the character 'p' occurs three times and the character 'z' occurs once. Note that there are many other valid strings such as "ohhh" and "love".

Example 2:

Input: n = 2
Output: "xy"
Explanation: "xy" is a valid string since the characters 'x' and 'y' occur once. Note that there are many other valid strings such as "ag" and "ur".

Example 3:

Input: n = 7
Output: "holasss"

Constraints:

1 <= n <= 500

Solution: Greedy

if n is odd, return n ‘a’s.
otherwise, return n -1 ‘a’s and 1 ‘b’

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

C++

// Author: Huahua

class Solution {

public:

string generateTheString(int n) {

return n & 1 ? string(n, 'a') : string(n - 1, 'a') + "b";

}

};

Python3

# Author: Huahua

class Solution:

def generateTheString(self, n: int) -> str:

ans = ['a'] * n if n % 2 == 1 else ['a'] * (n - 1) + ['b']

return ''.join(ans)

花花酱 LeetCode 1370. Increasing Decreasing String

By zxi on March 7, 2020

Given a string s. You should re-order the string using the following algorithm:

Pick the smallest character from s and append it to the result.
Pick the smallest character from s which is greater than the last appended character to the result and append it.
Repeat step 2 until you cannot pick more characters.
Pick the largest character from s and append it to the result.
Pick the largest character from s which is smaller than the last appended character to the result and append it.
Repeat step 5 until you cannot pick more characters.
Repeat the steps from 1 to 6 until you pick all characters from s.

In each step, If the smallest or the largest character appears more than once you can choose any occurrence and append it to the result.

Return the result string after sorting s with this algorithm.

Example 1:

Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
Explanation: After steps 1, 2 and 3 of the first iteration, result = "abc"
After steps 4, 5 and 6 of the first iteration, result = "abccba"
First iteration is done. Now s = "aabbcc" and we go back to step 1
After steps 1, 2 and 3 of the second iteration, result = "abccbaabc"
After steps 4, 5 and 6 of the second iteration, result = "abccbaabccba"

Example 2:

Input: s = "rat"
Output: "art"
Explanation: The word "rat" becomes "art" after re-ordering it with the mentioned algorithm.

Example 3:

Input: s = "leetcode"
Output: "cdelotee"

Example 4:

Input: s = "ggggggg"
Output: "ggggggg"

Example 5:

Input: s = "spo"
Output: "ops"

Constraints:

1 <= s.length <= 500
s contains only lower-case English letters.

Solution: Counting frequency of each character

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

C++

// Author: Huahua

class Solution {

public:

string sortString(string s) {

vector<int> count(26);

for (char c : s) ++count[c - 'a'];

string ans;

while (ans.length() != s.length()) {

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

if (--count[c - 'a'] >= 0)

ans += c;

for (char c = 'z'; c >= 'a'; --c)

if (--count[c - 'a'] >= 0)

ans += c;

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def sortString(self, s: str) -> str:

seq = string.ascii_lowercase + string.ascii_lowercase[::-1]

count = collections.Counter(s)

ans = ""

while len(ans) != len(s):

for c in seq:

if count[c] > 0:

ans += c

count[c] -= 1

return ans

花花酱 LeetCode 1360. Number of Days Between Two Dates

By zxi on February 23, 2020

Write a program to count the number of days between two dates.

The two dates are given as strings, their format is YYYY-MM-DD as shown in the examples.

Example 1:

Input: date1 = "2019-06-29", date2 = "2019-06-30"
Output: 1

Example 2:

Input: date1 = "2020-01-15", date2 = "2019-12-31"
Output: 15

Constraints:

The given dates are valid dates between the years 1971 and 2100.

Solution: Convert to days since epoch

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

C++

// Author: Huahua

class Solution {

public:

int daysBetweenDates(string date1, string date2) {

array<int, 12> m{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

auto isLeap = [](int year) {

return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);

};

auto daysFromEpoch = [&](const string& date) {

int year = stoi(date.substr(0, 4));

int month = stoi(date.substr(5, 2));

int days = stoi(date.substr(8, 2));

for (int i = 1970; i < year; ++i)

days += 365 + isLeap(i);

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

days += m[i - 1];

days += month > 2 && isLeap(year);

return days;

};

return abs(daysFromEpoch(date1) - daysFromEpoch(date2));

}

};

花花酱 1358. Number of Substrings Containing All Three Characters

By zxi on February 22, 2020

Given a string s consisting only of characters a, b and c.

Return the number of substrings containing at least one occurrence of all these characters a, b and c.

Example 1:

Input: s = "abcabc"
Output: 10
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "abc", "abca", "abcab", "abcabc", "bca", "bcab", "bcabc", "cab", "cabc" and "abc" (again).

Example 2:

Input: s = "aaacb"
Output: 3
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "aaacb", "aacb" and "acb".

Example 3:

Input: s = "abc"
Output: 1

Constraints:

3 <= s.length <= 5 x 10^4
s only consists of a, b or c characters.

Solution

Record the last index of each character.

At each index i, we can choose any index j that j <= min(last_a, last_b, last_c) as the starting point, and there will be min(last_a, last_b, last_c) + 1 valid substrings.

e.g. aabbabcc…
last_a = 4
last_b = 5
last_c = 7
min(last_a, last_b, last_c) = 4
aabba | bcc
We can choose any char with index <= 4 as string point, there are 5 of them:
aabbabcc
abbabcc
bbabcc
babcc
abcc

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

C++

// Author: Huahua

class Solution {

public:

int numberOfSubstrings(string s) {

vector<int> l{-1,-1,-1};

int ans = 0;

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

l[s[i] - 'a'] = i;

ans += 1 + *min_element(begin(l), end(l));

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def numberOfSubstrings(self, s: str) -> int:

l = {'a':-1, 'b':-1, 'c':-1}

ans = 0

for i, ch in enumerate(s):

l[ch] = i

ans += min(l.values()) + 1

return ans

花花酱 LeetCode 1316. Distinct Echo Substrings

By zxi on January 11, 2020

Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself.

Example 1:

Input: text = "abcabcabc"
Output: 3
Explanation: The 3 substrings are "abcabc", "bcabca" and "cabcab".

Example 2:

Input: text = "leetcodeleetcode"
Output: 2
Explanation: The 2 substrings are "ee" and "leetcodeleetcode".

Constraints:

1 <= text.length <= 2000
text has only lowercase English letters.

Solution 1: Brute Force + HashSet

Try all possible substrings

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

int distinctEchoSubstrings(string text) {

const int n = text.length();

string_view t(text);

unordered_set<string_view> s;

for (int k = 1; k <= n / 2; ++k)

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

if (t.substr(i, k) == t.substr(i + k, k))

s.insert(t.substr(i, 2 * k));

return s.size();

}

};

Posts published in “String”

花花酱 LeetCode 1374. Generate a String With Characters That Have Odd Counts

Solution: Greedy

C++

Python3

花花酱 LeetCode 1370. Increasing Decreasing String

Solution: Counting frequency of each character

C++

Python3

花花酱 LeetCode 1360. Number of Days Between Two Dates

Solution: Convert to days since epoch

C++

花花酱 1358. Number of Substrings Containing All Three Characters

Solution

C++

Python3

花花酱 LeetCode 1316. Distinct Echo Substrings

Solution 1: Brute Force + HashSet

C++