Posts tagged as “string”

花花酱 LeetCode 3029. Minimum Time to Revert Word to Initial State I

By zxi on February 5, 2024

You are given a 0-indexed string word and an integer k.

At every second, you must perform the following operations:

Remove the first k characters of word.
Add any k characters to the end of word.

Note that you do not necessarily need to add the same characters that you removed. However, you must perform both operations at every second.

Return the minimum time greater than zero required for word to revert to its initial state.

Example 1:

Input: word = "abacaba", k = 3
Output: 2
Explanation: At the 1st second, we remove characters "aba" from the prefix of word, and add characters "bac" to the end of word. Thus, word becomes equal to "cababac".
At the 2nd second, we remove characters "cab" from the prefix of word, and add "aba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 2 seconds is the minimum time greater than zero required for word to revert to its initial state.

Example 2:

Input: word = "abacaba", k = 4
Output: 1
Explanation: At the 1st second, we remove characters "abac" from the prefix of word, and add characters "caba" to the end of word. Thus, word becomes equal to "abacaba" and reverts to its initial state.
It can be shown that 1 second is the minimum time greater than zero required for word to revert to its initial state.

Example 3:

Input: word = "abcbabcd", k = 2
Output: 4
Explanation: At every second, we will remove the first 2 characters of word, and add the same characters to the end of word.
After 4 seconds, word becomes equal to "abcbabcd" and reverts to its initial state.
It can be shown that 4 seconds is the minimum time greater than zero required for word to revert to its initial state.

Constraints:

1 <= word.length <= 50
1 <= k <= word.length
word consists only of lowercase English letters.

Solution: Suffix ==? Prefix

Compare the suffix with prefix.

word = “abacaba”, k = 3
ans = 1, “~~aba~~caba” != “abac~~aba~~“
ans = 2, “~~abacab~~a” == “a~~bacaba~~“, we find it.

Time complexity: O(n * n / k)
Space complexity: O(1)

C++

class Solution {

public:

int minimumTimeToInitialState(string_view word, int k) {

const int n = word.length();

for (int i = 1; i * k < n; ++i) {

string_view t = word.substr(i * k);

if (t == word.substr(0, t.length())) return i;

}

return (n + k - 1) / k;

}

};

花花酱 LeetCode 2828. Check if a String Is an Acronym of Words

By zxi on August 26, 2023

Given an array of strings words and a string s, determine if s is an acronym of words.

The string s is considered an acronym of words if it can be formed by concatenating the first character of each string in words in order. For example, "ab" can be formed from ["apple", "banana"], but it can’t be formed from ["bear", "aardvark"].

Return true if s is an acronym of words, and false otherwise.

Example 1:

Input: words = ["alice","bob","charlie"], s = "abc"
Output: true
Explanation: The first character in the words "alice", "bob", and "charlie" are 'a', 'b', and 'c', respectively. Hence, s = "abc" is the acronym.

Example 2:

Input: words = ["an","apple"], s = "a"
Output: false
Explanation: The first character in the words "an" and "apple" are 'a' and 'a', respectively. 
The acronym formed by concatenating these characters is "aa". 
Hence, s = "a" is not the acronym.

Example 3:

Input: words = ["never","gonna","give","up","on","you"], s = "ngguoy"
Output: true
Explanation: By concatenating the first character of the words in the array, we get the string "ngguoy". 
Hence, s = "ngguoy" is the acronym.

Constraints:

1 <= words.length <= 100
1 <= words[i].length <= 10
1 <= s.length <= 100
words[i] and s consist of lowercase English letters.

Solution: Check the first letter of each word

No need to concatenate, just check the first letter of each word.

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

C++

// Author: Huahua

class Solution {

public:

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

if (words.size() != s.size()) return false;

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

if (words[i][0] != s[i]) return false;

return true;

}

};

花花酱 LeetCode 2825. Make String a Subsequence Using Cyclic Increments

By zxi on August 23, 2023

You are given two 0-indexed strings str1 and str2.

In an operation, you select a set of indices in str1, and for each index i in the set, increment str1[i] to the next character cyclically. That is 'a' becomes 'b', 'b' becomes 'c', and so on, and 'z' becomes 'a'.

Return true if it is possible to make str2 a subsequence of str1 by performing the operation at most once, and false otherwise.

Note: A subsequence of a string is a new string that is formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.

Example 1:

Input: str1 = "abc", str2 = "ad"
Output: true
Explanation: Select index 2 in str1.
Increment str1[2] to become 'd'. 
Hence, str1 becomes "abd" and str2 is now a subsequence. Therefore, true is returned.

Example 2:

Input: str1 = "zc", str2 = "ad"
Output: true
Explanation: Select indices 0 and 1 in str1. 
Increment str1[0] to become 'a'. 
Increment str1[1] to become 'd'. 
Hence, str1 becomes "ad" and str2 is now a subsequence. Therefore, true is returned.

Example 3:

Input: str1 = "ab", str2 = "d"
Output: false
Explanation: In this example, it can be shown that it is impossible to make str2 a subsequence of str1 using the operation at most once. 
Therefore, false is returned.

Constraints:

1 <= str1.length <= 10⁵
1 <= str2.length <= 10⁵
str1 and str2 consist of only lowercase English letters.

Solution: Two pointers

s1[i] and s2[j] can match if
s1[i] == s2[j] or inc(s1[i]) == s2[j]

If matched: ++i; ++j else ++i.

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

C++

class Solution {

public:

bool canMakeSubsequence(string_view s1, string_view s2) {

for (int i = 0, j = 0, n = s1.size(), m = s2.size(); i < n; ++i)

if (s1[i] == s2[j] || (s1[i] - 'a' + 1) % 26 == s2[j] - 'a')

if (++j == m) return true;

return false;

}

};

Iterator version

C++

class Solution {

public:

bool canMakeSubsequence(string_view s1, string_view s2) {

for (auto i = begin(s1), j = begin(s2); i != end(s1); ++i)

if (*i == *j || (*i - 'a' + 1) % 26 == *j - 'a')

if (++j == end(s2)) return true;

return false;

}

};

花花酱 LeetCode 2586. Count the Number of Vowel Strings in Range

By zxi on March 11, 2023

You are given a 0-indexed array of string words and two integers left and right.

A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are 'a', 'e', 'i', 'o', and 'u'.

Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].

Example 1:

Input: words = ["are","amy","u"], left = 0, right = 2
Output: 2
Explanation: 
- "are" is a vowel string because it starts with 'a' and ends with 'e'.
- "amy" is not a vowel string because it does not end with a vowel.
- "u" is a vowel string because it starts with 'u' and ends with 'u'.
The number of vowel strings in the mentioned range is 2.

Example 2:

Input: words = ["hey","aeo","mu","ooo","artro"], left = 1, right = 4
Output: 3
Explanation: 
- "aeo" is a vowel string because it starts with 'a' and ends with 'o'.
- "mu" is not a vowel string because it does not start with a vowel.
- "ooo" is a vowel string because it starts with 'o' and ends with 'o'.
- "artro" is a vowel string because it starts with 'a' and ends with 'o'.
The number of vowel strings in the mentioned range is 3.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 10
words[i] consists of only lowercase English letters.
0 <= left <= right < words.length

Solution:

Iterator overs words, from left to right. Check the first and last element of the string.

Time complexity: O(|right – left + 1|)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int vowelStrings(vector<string>& words, int left, int right) {

auto isVowel = [](char c) {

return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';

};

return count_if(begin(words) + left, begin(words) + right + 1, [&](const string& w) {

return isVowel(w.front()) && isVowel(w.back());

});

}

};

花花酱 LeetCode 2564. Substring XOR Queries

By zxi on February 12, 2023

You are given a binary string s, and a 2D integer array queries where queries[i] = [first_i, second_i].

For the i^th query, find the shortest substring of s whose decimal value, val, yields second_i when bitwise XORed with first_i. In other words, val ^ first_i == second_i.

The answer to the i^th query is the endpoints (0-indexed) of the substring [left_i, right_i] or [-1, -1] if no such substring exists. If there are multiple answers, choose the one with the minimum left_i.

Return an array ans where ans[i] = [left_i, right_i] is the answer to the i^th query.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "101101", queries = [[0,5],[1,2]]
Output: [[0,2],[2,3]]
Explanation: For the first query the substring in range [0,2] is "101" which has a decimal value of 5, and 5 ^ 0 = 5, hence the answer to the first query is [0,2]. In the second query, the substring in range [2,3] is "11", and has a decimal value of 3, and 3 ^ 1 = 2. So, [2,3] is returned for the second query.

Example 2:

Input: s = "0101", queries = [[12,8]]
Output: [[-1,-1]]
Explanation: In this example there is no substring that answers the query, hence [-1,-1] is returned.

Example 3:

Input: s = "1", queries = [[4,5]]
Output: [[0,0]]
Explanation: For this example, the substring in range [0,0] has a decimal value of 1, and 1 ^ 4 = 5. So, the answer is [0,0].

Constraints:

1 <= s.length <= 10⁴
s[i] is either '0' or '1'.
1 <= queries.length <= 10⁵
0 <= first_i, second_i <= 10⁹

Solution: Pre-compute

We can pre-compute all possible substrings

Time complexity: O(n*32 + m)
Space complexity: O(n*32)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> substringXorQueries(string s, vector<vector<int>>& queries) {

unordered_map<int, vector<int>> m;

const int l = s.length();

for (int i = 0; i < l; ++i) {

if (s[i] == '0') {

if (!m.count(0)) m[0] = {i, i};

continue;

}

for (int j = 0, cur = 0; j < 32 && i + j < l; ++j) {

cur = (cur << 1) | (s[i + j] - '0');

if (!m.count(cur))

m[cur] = {i, i + j};

}

vector<vector<int>> ans;

for (const auto& q : queries) {

int x = q[0] ^ q[1];

if (m.count(x))

ans.push_back(m[x]);

else

ans.push_back({-1, -1});

}

return ans;

}

};