# Posts tagged as “string”

A string is called a happy prefix if is a non-empty prefix which is also a suffix (excluding itself).

Given a string s. Return the longest happy prefix of s .

Return an empty string if no such prefix exists.

Example 1:

Input: s = "level"
Output: "l"
Explanation: s contains 4 prefix excluding itself ("l", "le", "lev", "leve"), and suffix ("l", "el", "vel", "evel"). The largest prefix which is also suffix is given by "l".


Example 2:

Input: s = "ababab"
Output: "abab"
Explanation: "abab" is the largest prefix which is also suffix. They can overlap in the original string.


Example 3:

Input: s = "leetcodeleet"
Output: "leet"


Example 4:

Input: s = "a"
Output: ""


Constraints:

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

## Solution: Rolling Hash

Time complexity: O(n) / worst case: O(n^2)
Space complexity: O(1)

## C++

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

Given a string containing only digits '0'-'9', write a function to determine if it’s an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Example 1:

Input: "112358"
Output: true
Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8


Example 2:

Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199


Constraints:

• num consists only of digits '0'-'9'.
• 1 <= num.length <= 35

## Solution: DFS

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

## Python3

Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.

The . character does not represent a decimal point and is used to separate number sequences.

For instance, 2.5 is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
2. Version strings do not start or end with dots, and they will not be two consecutive dots.

Solution: String

Split the version string to a list of numbers, and compare two lists.

Time complexity: O(l1 + l2)
Space complexity: O(l1 + l2)

## C++

Given an integer nreturn 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)

## Python3

Given the string s, return the size of the longest substring containing each vowel an even number of times. That is, ‘a’, ‘e’, ‘i’, ‘o’, and ‘u’ must appear an even number of times.

Example 1:

Input: s = "eleetminicoworoep"
Output: 13
Explanation: The longest substring is "leetminicowor" which contains two each of the vowels: e, i and o and zero of the vowels: a and u.


Example 2:

Input: s = "leetcodeisgreat"
Output: 5
Explanation: The longest substring is "leetc" which contains two e's.


Example 3:

Input: s = "bcbcbc"
Output: 6
Explanation: In this case, the given string "bcbcbc" is the longest because all vowels: a, e, i, o and u appear zero times.


Constraints:

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

## Solution: HashTable

Record the first index when a state occurs. index – last_index is the length of the all-even-vowel substring.

State: {a: odd|even, e: odd|even, …, u:odd|even}.

There are total 2^5 = 32 states that can be represented as a binary string.

whenever a vowel occurs, we flip the bit, e.g. odd->even, even->odd using XOR.

Time complexity: O(5*n)
Space complexity: O(32)

## Python3

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