# Posts published in “String”

Problem:

https://leetcode.com/problems/rotate-string/description/

We are given two strings, A and B.

shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts on A.

Note:

• A and B will have length at most 100.

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

C++

Problem:

https://leetcode.com/problems/palindromic-substrings/description/

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Example 2:

Note:

1. The input string length won’t exceed 1000.

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

C++

Problem:

https://leetcode.com/problems/longest-uncommon-subsequence-i/description/

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn’t exist, return -1.

Example 1:

Note:

1. Both strings’ lengths will not exceed 100.
2. Only letters from a ~ z will appear in input strings.

Idea:

If two strings are the same, then the longest uncommon sequence does not exist, return -1.

e.g. aaa vs aaa, return -1

Otherwise, the longer string is always a uncommon sequence of the shorter one.

e.g. aaab vs aaa, return 4

Solution 1:

Time complexity: O(n)

Space complexity: O(1)

C++

Java

Python3

Problem:

https://leetcode.com/problems/count-binary-substrings/description/

Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0’s and 1’s, and all the 0’s and all the 1’s in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

Example 1:

Example 2:

Note:

• s.length will be between 1 and 50,000.
• s will only consist of “0” or “1” characters.

Solution 0: Search

Try all possible substrings and check whether it’s a valid one or not.

Time complexity O(2^n * n) TLE

Space complexity: O(n)

Solution 1: Running Length

For S = “000110”, there are two virtual blocks “[00011]0” and “000[110]” which contains consecutive 0s and 1s or (1s and 0s)

Keep tracking of the running length of 0, 1 for each block

“00011” => ‘0’: 3, ‘1’:2

ans += min(3, 2) => ans = 2 (“[0011]”, “0[01]1”)

“110” => ‘0’: 1, ‘1’: 2

ans += min(1, 2) => ans = 3 (“[0011]”, “0[01]1”, “1[10]”)

We can reuse part of the running length from last block.

Time complexity: O(n)

Space complexity: O(1)

C++

https://leetcode.com/problems/number-of-matching-subsequences/description/

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

Note:

• All words in words and S will only consists of lowercase letters.
• The length of S will be in the range of [1, 50000].
• The length of words will be in the range of [1, 5000].
• The length of words[i] will be in the range of [1, 50].

Solution 1: Brute Force

Time complexity: O((S + L) * W)

C++ w/o cache TLE

Space complexity: O(1)

C++ w/ cache 155 ms

Space complexity: O(W * L)

Solution 2: Indexing+ Binary Search

Time complexity: O(S + W * L * log(S))

Space complexity: O(S)

S: length of S

W: number of words

L: length of a word

C++

Java

Python 3:

w/o cache

w/ cache

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