A string is a sequence of characters. Many tips that apply to arrays also apply to strings. You're recommended to read the page on [Arrays](./array.md) before reading this page.
| Find substring | O(n.m) | This is the most naive case. There are more efficient algorithms for string searching such as the [KMP algorithm](https://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm) |
Often you will need to count the frequency of characters in a string. The most common way of doing that is by using a hash table/map in your language of choice. If your language has a built-in Counter class like Python, ask if you can use that instead.
If you need to keep a counter of characters, a common mistake is to say that the space complexity required for the counter is O(n). The space required for a counter of a string of latin characters is O(1) not O(n). This is because the upper bound is the range of characters, which is usually a fixed constant of 26. The input set is just lowercase Latin characters.
A neat trick to count the characters in a string of unique characters is to use a 26-bit bitmask to indicate which lower case latin characters are inside the string.
To determine if two strings have common characters, perform `&` on the two bitmasks. If the result is non-zero, ie. `mask_a & mask_b > 0`, then the two strings have common characters.
An anagram is word switch or word play. It is the result of rearranging the letters of a word or phrase to produce a new word or phrase, while using all the original letters only once. In interviews, usually we are only bothered with words without spaces in them.
- Sorting both strings should produce the same resulting string. This takes O(n.log(n)) time and O(log(n)) space.
- If we map each character to a prime number and we multiply each mapped number together, anagrams should have the same multiple (prime factor decomposition). This takes O(n) time and O(1) space. Examples: [Group Anagram](https://leetcode.com/problems/group-anagrams/)
- Have two pointers at the start and end of the string. Move the pointers inward till they meet. At every point in time, the characters at both pointers should match.
When a question is about counting the number of palindromes, a common trick is to have two pointers that move outward, away from the middle. Note that palindromes can be even or odd length. For each middle pivot position, you need to check it twice - once that includes the character and once without the character. This technique is used in [Longest Palindromic Substring](https://leetcode.com/problems/longest-palindromic-substring/).
- For substrings, you can terminate early once there is no match
- For subsequences, use dynamic programming as there are overlapping subproblems. Check out [this question](https://leetcode.com/problems/longest-palindromic-subsequence/)
