grid47 Exploring patterns and algorithms
Sep 24, 2024
5 min read
Solution to LeetCode 438: Find All Anagrams in a String Problem
You are given two strings, s and p. Return the list of start indices of all the anagrams of string p in string s. An anagram of p is a rearrangement of the characters of p, and the start indices should be returned in any order.
Problem
Approach
Steps
Complexity
Input: The input consists of two strings, s and p, where s is the main string and p is the string whose anagrams are to be found.
Example: "cbaebabacd", "abc"
Constraints:
• 1 <= s.length, p.length <= 3 * 10^4
• s and p consist of lowercase English letters.
Output: Return an array of integers representing the starting indices of all the anagrams of p found in s.
Example: [0, 6]
Constraints:
• The result should be an array of integers, and the indices can be in any order.
Goal: The goal is to efficiently find all the starting indices where anagrams of string p appear in string s.
Steps:
• 1. Use a sliding window approach to check each substring of s that has the same length as p.
• 2. Compare the character frequencies of the current substring in s with the character frequencies of p.
• 3. If they match, store the starting index of the substring as an answer.
Goal: The solution should handle strings up to 30,000 characters in length efficiently.
Steps:
• 1 <= s.length, p.length <= 3 * 10^4
• s and p consist of lowercase English letters.
Assumptions:
• Both strings s and p consist only of lowercase English letters.
• Input: "cbaebabacd", "abc"
• Explanation: In this example, the substrings that are anagrams of 'abc' in 'cbaebabacd' are 'cba' at index 0 and 'bac' at index 6.
• Input: "abab", "ab"
• Explanation: Here, the substrings 'ab', 'ba', and 'ab' are all anagrams of 'ab' in 'abab', appearing at indices 0, 1, and 2 respectively.
Approach: The sliding window technique can be used to check substrings of s for anagrams of p. We will maintain a frequency count of characters in the current window and compare it with the frequency count of p.
Observations:
• We need to compare character frequencies in s and p efficiently to find anagrams.
• The sliding window approach with two frequency arrays can help us avoid recomputing the character frequencies for every substring from scratch.
Steps:
• 1. Create frequency arrays for both the string p and the current window in s.
• 2. Slide the window across s, updating the frequency counts for the current window as you go.
• 3. Compare the current window's frequency array with p's frequency array. If they match, add the index to the result list.
Empty Inputs:
• If either string s or p is empty, return an empty array.
Large Inputs:
• For very large strings, ensure that the solution runs efficiently within the given time constraints.
Special Values:
• Handle cases where p is larger than s (in which case no anagrams can exist).
Constraints:
• Handle edge cases such as empty strings or when p is longer than s.
Defines the function to find starting indices of substrings in 's' that are anagrams of 'p'.
2 : Variable Initialization
vector<int> pc(26, 0), sc(26, 0);
Initializes two frequency count vectors, one for the target string 'p' (pc) and another for the current window in 's' (sc).
3 : Loop Through Target
for(charx: p)
Iterates over each character in the string 'p' to build its frequency count vector.
4 : Frequency Update
pc[x-'a']++;
Increments the frequency of the character in the vector for 'p'.
5 : Result Storage
vector<int> ans;
Declares a vector to store the starting indices of anagram substrings.
6 : Sliding Window Loop
for(int i =0; i < s.size(); i++) {
Starts a loop to traverse through each character in the string 's' with a sliding window.
7 : Window Size Check
if(i >= p.size()) {
Checks if the current index exceeds the size of 'p' to adjust the sliding window.
8 : Update Frequency
sc[s[i - p.size()] -'a']--;
Decrements the frequency count of the character that is sliding out of the window.
9 : Update Frequency
sc[s[i]-'a']++;
Increments the frequency count of the character entering the sliding window.
10 : Comparison Check
if(sc == pc) ans.push_back(i - p.size() +1);
Checks if the frequency count of the current window matches 'p'. If true, adds the starting index to the result.
11 : Return Statement
return ans;
Returns the result vector containing the starting indices of all valid anagram substrings.
Best Case: O(n), where n is the length of string s. This occurs when the first window is already an anagram of p.
Average Case: O(n), where n is the length of s, because the sliding window approach allows us to process each character in s only once.
Worst Case: O(n), since the time complexity does not increase with the number of anagrams found.
Description: The sliding window approach ensures that we only need to scan each character of s once, making the time complexity O(n).
Best Case: O(k), since we always store the frequency of characters in the alphabet.
Worst Case: O(k), where k is the size of the alphabet (26 for lowercase English letters). This is the space required for the frequency arrays.
Description: The space complexity is O(k), where k is the constant number of lowercase English letters, so it is independent of the size of the input strings.
