grid47 Exploring patterns and algorithms
Oct 13, 2024
5 min read
Solution to LeetCode 242: Valid Anagram Problem
You are given two strings, s and t. Determine if t is an anagram of s. Two strings are considered anagrams if they contain the exact same characters, but possibly in a different order.
Problem
Approach
Steps
Complexity
Input: The input consists of two strings, s and t, both containing lowercase English letters.
Example: Input: s = "listen", t = "silent"
Constraints:
• 1 <= s.length, t.length <= 5 * 10^4
• s and t consist of lowercase English letters.
Output: Return true if t is an anagram of s, otherwise return false.
Example: Output: true
Constraints:
Goal: The goal is to verify if both strings contain the same frequency of characters.
Steps:
• 1. Initialize an array to track the frequency of characters in both strings.
• 2. Increment the count for characters in string s.
• 3. Decrement the count for characters in string t.
• 4. If all counts are zero, return true. Otherwise, return false.
Goal: Ensure the solution handles cases where the lengths of the strings vary and includes handling edge cases like empty strings.
Steps:
• Strings s and t must only contain lowercase English letters.
Assumptions:
• The input strings are case-sensitive and consist only of lowercase English letters.
• Input: Input: s = "listen", t = "silent"
• Explanation: Both strings contain the same characters with the same frequency. Therefore, they are anagrams, and the output is true.
• Input: Input: s = "rat", t = "car"
• Explanation: The characters in the two strings are different, so the output is false.
• Input: Input: s = "a", t = "a"
• Explanation: Both strings are identical, hence they are anagrams of each other. The output is true.
Approach: To check if two strings are anagrams, we can use a frequency count of characters from both strings. If both strings have identical frequency distributions for all characters, they are anagrams.
Observations:
• The solution involves comparing the character counts from both strings.
• A direct comparison of character counts is efficient and will help identify if the strings are anagrams.
Steps:
• 1. If the lengths of s and t are not equal, return false.
• 2. Create an array to track the frequency of characters for both strings.
• 3. Traverse the first string and increment the count for each character.
• 4. Traverse the second string and decrement the count for each character.
• 5. If any count is not zero after both traversals, return false. Otherwise, return true.
Empty Inputs:
• If both strings are empty, return true as they are trivially anagrams.
Large Inputs:
• The solution must efficiently handle strings of lengths up to 50,000 characters.
Special Values:
• Handle cases where the strings are identical or differ by only one character.
Constraints:
• Ensure that only lowercase letters are considered when counting characters.
Defines the function `isAnagram`, which takes two strings, `s` and `t`, and returns a boolean indicating whether the strings are anagrams of each other.
2 : Character Frequency Vector Initialization
vector<int> ch(26, 0);
Initializes a vector `ch` of size 26, which will be used to store the frequency count of each character in the alphabet (assuming the strings consist only of lowercase English letters).
3 : Increment Character Frequency for String s
for(charx: s) ch[x -'a']++;
Iterates through each character `x` in string `s` and increments the corresponding position in the `ch` vector based on the character's ASCII value (adjusted by subtracting `'a'`).
4 : Decrement Character Frequency for String t
for(charx: t) ch[x -'a']--;
Iterates through each character `x` in string `t` and decrements the corresponding position in the `ch` vector. This balances the frequency counts of characters in `s` and `t`.
5 : Check for Frequency Mismatch
for(intx: ch) if(x !=0) returnfalse;
Iterates through the `ch` vector to check if all character counts are zero. If any count is non-zero, it means the strings are not anagrams, so the function returns `false`.
6 : Return True for Anagrams
returntrue;
If all character counts are zero, the strings are anagrams, so the function returns `true`.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: The time complexity is linear in the size of the input strings, where n is the length of the strings.
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant since we only need a fixed-size array (of size 26) to count character frequencies.
