This approach involves sorting both strings and comparing them. If they are anagrams, both sorted strings will be identical since an anagram is defined as a rearrangement of letters. The time complexity mainly depends on the sorting step, which is O(n log n), where n is the length of the strings. Space complexity is O(1) if sorting is done in-place, otherwise O(n) with additional space for sorted copies.
Time Complexity: O(n log n), Space Complexity: O(n) due to sorting overhead.
1#include <iostream>
2#include <algorithm>
3
4bool isAnagram(std::string s, std::string t) {
5 if (s.size() != t.size()) {
6 return false;
7 }
8 std::sort(s.begin(), s.end());
9 std::sort(t.begin(), t.end());
10 return s == t;
11}
12
13int main() {
14 std::string s = "anagram";
15 std::string t = "nagaram";
16 std::cout << (isAnagram(s, t) ? "true" : "false") << std::endl;
17 return 0;
18}We check string lengths initially. Then, we use the std::sort function from the C++ STL to sort both strings and compare them. If they are identical post-sort, t is an anagram of s.
This approach uses two arrays (or hashmaps for more general cases) to count the frequency of each character in both strings. Since the problem constraints specify lowercase English letters, array indices (0-25) can be used to count character occurrences.
Time Complexity: O(n), Space Complexity: O(1) as the count array is fixed in size.
1using System;
2
3public class AnagramCheck {
4 public static bool IsAnagram(string s, string t) {
5 if (s.Length != t.Length) {
6 return false;
7 }
8 int[] count = new int[26];
9 for (int i = 0; i < s.Length; i++) {
10 count[s[i] - 'a']++;
11 count[t[i] - 'a']--;
12 }
13 foreach (int c in count) {
14 if (c != 0) {
15 return false;
16 }
17 }
18 return true;
19 }
20
21 public static void Main() {
22 string s = "anagram";
23 string t = "nagaram";
24 Console.WriteLine(IsAnagram(s, t));
25 }
26}Using an integer array to tally frequency differences, balanced zeroed counts after examination solidify the strings as anagrams.