Q.no:- Valid Anagram

🧠 Intuition

At first, my idea was simple:

If two strings are anagrams, they must contain the same characters with the same frequency, just in a different order.

So the plan was:

• Take each character from the first string.
• Try to find and remove that character from the second string.
• If all characters from the first string are found and removed successfully from the second string, and nothing is left, then it is a valid anagram.

🧩 Approach

To implement this idea:

1. Check if the lengths of both strings are equal.
   
   If not, return false right away — because anagrams must have the same number of characters.

2. Convert both strings into arrays of characters, so we can easily remove elements using splice().

3. Use a nested loop:

   • The outer loop goes through each character in the first string.
   • The inner loop tries to find that character in the second string.
   • If the character is found, remove it from the second string and break the inner loop.

4. After all loops:

   • If the second string is empty, return true (it's a valid anagram).
   • If not, return false.

⏱ Complexity

Time Complexity: O(n²)

We are using a nested loop:

• The outer loop runs n times (for each character in string s).
• The inner loop may also run up to n times (to find the matching character in t).

So in the worst case, it’s O(n * n) = O(n²).

❗ This is not the most efficient solution, but it works for small inputs and is easy to understand.

Space Complexity: O(1) (Constant)

We’re not using any additional data structures that grow with input size.

Only a few variables and the input arrays are used, so space complexity stays constant.

📌 Notes

• This approach is easy to understand but not optimal.
• For larger strings or performance-sensitive code, a better method would be to:
  • Use hash maps to count character frequency, or
  • Sort both strings and compare them directly.

Still, this method is a good starting point for beginners learning how to manipulate strings and arrays.

Code

/** * @param {string} s * @param {string} t * @return {boolean} */ var isAnagram = function(s, t) { s = s.split('') t = t.split('') if(s.length !== t.length) return false; let char = null; for(let j = 0 ; j < s.length ; j++){ char = s[j] for(let x = 0 ; x < s.length ; x++ ){ if(char === t[x]){ t.splice(x, 1); break; } } } if(t.length > 0){ return false } return true; };