Anagram Check Techniques in DSA Java - Time & Space Complexity

Time Complexity: Anagram Check Techniques

O(n)

Understanding Time Complexity

We want to understand how the time needed to check if two words are anagrams changes as the words get longer.

How does the work grow when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public boolean areAnagrams(String s1, String s2) {
    if (s1.length() != s2.length()) return false;
    int[] counts = new int[26];
    for (int i = 0; i < s1.length(); i++) {
        counts[s1.charAt(i) - 'a']++;
        counts[s2.charAt(i) - 'a']--;
    }
    for (int count : counts) {
        if (count != 0) return false;
    }
    return true;
}
    

This code checks if two strings are anagrams by counting character frequencies.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: One loop that runs through both strings simultaneously to count characters.
• How many times: Runs once for each character in the strings (n times).
• Secondary operation: Another loop that checks the counts array of fixed size 26.
• How many times: Always 26 times, independent of input size.

How Execution Grows With Input

As the strings get longer, the main loop runs more times, directly proportional to the string length.

Input Size (n)	Approx. Operations
10	About 10 steps for counting + 26 checks
100	About 100 steps for counting + 26 checks
1000	About 1000 steps for counting + 26 checks

Pattern observation: The work grows linearly with input size because the counting loop depends on string length, while the checking loop is constant.

Final Time Complexity

Time Complexity: O(n)

This means the time to check anagrams grows directly with the length of the strings.

Common Mistake

[X] Wrong: "Sorting both strings is always faster than counting characters."

[OK] Correct: Sorting takes more time (O(n log n)) compared to counting (O(n)), so counting is usually faster for anagram checks.

Interview Connect

Understanding how to check anagrams efficiently shows you can choose the best approach for a problem, a key skill in interviews and real projects.

Self-Check

"What if the input strings contain Unicode characters beyond 'a' to 'z'? How would the time complexity change?"

Practice

1. What does it mean if two strings are anagrams of each other?

easy

A. They end with the same character

B. They have the same length but different characters

C. They start with the same character

D. They contain the same characters in any order

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, using all original letters exactly once.

2. Step 2: Compare character sets

   If two strings have the same characters with the same frequency, they are anagrams regardless of order.

3. Final Answer:

   They contain the same characters in any order -> Option D

4. Quick Check:

   Anagrams = same characters any order [OK]

Hint: Check if sorted strings match to confirm anagrams [OK]

Common Mistakes:

• Thinking same length means anagrams
• Only comparing first or last characters
• Ignoring character frequency

2. Which Java code snippet correctly checks if two strings s1 and s2 are anagrams using sorting?

easy

A. char[] a1 = s1.toCharArray(); char[] a2 = s2.toCharArray(); Arrays.sort(a1); Arrays.sort(a2); return Arrays.equals(a1, a2);

B. Arrays.sort(s1.toCharArray()); Arrays.sort(s2.toCharArray()); return s1.equals(s2);

C. return s1.equalsIgnoreCase(s2);

D. return s1.length() == s2.length();

Solution

1. Step 1: Convert strings to char arrays and sort

   We must convert strings to char arrays because Arrays.sort works on arrays, not strings.

2. Step 2: Compare sorted arrays for equality

   After sorting, if arrays are equal, strings are anagrams. Use Arrays.equals to compare arrays.

3. Final Answer:

   char[] a1 = s1.toCharArray(); char[] a2 = s2.toCharArray(); Arrays.sort(a1); Arrays.sort(a2); return Arrays.equals(a1, a2); -> Option A

4. Quick Check:

   Sort char arrays and compare arrays [OK]

Hint: Sort char arrays, then compare with Arrays.equals [OK]

Common Mistakes:

• Sorting strings directly without converting to char arrays
• Using equals() on unsorted strings
• Comparing only lengths

3. What will be the output of the following Java code?

String s1 = "listen";
String s2 = "silent";
char[] a1 = s1.toCharArray();
char[] a2 = s2.toCharArray();
Arrays.sort(a1);
Arrays.sort(a2);
System.out.println(Arrays.equals(a1, a2));

medium

A. Runtime exception

B. false

C. true

D. Compilation error

Solution

1. Step 1: Convert and sort both strings

   Both strings "listen" and "silent" are converted to char arrays and sorted. Sorted arrays become ['e','i','l','n','s','t'] for both.

2. Step 2: Compare sorted arrays

   Since sorted arrays are identical, Arrays.equals(a1, a2) returns true.

3. Final Answer:

   true -> Option C

4. Quick Check:

   Sorted arrays match = true [OK]

Hint: Sort and compare arrays to check anagrams [OK]

Common Mistakes:

• Assuming order matters without sorting
• Confusing equals() on strings with arrays
• Forgetting to sort arrays

4. Identify the error in this Java method that checks if two strings are anagrams:

public boolean isAnagram(String s1, String s2) {
  if (s1.length() != s2.length()) return false;
  char[] a1 = s1.toCharArray();
  char[] a2 = s2.toCharArray();
  Arrays.sort(a1);
  Arrays.sort(a2);
  return a1.equals(a2);
}

medium

A. Using a1.equals(a2) compares references, not content

B. Missing length check before sorting

C. Sorting arrays is incorrect for anagram check

D. Method should return void, not boolean

Solution

1. Step 1: Understand array equality in Java

   Using a1.equals(a2) compares if both arrays are the same object, not their contents.

2. Step 2: Use correct method to compare arrays

   To compare contents, use Arrays.equals(a1, a2) instead of a1.equals(a2).

3. Final Answer:

   Using a1.equals(a2) compares references, not content -> Option A

4. Quick Check:

   Use Arrays.equals() for array content comparison [OK]

Hint: Use Arrays.equals() to compare array contents [OK]

Common Mistakes:

• Using equals() on arrays instead of Arrays.equals()
• Ignoring length check
• Returning wrong data type

5. You want to check if two strings are anagrams but ignoring case and spaces. Which approach is best?

hard

A. Compare strings directly with equalsIgnoreCase()

B. Remove spaces, convert to lowercase, then sort and compare char arrays

C. Count characters without removing spaces or changing case

D. Check if lengths are equal and first characters match

Solution

1. Step 1: Normalize strings by removing spaces and converting to lowercase

   Removing spaces and converting to lowercase ensures uniform comparison ignoring case and spaces.

2. Step 2: Sort char arrays and compare

   After normalization, convert to char arrays, sort them, and compare to check anagram status.

3. Final Answer:

   Remove spaces, convert to lowercase, then sort and compare char arrays -> Option B

4. Quick Check:

   Normalize then sort and compare [OK]

Hint: Normalize strings before sorting to ignore case and spaces [OK]

Common Mistakes:

• Not removing spaces before comparison
• Ignoring case differences
• Only comparing lengths or first characters