Anagram Check Techniques in DSA C - 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 input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <string.h>
#include <stdbool.h>
#include <stdio.h>

bool areAnagrams(char *str1, char *str2) {
    int count[256] = {0};
    int i;
    for (i = 0; str1[i] && str2[i]; i++) {
        count[(unsigned char)str1[i]]++;
        count[(unsigned char)str2[i]]--;
    }
    if (str1[i] || str2[i]) return false;
    for (i = 0; i < 256; i++) {
        if (count[i] != 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: Two loops: one to count characters in both strings, one to check counts.
• How many times: First loop runs once per character until string ends (n times), second loop runs 256 times (constant).

How Execution Grows With Input

The first loop grows with the length of the strings, doing work for each character. The second loop always does the same amount of work (256 checks), no matter the input size.

Input Size (n)	Approx. Operations
10	~10 + 256 = 266
100	~100 + 256 = 356
1000	~1000 + 256 = 1256

Pattern observation: The work grows mostly in a straight line with input size because the first loop depends on string length, while the second loop is constant.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "The second loop over 256 characters makes the algorithm O(n * 256) which is quadratic."

[OK] Correct: The second loop runs a fixed 256 times regardless of input size, so it is constant time and does not multiply with n.

Interview Connect

Understanding how counting characters leads to linear time helps you explain efficient solutions clearly and confidently in interviews.

Self-Check

"What if we sorted both strings first and then compared them? How would the time complexity change?"

Practice

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

easy

A. They contain the same characters in any order

B. They have the same length but different characters

C. They start with the same character

D. They end with the same character

Solution

1. Step 1: Understand the definition of anagrams

   Anagrams are words or phrases made by rearranging the letters of another, so they must have the same characters.

2. Step 2: Compare the character sets

   If two strings have exactly the same characters, just in different orders, they are anagrams.

3. Final Answer:

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

4. Quick Check:

   Anagrams = same characters [OK]

Hint: Anagrams have identical letters, order doesn't matter [OK]

Common Mistakes:

• Thinking length alone defines anagrams
• Assuming same start or end means anagram
• Ignoring character counts

2. Which of the following C code snippets correctly checks if two strings are anagrams by sorting?

easy

A. Check only first and last characters

B. Compare lengths only

C. Use strcmp directly without sorting

D. Sort both strings and compare with strcmp

Solution

1. Step 1: Understand sorting approach

   Sorting both strings puts their characters in order, so if they are anagrams, sorted strings will be identical.

2. Step 2: Use strcmp to compare sorted strings

   strcmp returns 0 if strings are equal, confirming anagram status.

3. Final Answer:

   Sort both strings and compare with strcmp -> Option D

4. Quick Check:

   Sort + strcmp = correct anagram check [OK]

Hint: Sort strings then compare to check anagrams [OK]

Common Mistakes:

• Comparing strings without sorting
• Ignoring length check before sorting
• Using strcmp on unsorted strings

3. What will be the output of the following C code snippet?

#include 
#include 
#include 

bool areAnagrams(char *str1, char *str2) {
    int count[256] = {0};
    int i;
    for (i = 0; str1[i] && str2[i]; i++) {
        count[(unsigned char)str1[i]]++;
        count[(unsigned char)str2[i]]--;
    }
    if (str1[i] || str2[i]) return false;
    for (i = 0; i < 256; i++) {
        if (count[i] != 0) return false;
    }
    return true;
}

int main() {
    if (areAnagrams("listen", "silent"))
        printf("Anagrams\n");
    else
        printf("Not Anagrams\n");
    return 0;
}

medium

A. Anagrams

B. Not Anagrams

C. Compilation error

D. Runtime error

Solution

1. Step 1: Analyze character counting logic

   The function increments count for each char in str1 and decrements for str2. If strings are anagrams, counts balance to zero.

2. Step 2: Check string lengths and counts

   Strings "listen" and "silent" have same length and characters, so count array ends with all zeros, returning true.

3. Final Answer:

   Anagrams -> Option A

4. Quick Check:

   Balanced counts = anagrams [OK]

Hint: Count chars increment/decrement; zero counts mean anagrams [OK]

Common Mistakes:

• Not checking string length equality
• Ignoring unsigned char cast for indexing
• Returning true without count check

4. Identify the bug in this C function intended to check anagrams:

bool isAnagram(char *s1, char *s2) {
    int count[26] = {0};
    for (int i = 0; s1[i]; i++) {
        count[s1[i] - 'a']++;
        count[s2[i] - 'a']--;
    }
    for (int i = 0; i < 26; i++) {
        if (count[i] != 0) return false;
    }
    return true;
}

medium

A. Missing semicolon in the first for loop

B. Uses int for loop variable instead of size_t

C. Does not check if strings have same length

D. Second loop starts from i=1 instead of i=0

Solution

1. Step 1: Check loop and length conditions

   The loop runs only while s1[i] is valid, but s2 may be shorter or longer, causing mismatch.

2. Step 2: Identify missing length check

   Without verifying s1 and s2 have same length, function may incorrectly return true or access invalid memory.

3. Final Answer:

   Does not check if strings have same length -> Option C

4. Quick Check:

   Length check missing = bug [OK]

Hint: Always check string lengths before counting chars [OK]

Common Mistakes:

• Assuming strings are always same length
• Ignoring null terminator in second string
• Not handling uppercase letters (less critical here)

5. Given two strings that may contain uppercase, lowercase, and spaces, which approach best checks if they are anagrams ignoring case and spaces?

hard

A. Sort strings directly including spaces and compare

B. Convert both to lowercase, remove spaces, then count characters

C. Compare lengths including spaces only

D. Check only first and last characters ignoring case

Solution

1. Step 1: Normalize strings by case and spaces

   Convert all letters to lowercase and remove spaces to compare only meaningful characters.

2. Step 2: Count characters or sort normalized strings

   After normalization, counting characters or sorting will correctly identify anagrams ignoring case and spaces.

3. Final Answer:

   Convert both to lowercase, remove spaces, then count characters -> Option B

4. Quick Check:

   Normalize case and spaces before checking [OK]

Hint: Normalize case and remove spaces before checking anagrams [OK]

Common Mistakes:

• Comparing strings with spaces included
• Ignoring case differences
• Assuming length check includes spaces