Anagram Check Techniques in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an anagram?

An anagram is a word or phrase formed by rearranging the letters of another word or phrase, using all the original letters exactly once.

beginner

How can sorting be used to check if two strings are anagrams?

Sort both strings alphabetically and compare them. If they are identical after sorting, they are anagrams.

intermediate

Explain the frequency counting method for anagram checking.

Count the frequency of each character in both strings using an array or map. If all counts match, the strings are anagrams.

intermediate

Why is the frequency counting method more efficient than sorting for anagram checking?

Frequency counting runs in linear time O(n), while sorting takes O(n log n). So frequency counting is faster for large strings.

beginner

What is a limitation of simple anagram checks when dealing with case sensitivity and spaces?

Simple checks may fail if strings differ in letter case or contain spaces. Preprocessing like converting to lowercase and removing spaces is needed.

Which method involves sorting both strings to check for anagrams?

A Frequency counting method

B Sorting method

C Hashing method

D Brute force method

Sorting both strings and comparing them is the sorting method for anagram checking.

What is the time complexity of frequency counting for anagram checking?

A O(n^2)

B O(n log n)

C O(n)

D O(1)

Frequency counting runs in linear time O(n), where n is the length of the strings.

What should you do before checking anagrams if the strings have different cases or spaces?

A Convert to lowercase and remove spaces

B Ignore case and spaces

C Sort without changes

D Use brute force

Preprocessing by converting to lowercase and removing spaces ensures accurate anagram checking.

If two strings have different lengths, can they be anagrams?

A Yes, always

B Only if one is a substring

C Only if sorted strings match

D No, never

Anagrams must have the same length because they use the same letters exactly once.

Which data structure is commonly used for frequency counting in anagram checks?

A Array or Hash Map

B Queue

C Linked List

D Stack

Arrays or hash maps are used to count character frequencies efficiently.

Describe two common techniques to check if two strings are anagrams.

Think about rearranging letters and counting letters.

Explain why preprocessing strings might be necessary before checking for anagrams.

Consider differences in letter case and extra spaces.

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