Anagram Check Techniques in DSA C - Execution Trace

Concept Flow - Anagram Check Techniques

Input two strings

↓

Check length equal?

No→Return False

Yes↓

Choose technique

↓

Sort both

↓

Compare results

↓

Return True if match, else False

Start by checking if strings have equal length. Then use one of the techniques: sorting, counting characters, or using a hash map to compare. Return true if they match.

Execution Sample

DSA C

bool areAnagrams(char* s1, char* s2) {
  if (strlen(s1) != strlen(s2)) return false;
  int count[256] = {0};
  for (int i = 0; s1[i]; i++) {
    count[(unsigned char)s1[i]]++;
    count[(unsigned char)s2[i]]--;
  }
  for (int 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 and comparing counts.

Execution Table

Step	Operation	Index i	Character s1[i]	Character s2[i]	Count Array Change	Count Array Snapshot	Result
1	Check length	-	-	-	-	-	Lengths equal, continue
2	Initialize count array	-	-	-	All zeros	[0,0,0,...0]	-
3	Process i=0	0	a	b	count['a']++, count['b']--	count['a']=1, count['b']=-1, others=0	-
4	Process i=1	1	b	a	count['b']++, count['a']--	count['a']=0, count['b']=0, others=0	-
5	Process i=2	2	c	c	count['c']++, count['c']--	all zeros	-
6	Check count array	-	-	-	No non-zero values	all zeros	Return True (anagrams)
7	End	-	-	-	-	-	Finished

💡 All counts zero means strings are anagrams; function returns True

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	Final
count['a']	0	1	0	0	0
count['b']	0	-1	0	0	0
count['c']	0	0	0	0	0
i	-	0	1	2	-

Key Moments - 3 Insights

Why do we increment count for s1[i] but decrement for s2[i] in the same loop?

Because we want to balance character counts. Incrementing for s1[i] and decrementing for s2[i] in the same loop helps detect mismatches quickly. See execution_table rows 3-5 where counts cancel out.

What if strings have different lengths?

The function immediately returns false at step 1 in execution_table because anagrams must have equal length.

Why check all counts after the loop instead of during?

Checking after ensures all characters are balanced. During the loop, partial counts may not reflect final state. See execution_table step 6 where final check confirms all zeros.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the count value for character 'a'?

A 1

B 0

C -1

D 2

💡 Hint

Check the 'Count Array Snapshot' column at step 4 in execution_table

At which step does the function decide the strings are anagrams?

A Step 3

B Step 5

C Step 6

D Step 7

💡 Hint

Look at the 'Result' column in execution_table for the step returning True

If the strings were 'abc' and 'abd', what would happen at step 5?

A Count for 'c' would be 1 and for 'd' would be -1

B Count for 'c' would be incremented and decremented

C Count array would be all zeros

D Function would return True immediately

💡 Hint

Think about how counts change when characters differ at the same index

Concept Snapshot

Anagram Check Techniques:
- Check if strings have equal length first
- Use character count array to track frequency differences
- Increment count for s1 chars, decrement for s2 chars
- After processing, if all counts are zero, strings are anagrams
- Returns true if anagrams, false otherwise

Full Transcript

This visualization shows how to check if two strings are anagrams by comparing character counts. First, the code checks if the strings have the same length. If not, it returns false immediately. Then it uses an array to count characters: for each position, it increments the count for the character in the first string and decrements for the character in the second string. After processing all characters, it checks if all counts are zero. If yes, the strings are anagrams and the function returns true. The execution table traces each step, showing how counts change and how the final decision is made.