Question

The following C function takes two ASCII strings and determines whether one is an anagram of the other. An anagram of a string s is a string obtained by permuting the letters in s.

int anagram (char *a, char *b)
{
int count [128], j;
for (j = 0; j < 128; j++)
count[j] = 0;
j=0;
while (a[j] && b[j])
{
A;
B;
}
for (j = 0; j < 128; j++)
if (count [j]) return 0;
return 1;
}

Choose the correct alternative for statements A and B

• A. A : count [a[j++]]++ and
  B : count [b[j]]--
• B. A : count [a[j]]++ and
  B : count [b[j]]--
• C. A : count [a[j]]++ and
  B : count [b[j]]++
• D. A : count [a[j]]++ and
  B : count [b[j++]]--

Answer 1

The correct option is D A : count [a[j]]++ and

B : count [b[j++]]--
Definition of anagram: a string s is a string obtained by permuting the letter in s.

1${}^{st}$ for loop just create an array count and n initialize to 0 for each element i.e. for 128 element. While loop run till string reaches to end of strings. In while loop, for every element a[i] count is increased which is decreased by array b[i] when same element arrive in array b[i] and increment loop with j++.

If both strings are anagram of each other then the count value will be zero for each elements.

So, count[a[j]]++; and count[b[j++]]--;

Example:
Inside while loop

$\begin{array}{cc}\left(1\right)count\left[g\right]++;& \left(2\right)count\left[a\right]++;\\ count\left[a\right]--;& count\left[t\right]--;\\ j++;& j++;\\ \left(3\right)count\left[t\right]++;& \left(4\right)count\left[e\right]++;\\ count\left[g\right]--;& count\left[e\right]--;\\ j++;& j++;\end{array}$

While loop fail for (5)
Since atlast count will contain all zero. So both are anagram of each other.