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GATE | Gate IT 2005 | Question 53

  • Last Updated : 28 Jun, 2021

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++]]–

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Answer: (D)
Explanation:




#include <stdio.h>
char a[100], b[100];
int main(void) {
 int flag;
  printf("Enter first string\n");
 gets(a);
  printf("Enter second string\n");
 gets(b);
  flag = anagram(a, b);
  if (flag == 1)
 printf("\"%s\" and \"%s\" are anagrams.\n", a, b);
 else
 printf("\"%s\" and \"%s\" are not anagrams.\n", a, b);
  return 0;
return 0;
}
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]) {
count [a[j]]++;
count[b[j++]]--;
}
for (j = 0; j < 128; j++) if (count [j]) return 0;
return 1;
}

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