Before writing code to check if two strings are anagrams in C, let’s understand- what is the anagram of the string?
Table of Contents
An anagram of the string is the string, obtained by rearranging the letters of the source string.
Example:
Source String: csestack Anagram of source String: cesstack
The second and third letters of the source string are altered, giving an anagram of the string.
Properties of Anagram String:
In C, you can check the length of the string using the strlen() function.
There are two approaches to check if the two strings are anagrams of each other or not.
#include<stdio.h>
#include<stdbool.h>
#include<string.h>
/* Function prototype
for sorting a given string using quick sort */
void quickSort(char *nArr, int nSI, int nEI);
/* function to check
whether two strings are anagram of each other */
bool checkAnagram(char *strTest1, char *strTest2)
{
int i =0;
// Get lengths of both strings
int nLen1 = strlen(strTest1);
int nLen2 = strlen(strTest2);
// If length of two strings are different,
// these are not anagram
if (nLen1 != nLen2)
return false;
// Sort both strings using quick sort
quickSort (strTest1, 0, nLen1 - 1);
quickSort (strTest2, 0, nLen2 - 1);
// Compare both the sorted strings
for (i = 0; i < nLen1; i++)
if (strTest1[i] != strTest2[i])
return false;
return true;
}
// Following functions
//(exchange and partition are needed for quickSort)
void swapElement(char *a, char *b)
{
char temp;
temp = *a;
*a = *b;
*b = temp;
}
//Partition string into two string using pivot element
int partitionPivot(char nArr[], int nSI, int nEI)
{
char x = nArr[nEI];
int i = (nSI - 1);
int j;
for (j = nSI; j <= nEI - 1; j++)
{
if(nArr[j] <= x)
{
i++;
swapElement(&nArr[i], &nArr[j]);
}
}
swapElement(&nArr[i + 1], &nArr[nEI]);
return (i + 1);
}
/* Quick Sort Implementation
nArr[] : int Array to be sorted
nSI : Starting index of String
nEI : Ending index of String
nPI : Partitioning index (Pivot Index)
*/
void quickSort(char nArr[], int nSI, int nEI)
{
int nPI;
if(nSI < nEI)
{
nPI = partitionPivot(nArr, nSI, nEI);
quickSort(nArr, nSI, nPI - 1);
quickSort(nArr, nPI + 1, nEI);
}
}
int main()
{
char strTest1[] = "csestack";
char strTest2[] = "cesstack";
if (checkAnagram(strTest1, strTest2) )
printf("The two strings are anagram of each other");
else
printf("The two strings are not anagram of each other");
return 0;
}
Output:
The two strings are an anagram of each other
For practice, you can solve the problem using any other popular sorting algorithms.
#include<stdio.h>
#include<stdbool.h>
#include<string.h>
# define NO_OF_CHARS 26
/* function to check whether
two strings are anagram of each other */
bool checkAnagram(char *strTest1, char *strTest2)
{
// Create two count arrays and initialize all values as 0
int nCount1[NO_OF_CHARS] = {0};
int nCount2[NO_OF_CHARS] = {0};
int i=0;
if (strlen(strTest1) != strlen(strTest2))
return false;
// For each character in input strings,
// increment count in
// corresponding count array
//97 is ascii value of 'a'
for (i = 0; strTest1[i] && strTest2[i]; i++)
{
nCount1[strTest1[i]-97]++;
nCount2[strTest2[i]-97]++;
}
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (nCount1[i] != nCount2[i])
return false;
return true;
}
/* Driver program to test to pront printDups*/
int main()
{
char strTest1[] = "csestack";
char strTest2[] = "stackcse";
if (checkAnagram(strTest1, strTest2))
printf("The two strings are anagram of each other");
else
printf("The two strings are not anagram of each other");
return 0;
}
Output:
The two strings are anagram of each other.
This is one of the coding questions asked in the interview. If you are preparing for the job, practice solving these coding interview questions.
This is all about the program to check if two strings are anagrams in C. If you find any other better way of solving the same problem, let’s discuss it in the comment.
I am complete Python Nut, love Linux and vim as an editor. I hold a Master of Computer Science from NIT Trichy. I dabble in C/C++, Java too. I keep sharing my coding knowledge and my own experience on <b>CSEstack.org</b> portal.
The Coding style is absolutely awesome ans easy to understand..I like this code ..
Thanks, Anand.