CareerCup

int isAnagram(char *s1,char *s2)
{
char characterCounter[256]={0};
int i;
for(;*s1;s1++)
{
 if(*s1==' ') continue;
 characterCounter[*s1]+=1;
}
for(;*s2;s2++)
{
 if(*s2==' ') continue;
   characterCounter[*s2]-=1;
   if(  characterCounter[*s2] <0)
      return 0; //false
}
for(i=0;i<256;i++)
    if(characterCounter[0]!=0)
      return 0; //false

return 1; // True, as both are anagram
}

solution 1:
use sorting and compare the sorted strings
solution 2:
Use buckets and then pop elements from buckets

public static boolean isAnagrams(String str1, String str2)
	{

		if(str1 == null && str2 == null)
			return true;

		if(str1 == null || str2 == null)
			return false;

		char[] charArray1 = str1.toCharArray();
		char[] charArray2 = str2.toCharArray();

		Map<Character, Integer> charCount1 = new HashMap<Character, Integer>();

		for(int i = 0; i < charArray1.length; i++)
		{
			if(charArray1[i] != ' ')
			{
				Integer value = charCount1.get(charArray1[i]);	
				if(value == null)
					value = 0;
				charCount1.put(charArray1[i], ++value); 
			}
		}

		Map<Character, Integer> charCount2 = new HashMap<Character, Integer>();
		for(int i = 0; i < charArray2.length; i++)
		{
			if(charArray2[i] != ' ')
			{
				Integer value = charCount2.get(charArray2[i]);	
				if(value == null)
					value = 0;
				charCount2.put(charArray2[i], ++value); 
			}
		}
		
		int len1 = charCount1.keySet().size();
		int len2 = charCount2.keySet().size();
		if(len1 != len2)
			return false;
		Iterator<Character> it = charCount1.keySet().iterator();
		while(it.hasNext())
		{
			Character current = it.next();
			Integer count1 = charCount1.get(current);
			Integer count2 = charCount2.get(current);
			
			if(count1 != count2)	
				return false;
		}

		return true;

	}

this is a solution for this question

#include<stdio.h>
#include<string.h>
main()
{
char str1[20],str2[20];
printf("enter 2 strings..\n");
gets(str1);
gets(str2);
if(strlen(str1)==strlen(str2))
{
char ch;
int i=0;
for(i=0;str1[i];i++)
{
int j=0,k=0,l=0;
ch=str1[i];
for(j=0;str1[j];j++)
if(ch==str1[j])
k++;
for(j=0;str2[j];j++)
if(str2[j]==ch)
l++;
if(k!=l)
break;
}
if(i==strlen(str1))
printf("both strings are with same alphabets..\n");
}
}

Possible solution in python:

def is_anagram(x1,x2)
    x1 = ''".join(x1.split(' '))
    x2 = "".join(x2.split(' '))
    return sorted(x1) == sorted(x2)

#include <stdio.h>

int main()
{
int i=0,j=0;
char *first = (char*)malloc(sizeof(50));
char *second = (char*)malloc(sizeof(50));
scanf("%[^\n]s\n",first);
scanf("%[^\n]s\n",second);
while(*first != '\0')
{
if(*first != ' ')
{
i = i + *first;
}
first++;
}
while(*second != '\0')
{
if(*second != ' ')
{
j = j + *second;
}
second++;
}
if(i == j)
{
printf("Anagram");
}
else
{
printf("not anagram");
}
return 0;
}

#include<stdio.h>
#include<string.h>
main()
{
char str1[20],str2[20];
printf("enter 2 strings..\n");
gets(str1);
gets(str2);
if(strlen(str1)==strlen(str2))
{
char ch;
int i=0;
for(i=0;str1[i];i++)
{
int j=0,k=0,l=0;
ch=str1[i];
for(j=0;str1[j];j++)
if(ch==str1[j])
k++;
for(j=0;str2[j];j++)
if(str2[j]==ch)
l++;
if(k!=l)
break;
}
if(i==strlen(str1))
printf("both strings are with same alphabets..\n");
}
}

#include<stdio.h>
#include<string.h>
main()
{
char str1[20],str2[20];
printf("enter 2 strings..\n");
gets(str1);
gets(str2);
if(strlen(str1)==strlen(str2))
{
char ch;
int i=0;
for(i=0;str1[i];i++)
{
int j=0,k=0,l=0;
ch=str1[i];
for(j=0;str1[j];j++)
if(ch==str1[j])
k++;
for(j=0;str2[j];j++)
if(str2[j]==ch)
l++;
if(k!=l)
break;
}
if(i==strlen(str1))
printf("both strings are with same alphabets..\n");
}
}

short arr[26] = {0};
bool isAnagram(const char * str1, const char* str2){
	bool retVal = true;
	for(int ii = 0; ii < strlen(str1); ii++){
		if(str1[ii] == ' ') continue;
		arr[str1[ii] - 'a']++;
	}
	for(int ii = 0; ii < strlen(str2); ii++){
		if(str2[ii] == ' ') continue;
		arr[str2[ii] - 'a']--;
	}
	for(int ii = 0; ii < 26; ii++){
		if(arr[ii]) {
			retVal = false;
			break;
		}
	}
	return retVal;
}

#include <algorithm>

bool IsAnagram(string inString1, string inString2)
{ 
 //remove whitespace
 inString1.erase(remove(inString1.begin(), inString1.end(), ' '), inString1.end());
 inString2.erase(remove(inString2.begin(), inString2.end(), ' '), inString2.end());
 
 //sort
 sort(inString1.begin(), inString1.end());
 sort(inString2.begin(), inString2.end());
 
 return(inString1 == inString2);
}