CareerCup

You could use a char array[256] indexed by character.

For instance,

abcdd
adcbd

array['a']+=1
array['b']+=1
array['c']+=1
array['d']+=1
array['d']+=1

array['a']-=1
array['d']-=1
array['c']-=1
array['b']-=1
array['d']-=1

for(int i=0;i<string1.length();i++)
{
if(array[string.charAt(i)]>0)
return false;
}

Arrays have good locality so you could have impressed them by saying it would utilize cache better. :o)

No. What if provided strings are both very long string. if each string is n characters long, then ur algo is talking 'n' extra memory for each string. Secondly, It will not work in case of Chinese or Japense strings ...

In an interview, you could stick with ASCII. The unicode-equivalent isn't much different, although it's worth mentioning...

Anyhow, it's O(1) memory and O(n) traversal for ASCII. This is more efficient than hashtable for English locale.

Thanks Jack, First I also thought like you and gave following solution, which is efficient I think ..

-Iterate through each character in string, and add the ASCII code value.
-No matter How characters in strings are arrange, if they contain same characters, then the final sum of ascii values of both of these strings would be same.
- This solution not requires '2n' extra space, nor it has expensive 'If Statement' n times.

But the thing was, they wanted generic solution and i think my answer was Good enuff.

Altaf your approach will not work. For example consider the scenario

char string1[] = "ZZ"; //ascii total 90 + 90 = 180
char string2[] = "Pd"; //ascii total 80 + 100 = 180

However you could fix the summing routine as follows

int sum(char * string){
. int sigma = 0;
. for(i = 0; i < strlen(string); i++){
. . sigma += string[i] * i;
. }
. return sigma;
}

Note: Jacks method does not require 2n memory but constant memory of 256 ints irrespective of the size of the input.

Ops a bug, quick fix:

int sum(char * string){
. int sigma = 0;
. for(i = 0; i < strlen(string); i++){
. . sigma += string[i] * (i+1);
. }
. return sigma;
}

On a side note...

If the machine is byte-addressable, than a char will take up one address location. If sizeof(int)=16, then storing ints would take up double that. Assuming a word is at least 2 bytes(1byte=8bits).

In general, when you make a claim, make sure you can prove it formally. In an interview, this would be especially worth it.

Oops bug again, I have been cranky all morning:

int sum(char * string){
. int sigma = 0;
. for(i = 0; i < strlen(string); i++){
. . sigma += string[i] * string[i];
. }
. return sigma;
}

Why are u guys giving another solution. I think Altaf's algo was the correct choice. He was right on spot. We can create all sorts of solutions but why waste time when the original solution posted is the right solution. And even the lexographical sorting can be done with O(n) time so this is a pretty solid algo.

As an interviewer I would say Jack's solution is pretty good. Though Altaf's solution is also correct it is not 'efficient'. Now Altaf was right at pointing out that there are problems with Jack's solution (especially for unicode) and very long strings but the catch here is that this solution is 'interesting'. It tells the interviewer that you are not thinking like an 'average' person. In fact Jack can optimize his solution by just using a bit instead of whole char or int. Doing so you can extend this even to UNICODE with not a huge amount of memory (8 KB).

Now if the interviewer had problems with the algorithm you could consider using as HASH tables. To be on a safer side you should also mention the O(n^2) solution and make a comment saying its simple but not very efficient - what to choose depends on what the algorithm is targeted for.

Jack made yet another great point - make sure you can prove formally what you claim. Giving quick but average solutions can do more harm than good.

There are a lot of problems in Altaf's code. Let me highlight. I am not trying to make anybody feel bad. I am just hoping you learn from your mistakes.

Given answer:
- If (string 1 == null || string2 == null ) then return false;
- if string1.length != string2.length then return false;
- Sort String1 and String2 in Lexographical Order (Alphabetical Order), If they both are anagrams (i-e if they both are made up of the same characters) they will become same after sorting.
- if SortedString1 == SortedString2 then return true else false;

Problems:
1. What if both are null - are they anagrams? Did you ask the interviewer? Asking good questions helps. Trust me!
2. string1.length if not precomputed can be really expensive. This takes additional O(n). If you are using java you might want to say that this is precomputed so its cheap.
3. Sorting - why sort??? As I said sorting is bad - costs O(nlogn). Try avoiding it as much as you can. If you have to use it - justify why it has to be done so.

You might find it really hard to accept some of my comments here. What you must understand is that the interviewer is simplfying a huge problem and putting it into something tangible. Now you show that you are smart by offering many solutions and telling him/her when one solution is better than the other.

Hope this helps!

Great Job Amod...Your reply is one of the most helpful replies on this website....

Hemant, try out this string with "a\0" and "b\0", your algo will return same sigma for both of these strings.

Hemant, try out this with strings "a\0" and "b\0", your algo will return same sigma for both of these strings.

Amod, Thanks. Your feedback is really helpful.

I can understand sum can be same for some values but how about we use multiplication. Let me know if this is not a right solution

int sum(char * string){

int sigma = 0,i;
for(i = 0; i < strlen(string); i++){
sigma += (string[i] - '0') * (string[i] - '0');
}

return sigma;
}

There was a small mistake in Jack's solution:
if(array[string.charAt(i)]>0) should be replaced by
if(array[string.charAt(i)]!= 0)
Because second string can have characters which first string does not and hence array can have -ve values.

---Corrected Version-----
for(int i=0;i<string1.length();i++)
{
if(array[string.charAt(i)] != 0)
return false;
}

Amod,
I could not understand how can we use bits instead of int's to store the count. A bit can only tell if the character occurs or not while clearly we need to know how many times it occurs.

Sorting-matching and occurence-counting seem to be the correct solutions to me.

Say
string1="abcd"
string2="abcde"

I presume the for loop mentioned above will be executed for string1 as well as string2. Otherwise we cannot say with surity that the two strings are anagrams.

Correct me if I am wrong.

yes, the for loop above needs to run for both strings.

To further optimize it:
if( string1.length > 256 || string2.length > 256 or string1.length + string2.length > 256 )
run the for loop for the array instead of strings.

(256 is used assuming they have only ASCII chars)

Answer : I gave my answer in following PseudoCode
- If (string 1 == null || string2 == null ) then return false;
- if string1.length != string2.length then return false;
- Sort String1 and String2 in Lexographical Order (Alphabetical Order), If they both are anagrams (i-e if they both are made up of the same characters) they will become same after sorting.
- if SortedString1 == SortedString2 then return true else false;

Jack's code will work except for two things:

1. it is important to initialize the "Array" to all zeroes
2. secondly check for "Array[s[i]] != 0", to return a negative answer
- this is because if the second string has duplicates of a character, then Array[character] will be negative

On Amod's solution:
I suspect you cannot use a bitmap representation because it cannot differentiate between 'abb' and 'aab' (and they are not anagrams). I guess you need an int array (or a byte array if one is stingy)

I dont see jack's code working because
if
s1="abcdef";
s2="rstvuw";

Jack only checks s1, he has to check both s1 and s2 and make sure array[s1[i]]==0 and array[s2[i]==0 after the updating the ascii table.

The solution is O(2n) (provided n and m are of same length) and O(1) space.

The other solution of altaf, is basically generating a hash. If the two hash values are same then they are anagrams. But as mentioned earlier you can have different string generating the same hashvalue. So the only problem with this solution is to come with a good hash function.

Cheers

Check .Net 2.0 GetHashCode function

fixed (char* text = ((char*) this))
{
char* chPtr = text;
int num = 0x15051505;
int num2 = num;
int* numPtr = (int*) chPtr;
for (int i = this.Length; i > 0; i -= 4)
{
num = (((num << 5) + num) + (num >> 0x1b)) ^ numPtr[0];
if (i <= 2)
{
break;
}
num2 = (((num2 << 5) + num2) + (num2 >> 0x1b)) ^ numPtr[1];
numPtr += 2;
}
return (num + (num2 * 0x5d588b65));
}

First of all compare length, for unequal case, they cannot be anagrams...

For equal case :-
Take a bit map and start with reading 1st string character by character. Set bit corresponding to each alphabet (e.g., 1st bit for 'a', 2nd for 'b' and so on..).
Now start reading 2nd string and for each character check whether its corresponding bit is set or not. If it is not then they are not anagrams.

This algorithm is O(m+n) in worst case and O(n+1) in best case.
Additional memory required is just a 32 bit integer.

guys u can add the ascii value multiplied by a prime number..which will be unqiue for every string....

say adz=1*1+4*5+26*(26th prime)

Using prime numbers is no doubt a creative idea. But, it has some disadvantages:

1. generating prime numbers is difficult
2. we could use generated/hard-coded prime numbers in the code, but its cumbersome (especially when the character set is unicode)
3. a separate buffer would be required to store prime numbers

Altaf's solution is most inefficient. 2 shorting effort + 1 string comparison and here to compare you have to traverse all the way of the string!

Jack's solution is best so far. 2 single pass + 1 single pass to determine if each character counter in the string is 0, may be partial if you are lucky enough! To accomodate counter for string characters, big size data type can be used in place of char in array[256]. We are not going to solve for the strings which wrap around the world.:) cheers Jack!

#include<stdio.h>
#include<stdlib.h>
void main()
{
	char str1[1000], str2[1000], len1 = 0, len2 =0, a[256];
	int flag = true;
	printf("Enter the first string\n");
	gets(str1);
	printf("Enter the second striing\n");
	gets(str2);

	for(int i =0;i<256;i++)
		a[i] = 0;

	while(str1[len1] != '\0')
	{
		a[str1[len1]] = a[str1[len1]] + 1;
		len1++;
	}

	while(str2[len2] != '\0')
	{
		a[str2[len2]] = a[str2[len2]] - 1;
		len2++;
	}
	if((len1 != len2) || str1 == 0 || str2 == 0)
	{
		printf("Not an Anagram\n");
		exit(0);
	}

	for(int i =0;i<256;i++)
		if(a[i] != 0)
		{
			printf("Not an Anagram\n");
			flag = false;
			break;
		}
	if(flag)
		printf("The two strings form an Anagram\n");	
}

Sort in nlogn time N then compare :)

Yea.. We can sort the strings and then compare them. A program for the same is below..

package myPrograms;

public class Anagrams {

public void checkAnagram(String str1, String str2) {

if (signature(str1).equals(signature(str2))) {
System.out.println(str1 + " and " + str2 + " are anagrams!");
} else {
System.out.println(str1 + " and " + str2 + " are not anagrams!");
}

}

public String signature(String str) {

if (str == null) {
return null;
}
return sortLetters(str);
}

private String sortLetters(String str) {
int length = str.length()-1;
char[] charArr = str.toCharArray();
for (int i = 0; i < length; i++) {
for (int j = 0; j < length-i; j++) {
if (charArr[j] > charArr[j + 1]) {
char temp = charArr[j];
charArr[j] = charArr[j + 1];
charArr[j + 1] = temp;
}
}
}
return new String(charArr);
}

public static void main(String[] args) {
// TODO Auto-generated method stub
Anagrams myStr = new Anagrams();
myStr.checkAnagram("leolh", "hello");
}

}

One another method I can suggest but it is little expensive.

Have each char in ascii be represented as a prime number. For example a = 2. b = 3, c = 5, d = 7, e = 11 ...etc.

str1 = CAT and str2= ACT. Both these strings always will have the same product since we are multiplying prime numbers. By checking the product of both the strings, we can determine whether it is an anagram or not.

I agree, that this is an expensive solution as multiplication of large numbers can be expensive. But it is one another approach to solve the problem.

My 2 cents.

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The function contains a histogram that stores the frequency of each
* character it encounters in the first string. For each character of the
* second string, it decrements the corresponding entry in the histogram. If
* before every decrement, the value of the histogram bucket reaches zero,
* then the two strings are not anagrams, as there is some character that has
* more occurrences in the second string than in the first string

Sort them based on ASCII value - O(NlogN). Compare them O(N). So total O(NlogN).

In python

#!/usr/bin/python

str1 = "hello"
str2 = "eollh"

for i in range(len(str1)):
    found = 0
    for k in range(len(str2)):
        #print "search ", str1[i], " in ", str2[k]
        if (str1[i] == str2[k]):
            #print "found ",  str1[i], " at ", k
            #print str2, " -> ",  str2[:k] , " + ", str2[k+1:], " k=", k
            str2 = str2[:k] + str2[k+1:]
            found = 1
            break 
    if (found == 0):
        break;
        
if (found and len(str2) == 0): print "palindrome!"
else: print "nope"