如何证明两个字符串是彼此的排列？

Question

I am doing cracking the code interview book and I came across the question in the arrays and strings chapter where they're asking to write a method that proves that two strings given as input are permutations of each other.

The answers in the book are pretty clean and clear. One is to sort, and then compare if they're identical, and the other is to check if the two strings have identical character counts.

However, I had a different approach for this problem, and I wanted to share it with you to see your opinion.

I am making the assumption that the characters are ASCII characters. So what I was thinking of is first check if the lengths are equal for both strings, if not we directly return false because obviously it opposes the definition of permutations.

If it's the case we proceed with the algorithm. First, we initialize:

int sum = 0;
int sum1 = 0;

Then we go through the character of each string adding the ASCII value of each character to the sum and comparing the sums in the end. If they're equal, then we got ourselves a permutation.

Does this approach work?

Answer 1

No, it doesn't work, because 12 is both the sum of 2 and 10 and the sum of 3 and 9 .

With your algorithm "ad" would be a permutation of "bc" .

In the general case, if you allow a reasonable range of characters and string lenght, there's no real shortcut. The best solution among the two you mention depends on the language.

Answer 2

dystroy is right

to get it work at 99.999% correctness (by your approach) you compute:

sum1 = sum (ASCII(i))
sum2 = sum (ASCII(i)^2)
sum3 = sum (ASCII(i)^3)

• for both strings and if all of the same powered sum is the same
• then you have most likely permutated string ...

to be sure compare histograms (as you mentioned in question) but that need more memory ...

Answer 3

Your approach won't work because there will be lots of collisions for sums ie basically what you are assuming is 5+3 = 8 and there is no other combination that would produce 8 but you are wrong example 4 + 4 is also 8 .

There are many Ad hoc methods to solve this problem i am going to describe two of them . You can use prime numbers instead to solve the problem by method similar to yours or simply allocate 2 arrays and keep record of the characters.

1. You can initialize 2 integer arrays of size 27 each say list1[27] and list2[27] initialized as 0 , read both the strings character by character say if you read 'c' from string 1 , increment the 3rd element of the list1 because 'c' is the third character and so on and when you are done reading both the string scan both the arrays for mismatch if there is any mismatch they are not permutations of each other.

A possible implementation can be

char str1[50]="permutation";
char str2[50]="importunate";

int list1[27]={0},list2[27]={0};


for(int i=0;i<11;i++){
    list1[(int)str1[i]-(int)'a'+1]++;
    list2[(int)str2[i]-(int)'a'+1]++;
}

for(int i=0;i<=27;i++){
    if(i==27){
        return true;
    }
    if(list1[i]!=list2[i])
    {
        return false;
    }
}

this method can be easily extended to consider spaces , different case characters and digits .

2. This method is similar to what you have done but instead of using ASCII values it uses prime numbers and instead of addition it uses multiplication .Problem with your method was lot's of possible collisions as dystroy pointed out if you chose to multiply instead you will again face the same problem but what if instead of multiplying ascii values we multiply prime numbers assigned to a particular character.

here we first allocate an array which stores first 26 prime numbers starting from 2 , and read strings character by character and multiply all the respective prime numbers assigned to each character of the string we finally compare the two large integer numbers and if those are equal then the strings are permutations of each other

A possible implementation could be

int arr[27]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103};
char str1[50]="permutation";
char str2[50]="importunate";

int prd1=1,prd2=1;


for(int i=0;i<11;i++){
    prd1=prd1*arr[(int)str1[i]-(int)'a'];
    prd2=prd2*arr[(int)str2[i]-(int)'a'];
}

if(prd1==prd2)
    return true;

else
    return false;

This method is not much extendable as the first one because numbers grow big with the length of string , we can

prd1=prd1*arr[(int)str1[i]-(int)'a']%1000000009;
prd2=prd2*arr[(int)str2[i]-(int)'a']%1000000009;//or some other large prime number 

Answer 4

This cannot be achieved using sum, since a number doesn't have unique sum-factors (as mentioned by previous answers)

This can be done by comparing character-histograms

Code Java

class Character_Histogram
{
    public Map<Character,Integer> histogram;

    public Character_Histogram ()
    {
        histogram = new TreeMap<Character,Integer> ();
    }

    public void count (Character c)
    {
        if (histogram.containsKey(c))
            histogram.put(c, histogram.get(c)+1);
        else
            histogram.put(c, 1);
    }

    public void count (String str)
    {
        for(char c : str.toCharArray())
            count(new Character(c));
    }
}