查找字符串中最常见的字符对

Question

i have written the following function

//O(n^2)
void MostCommonPair(char * cArr , char * ch1 , char * ch2 , int * amount)
{
    int count , max = 0;
    char cCurrent , cCurrent2;
    int i = 0 , j;
    while(*(cArr + i + 1) != '\0')
    {
        cCurrent = *(cArr + i);
        cCurrent2 = *(cArr + i + 1);
        for(j = i , count = 0 ; *(cArr + j + 1) != '\0' ; j++)
        {
            if(cCurrent ==  *(cArr + j) && cCurrent2 ==  *(cArr + j + 1))
            {
                count++;
            }
        }
        if(count > max)
        {
            *ch1 = cCurrent;
            *ch2 = cCurrent2;
            max = *amount = count;
        }
        i++;
    }
}

for the following input

"xdshahaalohalobscxbsbsbs"

ch1 = b ch2 = s amount = 4

but in my opinion the function is very un efficient , is there a way to go through the string only once or to reduce the run size to O(n)?

Answer 1

Since char can hold up to 256 values, you can set up a two-dimensional table of [256*256] counters, run through your string once, incrementing the counter that corresponds to each pair of character in the string. Then you can go through the table of 256x256 numbers, pick the largest count, and know to what pair it belongs by looking at its position in the 2D array. Since the size of the counter table is fixed to a constant value independent of the length of the string, that operation is O(1) , even though it requires two nested loops.

int count[256][256];
memset(count, 0, sizeof(count));
const char *str = "xdshahaalohalobscxbsbsbs";
for (const char *p = str ; *(p+1) ; p++) {
    count[(int)*p][(int)*(p+1)]++;
}
int bestA = 0, bestB = 0;
for (int i = 0 ; i != 256 ; i++) {
    for (int j = 0 ; j != 256 ; j++) {
        if (count[i][j] > count[bestA][bestB]) {
            bestA = i;
            bestB = j;
        }
    }
}
printf("'%c%c' : %d times\n", bestA, bestB, count[bestA][bestB]);

Here is a link to a demo on ideone .

Keep in mind that although this is the fastest possible solution asymptotically (ie it's O(N) , and you cannot make it faster than O(N) ) the performance is not going to be good for shorter strings. In fact, your solution will beat it hands-down on inputs shorter than approximately 256 characters, probably even more. There is a number of optimizations that you can apply to this code, but I decided against adding them on to keep the main idea of the code clearly visible in its purest and simplest form.

Answer 2

If you want O(n) runtime you can use a hashtable (For example, Java's HashMap )

• Iterate through your string exactly once, 1 character at a time O(n)
• For each character visited, look ahead by exactly 1 more character (This is thus your character pair - just concatenate them) O(1)
• For each such character pair found, first look for it in the hashtable: O(1)
  • If it's not in the hashtable yet, add it in with the character pair as the key, and int 1 as the value (this counts the number of times you've seen it in the string). O(1)
  • If it's already in the hashtable, increment its value O(1)
• After you are done looking through the string, check the hashtable for the pair with the highest count. O(m) (where m is the number of possible pairings; m <= n necessarily)

Answer 3

Yes, you can do this in approximately linear time by keeping a running count.

Does that help?

Answer 4

Assuming by most "common pair" you mean the most common set of two sequential characters

---

At pseudo-code level you want to

 Read the first character into the "second character" register
 while(there is data)
    store the old second character as the new first character
    read the next character as the second one
    increment the count associated with this pair
 Select the most common pair

So what you need is an efficient algorythm for storing and counts associated with character pairs and finding the most common one.