How to count the number of ways of choosing of k equal substrings from a List L(the list of All Substrings)

Question

Given a string S consisting of N lowercase English alphabets.

Suppose we have a list L consisting of all non empty substrings of the string S .

I need to count the number of ways to choose exactly K equal strings from the list L (note that length of substring is not necessary to be equal to k ). 1 ≤ N ≤ 5000 1 ≤ K ≤ 10^9

Exampple:

Let S=ababa. 

As List L = {"a", "b", "a", "b", "a", "ab", "ba", "ab", "ba", "aba", "bab", "aba", "abab", "baba", "ababa"}

let k=2

The number of ways will be 7:

("a", "a")
("a", "a") 
("a", "a") 
("b", "b") 
("ab", "ab") 
("ba", "ba") 
("aba", "aba")

Similarly:

let k=3

The no of ways will be 1:

("a", "a", "a")

Answer 1

"A list of all substrings". Why would you have a list of all substrings? Let's say you have a string of one million characters, there are 500 billion substrings. The list of all substrings is not at all needed to solve the problem.

If K = 0 then there is one way. If K = 1 then there are N ways.

For k = 1 to N, each substring of length k can start at an index from 0 to N - k, that's N - k + 1 substrings. Identify the different strings and count how many there are of each using a hash table. Then for each different string that occurs n times, n >= k, add (n over K) to your count.

That's it.

You can do it faster by looking at strings of length 1 first, ignore all those where you have less than K equal strings, count the number of ways, then add another character to each and repeat. Say K = 5, you had a million characters in the string, and only two substrings of length 6 that occurred five or more times, then you only need to add characters to these two substrings.

Answer 2

Build suffix array for given string.

Walk for this array, look for common starting symbols of (at least k) neighbour siffixes.

Answer 3

Here's something in JavaScript:

function choose(n,k){
 if(k>n)return 0;if(k==0||n==k)return 1;var p=n;for(var i=2;i<=k;i++)p*=(n+1-i)/i;return p;
}

function f(str,k){
  var n = str.length,
      h = {},
      count = 0;

  for (var i=0; i<n; i++){
    var s = "";
    for (var j=i; k <= n - j + i && j < n; j++){
      s += str.charAt(j);
      if (h[s])
        h[s]++;
      else
        h[s] = 1;
    }
  }

  for (var i in h)
    count += choose(h[i],k);

  return count;
}

Output:

console.log(f("ababa",2));
console.log(f("ababa",3));

7
1

Answer 4

As other people have noticed, you don't really need the list of substrings. Because you only care about equal substrings, you only need to count how many times a substring appears, and can use a hash/dictionary/map to keep track of that. Then, the number of ways to choose exactly k equal substrings when a substring appears n times is the binomial coefficient c(n,k) . You can add up all of those binomial coefficients for each different substring, and you have your answer.

Notice that if you are asking this question for multiple k values, you only need to build the hash/dictionary/map once.

Answer 5

Without any specifics about the languages you're learning in, I believe you can accomplish this with a simple nested loop. Just compare each value to all values in the array or list.