Find the longest common starting substring in a set of strings [closed]

Answer 1

Here's an efficient solution in ruby. I based the idea of the strategy for a hi/lo guessing game where you iteratively zero in on the longest prefix.

Someone correct me if I'm wrong, but I think the complexity is O(n log n), where n is the length of the shortest string and the number of strings is considered a constant.

def common(strings)
  lo = 0
  hi = strings.map(&:length).min - 1
  return '' if hi < lo

  guess, last_guess = lo, hi

  while guess != last_guess
    last_guess = guess
    guess = lo + ((hi - lo) / 2.0).ceil

    if strings.map { |s| s[0..guess] }.uniq.length == 1
      lo = guess
    else
      hi = guess
    end
  end

  strings.map { |s| s[0...guess] }.uniq.length == 1 ? strings.first[0...guess] : ''
end

And some checks that it works:

>> common %w{ interspecies interstelar interstate }
=> "inters"

>> common %w{ dog dalmation }
=> "d"

>> common %w{ asdf qwerty }
=> ""

>> common ['', 'asdf']
=> ""

Answer 2

It's a matter of taste, but this is a simple javascript version: It sorts the array, and then looks just at the first and last items.

//longest common starting substring in an array

function sharedStart(array){
    var A= array.concat().sort(), 
    a1= A[0], a2= A[A.length-1], L= a1.length, i= 0;
    while(i<L && a1.charAt(i)=== a2.charAt(i)) i++;
    return a1.substring(0, i);
}

DEMOS

sharedStart(['interspecies', 'interstelar', 'interstate'])  //=> 'inters'
sharedStart(['throne', 'throne'])                           //=> 'throne'
sharedStart(['throne', 'dungeon'])                          //=> ''
sharedStart(['cheese'])                                     //=> 'cheese'
sharedStart([])                                             //=> ''
sharedStart(['prefix', 'suffix'])                           //=> ''

Answer 3

Ruby one-liner:

l=strings.inject{|l,s| l=l.chop while l!=s[0...l.length];l}

Answer 4

My solution in Java:

public static String compute(Collection<String> strings) {
    if(strings.isEmpty()) return "";
    Set<Character> v = new HashSet<Character>();
    int i = 0;
    try {
        while(true) {
            for(String s : strings) v.add(s.charAt(i));
            if(v.size() > 1) break;
            v.clear();
            i++;
        }
    } catch(StringIndexOutOfBoundsException ex) {}
    return strings.iterator().next().substring(0, i);
}

Answer 5

Doesn't seem that complicated if you're not too concerned about ultimate performance:

def common_substring(data)
  data.inject { |m, s| s[0,(0..m.length).find { |i| m[i] != s[i] }.to_i] }
end

One of the useful features of inject is the ability to pre-seed with the first element of the array being interated over. This avoids the nil memo check.

puts common_substring(%w[ interspecies interstelar interstate ]).inspect
# => "inters"
puts common_substring(%w[ feet feel feeble ]).inspect
# => "fee"
puts common_substring(%w[ fine firkin fail ]).inspect
# => "f"
puts common_substring(%w[ alpha bravo charlie ]).inspect
# => ""
puts common_substring(%w[ fork ]).inspect
# => "fork"
puts common_substring(%w[ fork forks ]).inspect
# => "fork"

Update: If golf is the game here, then 67 characters:

def f(d)d.inject{|m,s|s[0,(0..m.size).find{|i|m[i]!=s[i]}.to_i]}end

Answer 6

Instead of sorting, you could just get the min and max of the strings.

To me, elegance in a computer program is a balance of speed and simplicity. It should not do unnecessary computation, and it should be simple enough to make its correctness evident.

I could call the sorting solution "clever", but not "elegant".