Common Characters

May 14, 2019

Today’s exercise comes from a coding challenge site:

Given a list of words containing only lower-case letters, return a list of characters that appear in all the words, with multiplicity. For instance, given the words BELLA, LABEL and ROLLER, the characters E, L and L appear in all three words.

Your task is to return a list of characters that appear in all words of an input list. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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Posted by programmingpraxis
Filed in Exercises
3 Comments »

3 Responses to “Common Characters”

  1. Globules said
    May 15, 2019 at 2:59 AM

    A Haskell version.

    import Data.List (foldr1, intersect)

common :: Eq a => [[a]] -> [a]
common []  = []
common xss = foldr1 intersect xss

main :: IO ()
main = do
  print $ common ([] :: [String])
  print $ common ["BELLA"]
  print $ common ["BELLA", "LABEL"]
  print $ common ["BELLA", "LABEL", "ROLLER"]

    $ ./commchar
""
"BELLA"
"BELLA"
"ELL"
  2. Paul said
    May 17, 2019 at 5:49 AM

    In Python.

    from collections import Counter
from functools import reduce
from operator import and_

def common(listofwords):
    return list(reduce(and_, (Counter(w) for w in listofwords)).elements())

print(common(["BELLA", "LABEL"]))  # -> ['B', 'E', 'L', 'L', 'A']
print(common(["BELLA", "LABEL", "ROLLER"]))  # -> ['E', 'L', 'L']
  3. Daniel said
    June 30, 2019 at 1:48 AM

    Here’s a solution in C.

    #include <stdio.h>
#include <stdlib.h>

// Returns the result in s1
void common(char* s1, char* s2) {
  int a1[26] = {0};
  int a2[26] = {0};
  char* _s1 = s1;
  for (; *s1; ++s1) {
    ++(a1[*s1 - 'a']);
  }
  for (; *s2; ++s2) {
    ++(a2[*s2 - 'a']);
  }
  int pos = 0;
  for (int i = 0; i < 26; ++i) {
    int count = a1[i] < a2[i] ? a1[i] : a2[i];
    for (int j = 0; j < count; ++j) {
      _s1[pos++] = 'a' + i;
    }
  }
  _s1[pos] = '\0';
}

int main(int argc, char* argv[]) {
  if (argc == 1) return EXIT_FAILURE;
  for (int i = 1; i < argc; ++i) {
    common(argv[1], argv[i]);
  }
  printf("%s\n", argv[1]);
  return EXIT_SUCCESS;
}

    Example Usage:

    $ ./a.out bella label roller
ell

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