Checking if a word is contained within an array - c++

Question

I want to check for a word contained within a bigger string, but not necessarily in the same order. Example: The program will check if the word "car" exists in "crqijfnsa". In this case, it does, because the second string contains c, a, and r.

Answer 1

You could build a map containing the letters "car" with the values set to 0. Cycle through the array with all the letters and if it is a letter in the word "car" change the value to 1. If all the keys in the map have a value greater than 0, than the word can be constructed. Try implementing this.

Answer 2

An anagram is a type of word play, the result of rearranging the letters of a word or phrase to produce a new word or phrase, using all the original letters exactly once;
So, actually what you are looking for is an algorithm to check if two words are "Anagrams" are not.
Following thread provides psuedocode that might be helpful
Finding anagrams for a given word

Answer 3

A very primitive code would be something like this:
for ( std::string::iterator it=str.begin(); it!=str.end(); ++it)
for ( std::string::iterator it2=str2.begin(); it2!=str2.end(); ++it2) {
if (*it == *it2) {
str2.erase(it);
break;
}
}
if (str2.empty())
found = true;

Answer 4

You could build up a table of count of characters of each letter in the word you are searching for, then decrement those counts as you work through the search string.
bool IsWordInString(const char* word, const char* str)
{
// build up table of characters in word to match
std::array<int, 256> cword = {0};
for(;*word;++word) {
cword[*word]++;
}
// work through str matching characters in word
for(;*str; ++str) {
if (cword[*str] > 0) {
cword[*str]--;
}
}
return std::accumulate(cword.begin(), cword.end(), 0) == 0;
}
It's also possible to return as soon as you find a match, but the code isn't as simple.
bool IsWordInString(const char* word, const char* str)
{
// empty string
if (*word == 0)
return true;
// build up table of characters in word to match
int unmatched = 0;
char cword[256] = {0};
for(;*word;++word) {
cword[*word]++;
unmatched++;
}
// work through str matching characters in word
for(;*str; ++str) {
if (cword[*str] > 0) {
cword[*str]--;
unmatched--;
if (unmatched == 0)
return true;
}
}
return false;
}
Some test cases
"" in "crqijfnsa" => 1
"car" in "crqijfnsa" => 1
"ccar" in "crqijfnsa" => 0
"ccar" in "crqijfnsac" => 1

Answer 5

I think the easiest (and probably fastest, test that youself :) ) implementation would be done with std::includes:
std::string testword {"car"};
std::string testarray {"crqijfnsa"};
std::sort(testword.begin(),testword.end());
std::sort(testarray.begin(),testarray.end());
bool is_in_array = std::includes(testarray.begin(),testarray.end(),
testword.begin(),testword.end());
This also handles all cases of duplicate letters correctly.
The complexity of this approach should be O(n * log n) where n is the length of testarray. (sort is O(n log n) and includes has linear complexity.