how to return longest words from an array? - objective-c

Question

given 2 arrays wrds[] , chars[] as an input to a function such that
wrds[] = [ "abc" , "baa" , "caan" , "an" , "banc" ]
chars[] = [ "a" , "a" , "n" , "c" , "b"]
Function should return the longest word from words[] which can be constructed from the chars in chars[] array.
for above example - "caan" , "banc" should be returned
Note: Once a character in chars[] array is used, it cant be used again.
eg: words[] = [ "aat" ]
characters[] = [ "a" , "t" ]
then word "aat" can't be constructed, since we've only 1 "a" in chars[].
There are kinds of anwers online but they are not written in Objective C.Can anyone help me solve this question in OC?

Answer 1

First, walk through the word array, one word at a time, throwing out all the words that can't be formed from the second array. To do that, for each word, walk through the characters of the word, throwing out that character from the second array. If we come to a character that's not in the second array, that word can't be formed from those characters.
Now we have an array consisting solely of the words that can be formed from those characters. Now sort that array by word length, longest first. Now start walking the array, looking at the length of each word. When that length value changes, stop; you have found all the longest words.

Answer 2

// Program do to do the same in C++
#include <iostream>
#include <string>
#include <map>
#include <vector>
using namespace std;
using std::vector;
vector<char*> match(char** words, int size, char* chars, map<char,int> &second)
{
vector<char*> res;
std::map<char,int> mapi = second;
int currsize = 0;
for(int i = 0; i < size ; i++){
char* wo;
wo = words[i];
int s= 0;
for( s=0; wo[s] != '\0'; s++){
}
if(s < currsize) {
//No need to iterate if already found a bigger word
//continue to see if the next word if bigger of the same size as currsize
continue;
}
// iterate through the map to see if all the letters present in the first array
bool found = true;
for(int j = 0; j <s ; j++){
map<char, int>::iterator it = mapi.find(wo[j]);
if(it == mapi.end()) {
found= false;
break;
}
}
if(!found) {
continue;
}
if(s > currsize) {
//remove the past res as found a bigger one
res.clear();
}
//Store this word in the vector as it is one of the biggest word so far
res.push_back(wo);
currsize = s;
}
return res;
}
int main()
{
map<char, int> leters;
char* words[5] = {"adc", "baa", "caan", "daanns", "banc"};
char ch1[]= {'a', 'a', 'n', 'c', 'b'};
int chsize = sizeof(ch1);
// put the ch1 chars in a map
for(int i = 0; i < chsize; i++) {
map<char,int>::iterator it =leters.find(ch1[i]);
if(it != leters.end()) {
it->second = it->second+1;
} else {
leters.insert(make_pair(ch1[i], 1));
}
}
char* chars = ch1;
vector<char*>v = match(words, 5, ch1, leters);
for(vector<char*>::iterator it = v.begin(); it != v.end(); it++) {
// it will print the result
cout << *it << endl;
}
return 0;
}