LeetCode Find Words That Can Be Formed by Characters Solution Explained - Java

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[diwakarsharma]

Your style of deliver the enlightenment coding wisdom is Thespian level
Thankyou

[jugsma]

This is my solution in Python:

def find_words_with_chars(words, chars):
if len(words)==0 or len(chars)==0:
return 0
max_count = 0
for word in words:
tmp_chars = copy.deepcopy(chars)
tmp_chars = list(tmp_chars)
cnt = 0
for char in word:
if char in tmp_chars:
idx = tmp_chars.index(char)
tmp_chars[idx] = ''
cnt += 1
else:
cnt = 0
break
max_count += cnt
return max_count

[amanjakhar]

thanks man! really helpful, can you help with medium level questions because the problem lies in the medium ones

[harshnist]

similar approach using array list
class Solution {
public int countCharacters(String[] words, String chars) {
int count=0;
ArrayList<Character>list =new ArrayList<>();
for(int i=0;i<chars.length();i++)
{
list.add(chars.charAt(i));
}

for(int i=0;i<words.length;i++)
{
String s=words[i];
int k=0;
List<Character> cloned_list
= new ArrayList<Character>();
cloned_list.addAll(list);
for(int j=0;j<s.length();j++)
{


{

k++;

}

}

if(k==s.length())
{
count =count+s.length();
}

}

return count;

}
}

[egorshalnev]

thanks man, this is helpful, but your hand clapping killed my earbuds.

[unknownman]

class Solution {
public int countCharacters(String[] words, String chars) {
int result = 0;
HashMap<Character, Integer> mymap = new HashMap<>();

for (char c : chars.toCharArray()) {
mymap.put(c, mymap.getOrDefault(c, 0) + 1);
}

for (String word : words) {
// Create a copy of the original frequency map for each word

if (canbeformed(word, new HashMap<Character, Integer>(mymap))) {
result += word.length();
}
}

return result;
}

private boolean canbeformed(String word, Map<Character, Integer> copymap) {
for (char ch : word.toCharArray()) {
if (!copymap.containsKey(ch) || copymap.get(ch) < 1) {
return false;
}
copymap.put(ch, copymap.get(ch) - 1);
}
return true;
}
}