LeetCode 1002 - Find Common Characters

https://leetcode.com/problems/find-common-characters/

Given an array A of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates).  For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.

You may return the answer in any order.

Example 1:

Input: ["bella","label","roller"]
Output: ["e","l","l"]

Example 2:

Input: ["cool","lock","cook"]
Output: ["c","o"]

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100
3. A[i][j] is a lowercase letter

https://leetcode.com/problems/find-common-characters/discuss/247573/C%2B%2B-O(n)-or-O(1)-two-vectors

For each string, we count characters in cnt1. Then, we track the minimum count cnt.

vector<string> commonChars(vector<string>& A) {
  vector<int> cnt(26, INT_MAX);
  vector<string> res;
  for (auto s : A) {
    vector<int> cnt1(26, 0);
    for (auto c : s) ++cnt1[c - 'a'];
    for (auto i = 0; i < 26; ++i) cnt[i] = min(cnt[i], cnt1[i]);
  }
  for (auto i = 0; i < 26; ++i)
    for (auto j = 0; j < cnt[i]; ++j) res.push_back(string(1, i + 'a'));
  return res;
}

Complexity Analysis

Runtime: O(n), where n is the total number of characters.

https://leetcode.com/problems/find-common-characters/discuss/247558/Java-15-liner-count-and-look-for-minimum.

Initialize count array with Integer.MAX_VALUE, loop through the input to count the chars in each string; then find out the minimum for each char.

    public List<String> commonChars(String[] A) {
        List<String> ans = new ArrayList<>();
        int[] count = new int[26]; 
        Arrays.fill(count, Integer.MAX_VALUE);
        for (String a : A) {
            int[] t = new int[26];
            for (char c : a.toCharArray()) { ++t[c - 'a']; }
            for (int i = 0; i < 26; ++i) {
                count[i] = Math.min(t[i], count[i]);
            }
        }
        for (int i = 0; i < 26; ++i) {
             if (count[i] == Integer.MAX_VALUE) continue;
             while (count[i]-- > 0) { ans.add("" + (char)(i + 'a')); }
        }
        return ans;
    }

X. https://leetcode.com/problems/find-common-characters/discuss/247637/Java-Two-HashMap-Solution-Easy-To-Understand

The idea here is to use two HashMaps. The first, which we call union, will contain all of the common chars found in the strings compared before our current string. Then, before we go through the chars in the current string, we declare a new HashMap, called temp. temp will contain the chars that are found in our current string which are also found in union.

We have to make sure that we don't add a common char too many times. For example, if we have 2 instances of the letter c in common strings before the current, but we have three instances of c in the current string, we have to make sure that we only count 2 instances of c in common.

We do this via the following line: Math.min(union.get(curr), temp.getOrDefault(curr, 0)+1). Then, when we finish looping through the current string, we just set union to temp.

 List<String> ans = new ArrayList<String>(); 
    if(A == null || A.length == 0) return ans; 
    
    HashMap<Character, Integer> union = new HashMap<>(); 
    for(int i = 0; i < A[0].length(); i++) union.put(A[0].charAt(i), union.getOrDefault(A[0].charAt(i), 0)+1); 
    
    for(int i = 1; i < A.length; i++){
        HashMap<Character, Integer> temp = new HashMap<>(); 
        for(int j = 0; j < A[i].length(); j++){
            char curr = A[i].charAt(j); 
            if(union.containsKey(curr)) temp.put(curr, Math.min(union.get(curr), temp.getOrDefault(curr, 0)+1)); 
        }
        union = temp; 
    }
    
    for(char c : union.keySet()){
        for(int i = 0; i < union.get(c); i++) ans.add(c + ""); 
    }
    
    return ans;

X. https://leetcode.com/problems/find-common-characters/discuss/247710/Java-Solution-%3A-Using-Map
We can use iterator to remove

List result = new ArrayList<>();

    String word = A[0];
    Map<Character, Integer> common =  getChars(word);

     for(int i=1; i < A.length; i++){
        Map<Character, Integer> chars = getChars(A[i]);
        Set<Character> keysToDelete = new HashSet<>();
        for(Character c : common.keySet()){
            if(chars.containsKey(c)){
                if(chars.get(c) < common.get(c)){
                    common.put(c, chars.get(c));
                }
            }else{
                keysToDelete.add(c);
            }
        }
        for(Character c : keysToDelete){
            common.remove(c);
        }
    }


    for(Character c : common.keySet()){
        int length = common.get(c);
        for(int i=0; i < length; i++){
            result.add(Character.toString(c));
        }
    }
    
    return result;
}

private Map<Character, Integer> getChars(String word) {
    Map<Character, Integer> map = new HashMap<>();
    for(char c : word.toCharArray()){
        map.put(c, map.getOrDefault(c, 0) + 1);
    }
    return map;
}