AlgoPlan0605

#count good strings
##key idea:
- use maps to store counts of different characters
- loop over the string

class Solution {
    public boolean isGoodString(String s){
        Map<Character,Integer> map = new HashMap();
        for (int k = 0; k < s.length();k++){
            if (!map.containsKey(s.charAt(k))){
                map.put(s.charAt(k),1);
            }else{
                map.put(s.charAt(k), map.get(s.charAt(k))+1);
            }
        }
        for (int k : map.values()){
            if (k >1){
                return false;
            }
        }
        return true;
    }
    public int countGoodSubstrings(String s) {
        String temp = "";
        int count = 0;
        for (int k = 0; k<= s.length()-3 ; k++){
            temp =  s.substring(k, k+3);
            if (isGoodString(temp)){
                count++;
            }
        }
        return count;
    }
}


#create k palindrome strings
##key idea: 
- if the # of characters <k, cannot;
- if the # of characters who have odd counts > k, which means some of the characters cannot be placed at centers or in pair, cannot;
- all the other cases, it is plausible.

class Solution {
    public boolean canConstruct(String s, int k) {
        if (s.length() <k){
            return false;
        }
        
        Map<Character, Integer> map = new HashMap();
        char temp;
        for (int i = 0;i < s.length();i++){
            temp = s.charAt(i);
            if (map.containsKey(temp)){
                map.put(temp,map.get(temp)+1);
            }else{
                map.put(temp,1);
            }
        }
        int notoddcount=0;
        for (int count: map.values()){
            if (count %2 !=0 ){
                notoddcount++;
            }
        }
        if (notoddcount >k){
            return false;
        }
        return true;
    }
}