(Easy) Can Make Palindrome - LeetCode Contest (in progress)

Description:

Given a string s, we make queries on substrings of s.

For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return an array answer[], where answer[i] is the result of the i-th query queries[i].

Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters. (Also, note that the initial string s is never modified by any query.)

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0] : substring = "d", is palidrome.
queries[1] : substring = "bc", is not palidrome.
queries[2] : substring = "abcd", is not palidrome after replacing only 1 character.
queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.

Constraints:

1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s only contains lowercase English letters.

Solution:

Attempt1 :

Failed one test case , time out exception:

class Solution {
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        
        if(queries==null||s==null ||s.length()==0||queries.length==0||queries[0].length==0){
            return null;
        }
        
        List<Boolean> list = new ArrayList<Boolean>();
        
        for(int i = 0; i<queries.length;i++){
            
            String tmp = "";
           
                int st = queries[i][0];
                int en = queries[i][1];
                int k = queries[i][2];
                
                boolean test = isPalindrome2(s.substring(st,en+1),k);
            
                list.add(test);
            
            
        }
        
        return list;
        
    }
    
    /*
    public boolean isPalindrome(String s, int k){
        
        int begin = 0; 
        int end = s.length()-1;
        int times =0;
        while(begin<end){
            if(s.charAt(begin)!= s.charAt(end)){
                if(times <k){
                    times = times+1;
                }
                else{
                    return false;
                }
            }
            begin++;
            end--;
        }
        return true;
    }
    */
    
    public boolean isPalindrome2(String s, int k){
        
        int count =0;
        
       HashMap<Character, Integer> map = new HashMap<>();

        
        for(int i = 0; i<s.length(); i++){
            
            if(!map.containsKey(s.charAt(i))){
                
                int tmp  = Count(s,s.charAt(i));
                
                if(tmp%2!=0){
                    map.put(s.charAt(i),tmp);
                }
            }
            
        }
        
       System.out.println(s+" "+ s.length()+" "+map.size());
    
        if(s.length()%2==0){
            
            if(map.size()/2<=k)
                return true;
            
        }
        else{
            
            if(map.size()/2-1<k){
                return true;
            }
        }
        
       
       return false; 
        
    }
    
    public int Count(String s, char a){
        int count = 0; 
        for(int i =0; i<s.length(); i++){
            if(s.charAt(i)==a){
                count++;
            }
        }
        return count;
    }
}

原文地址：https://www.cnblogs.com/codingyangmao/p/11441909.html