1657.Determine if Two Strings Are Close

Intuition

in this question we have to check 3 conditions

1. all the character of word1 should be present on word2 and vice-versa

2. the character frequency count should be equal

Approach

in this question we have to check 3 conditions

1. all the character of word1 should be present on word2 and vice-versa

2. the character frequency count should be equal example: w1 = abbccdde, w2 = eeaacbbd

Here for w1: 1 times of character = 2(a,e) 2 times of character = 3(b,c,d)

and in w2: 1 times of character = 2(c,d) 2 times of character = 3(e,a,b)

So In w1 and w2 both the freq of characters are equal, hence it is valid answer.

Complexity

• Time complexity:

First check : To check every character is present on both words will take O(n) time Second Check : checkAnotherCondition this method is also taking O(n) so the overall time complexity is O(n);

• Space complexity:

O(1)

Solution

class Solution {
    public boolean closeStrings(String word1, String word2) {
        if(word1.length()!=word2.length())
          return false;

        return checkAns(word1,word2)&&checkAns(word2,word1)&&checkAnotherCondition(word1,word2);
        
    }
    public boolean checkAns(String word1,String word2){
        HashMap<Character,Boolean> map = new HashMap<>();
        for(int i=0;i<word1.length();i++){
            map.put(word1.charAt(i),true);
        }
        for(int i=0;i<word2.length();i++){
            if(!map.containsKey(word2.charAt(i))){
                return false;
            }
        }
        return true;
    }
    public boolean checkAnotherCondition(String word1,String word2){
        HashMap<Integer,Integer> map1 = getValue(word1);
        HashMap<Integer,Integer> map2 = getValue(word2);
        for(Map.Entry<Integer,Integer> map : map1.entrySet()){
            if(map2.get(map.getKey())!=map.getValue())
               return false;
        }
        return true;
    }

    public HashMap<Integer,Integer> getValue(String word){
        HashMap<Character,Integer> freq = new HashMap<>();
        for(int i=0;i<word.length();i++){
           char c = word.charAt(i);
           freq.put(c,freq.getOrDefault(c,0)+1);
        }
        HashMap<Integer,Integer> values = new HashMap<>();
        for(Map.Entry<Character,Integer> ans:freq.entrySet()){
            int val = ans.getValue();
            values.put(val,values.getOrDefault(val,0)+1);
        }
        return values;
    }
}