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  4. Anagram
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Anagram

  • fcampinho + 9 comments

    C# code using dictionary

    int t = Convert.ToInt32(Console.ReadLine()); //int t = 1;

            for (int a0 = 0; a0 < t; a0++)
        {
            string s = Console.ReadLine();

            if (s.Length % 2 != 0) { Console.WriteLine(-1); continue; }
            Dictionary<char, int> l = new Dictionary<char, int>();

            int half = s.Length / 2;
            for (int i = 0; i < half; i++)
            {
                if (i < half)
                {
                    if (l.ContainsKey(s[i])) l[s[i]]++;
                    else l.Add(s[i], 1);
                }
            }

            int qtyDiff = 0;
            for (int j = half; j < s.Length; j++)
            {
                if (l.ContainsKey(s[j]) && l[s[j]] > 0) l[s[j]]--;
                else qtyDiff++;
            }

            Console.WriteLine(qtyDiff);
        }
    • simonzhu91 + 2 comments

      Interesting... I did it the other way: By checking the number of elements in the dictionary after the 2nd for loop, as opposed to counting the number of elements in the 2nd string that the dictionary does not contain.

      Also, I tried to experiment with a 1 loop solution, ie: count the total number of characters in the entire string, and count the number of odd occurences and divide that by 2. But that does not seem to work :(

      • DIVYANSHUCEG + 8 comments

        C solution

        int main() {

        /* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t,x;
scanf("%d",&t);
for(x=0;x<t;x++)
{
    int first[26]={0},second[26]={0};
    int count=0;
    char str[10000];
    scanf("%s",str);
    int i,k=strlen(str)-1;
     if(strlen(str)%2==1)
        count=-1;
    else
    {
        i=0;
        while(i<strlen(str)/2)
        {
            first[str[i++]-'a']++;
            second[str[k--]-'a']++;
        }
        for(i=0;i<26;i++)
        {

            if(first[i]>second[i]&&first[i]!=0)
                count+=abs(first[i]-second[i]);
        }
    }
     printf("%d\n",count);
}
return 0;

        }

        • udayshankar911 + 7 comments
              int diff = 0,t;
    t = s.nextInt();
    String str = s.next();
    int len = str.length();
    for(int k = 0; k < t;k++)
    {    
    if(len%2!=0)
        System.out.println("-1");
    else
    {
        String str1 = str.substring(0,(len/2));
        String str2 = str.substring(len/2);        
        for(int i=0;i<str1.length();i++)
        {

                if(!str2.contains(str1.charAt(i)+""))
                    diff++;                       

        }
                   System.out.println(diff);
    }
   }
          • udayshankar911 + 2 comments

            Guys TC 1 an 14 executed but others failed. when i tested in Netbeans IDE i dont find any prob. can anyone look at it and leem know the issue .

            • DanLuo + 1 comment

              if there is 2 "a" in str1, but only 1 "a" in str2, diff will not increase

              • cherry_9 + 1 comment

                can you please explain this further

                • omtonape + 1 comment

                  take second string into stringbuilder and if char from first string is present in second string remove char at that position from stringbuilder if char is not present in second string then increment count final answer is count.... thanks DanLuo

                  • mayurnagdev123 + 0 comments

                    I did something similar.Converted String to StringBuffers sb1 and sb2. Upon iterating characters through sb1: if the character is found in sb2,then change that character value in sb2 to some other value(say '/') so that it couldn't be found in further iterations. This is essential as sb1 may have repeating characters.

            • sivach + 3 comments

              Use stringbuilder for first string, remove the char from builder if exist and at the end, answer = string builder length.

              • udayshankar911 + 1 comment

                Got it. thnq :)

                • dsaxena + 0 comments

                  I understood the problem in your code but I a, little bit confused about the mentioned solution. actually I am not much familier with Java, I am currently learning, So can you please explain me the solution??

              • omtonape + 0 comments

                but in that case also mine are passing test cases 1 and 14 but rest are failing

              • andadeacu + 0 comments

                static int anagram(String s) { int result = 0; Map map = new HashMap();

                    if (s.length() % 2 != 0) {
        return -1;
    } else {
        int half = s.length() / 2;
        for (int k = 0; k < half; k++) {
            char letter = s.charAt(k);

            if (!map.containsKey(letter)) {
                map.put(letter, 1);
            } else {
                Integer frequency = map.get(letter);
                map.put(letter, ++frequency);
            }
        }

        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            System.out.println(entry.getKey() + " " + entry.getValue());
        }

        for (int k = half; k < s.length(); k++) {
            char letter = s.charAt(k);
            //System.out.println("a doua jumatate este:");
            //System.out.println(letter);

            if (!map.containsKey(letter)) {
                result++;
            } else {
                Integer freq = map.get(letter);
                freq--;
                map.put(letter, freq);

            }

        }



        for (Map.Entry<Character, Integer> entry : map.entrySet()) {
            System.out.println(entry.getKey() + " " + entry.getValue());
        }

        return result;

    }
}

                for String xaxbbbxx I obtain 0.

                Can someone one say me why?

          • 19soumikrakshi96 + 0 comments

            you should take the string input inside the testcase loop.

          • 13h61a05f3 + 0 comments

            Your code don't work for input aabacb

          • deepa_navaneeth + 1 comment

            The last test case fails. Try with xaxbbbxx.

            • iainws + 0 comments

              I failed with c++ because of map.count(element). when the element count is 0, it still registers as a hit. Somewhat confusing if you choose to use this language reference. its better to check with if(map.count(element) && map.count(element) != 0). good problem.

          • shaik_jabivulla + 0 comments

            its not full solution

          • mayurnagdev123 + 0 comments

            This wont work for: "fdhlvosfpafhalll" which is the input string in the first test case

          • knaman1009 + 0 comments

            Plz send the modified code

        • anuj19 + 0 comments

          If your if condition is first[i] >second[i] then what is the need to use abs func ?

        • saravanakumars + 1 comment

          first[str[i++]-'a']++; what happen in this step i dont understand can u please help

          • DIVYANSHUCEG + 1 comment

            converting the characters in the range of 0-26 because my array size is 26.

            • SHUBH778 + 0 comments

              So basically solution will be located at different positions ,so how did you go through it

        • rwan7727 + 1 comment

          C++:

          int q;
cin >> q;

for(int a0 = 0; a0 < q; a0++){
    string s;
    cin >> s;

    if (s.length()%2) cout << -1 << endl;
    else {
        vector <int> s1(26);
        vector <int> s2(26);
        for (int i=0;i<s.length()/2;i++) {
            s1[s[i]-'a']++;
            s2[s[i+s.length()/2]-'a']++;
        } 

        int min=0;
        for (int i=0;i<26;i++) if (s2[i]>s1[i]) min+=(s2[i]-s1[i]);
        cout << min << endl;
    }
}
          • ishujain18123 + 0 comments

            include

            include

            using namespace std; int main() {int t; char s[100000]; cin>>t; for(int i=1;i<=t;i++) {cin>>s; int coun=0,l=strlen(s); if(l%2==1){cout<<"-1"<

            } cout<

            }

            }

            what is wrong with this code?

        • hh170031061 + 2 comments

          abs was not described in the scope. please tell the meaning of that line.

          • malawadkarrahul + 0 comments
            [deleted]
          • malawadkarrahul + 0 comments

            You have not included the header file which includes the library function abs!! Which is cmath or math.h

        • s05_31468 + 0 comments

          can you explain the last if condition

        • akshar36 + 0 comments

          Really nice code..but why first[i]>second[i] .. i couldnt understand that part.. can you explain :)

        • 2019201057_anur + 0 comments

          can you explain how did you calculate count

      • jake_griffin + 0 comments

        That doesn't work because of something like "aabb". Both characters need to change, but your solution would find that there are an even number of each, so nothing needs to change. If you increment the character counts for the first half of the string and decrement for the second half, then discard the negatives and sum the positives, I think that would work.

    • souravsachdeva + 0 comments
      [deleted]
    • souravsachdeva + 0 comments
      [deleted]
    • souravsachdeva + 0 comments
      [deleted]
    • souravsachdeva + 0 comments
      [deleted]
    • raviloga316 + 0 comments

      What is this "&& l[s[j]] > 0" for?

    • sungmkim80 + 0 comments

      It seems like my C# solution is sloppier compared to this dictionary version... I am removing each character that doesn't match. Removing would decrease the length, that's where "i--" comes into play in my code.

      private static int GetChangeCountForAnagram2(string testCase)
{
	if (IsLengthOdd(testCase)) return NOT_POSSIBLE;

	// Parse test case for left and right texts
	int middleIndex = testCase.Length / 2;
	var left = testCase.Substring(0, middleIndex);
	var right = testCase.Substring(middleIndex);

	int total = 0;
	for (int i = 0; i < right.Length; i++)
	{
		var c = right[i];

		int index = left.IndexOf(c);
		if (index < 0)
		{
			right = right.Remove(i, 1);
			total++;
		}
		else
		{
			left = left.Remove(index, 1);
			right = right.Remove(i, 1);
		}

		i--;
	}

	return total;
}
    • juanmf + 1 comment

      java by counting: https://www.hackerrank.com/challenges/anagram/forum/comments/314333

      • Rishabh094 + 0 comments
        int anagram(string s){
   
    vector<int> count(26,0);
      
      if(s.length()%2==0){
        
        for(int i=0;i<s.size()/2;i++){
            count[s[i]-'a']++;
        }
    
        for(int i=s.size()/2;i<s.size();i++){
            count[s[i]-'a']--;
        }
        
        int sum = 0;
        
        for(int i=0;i<26;i++){
            sum += abs(count[i]);
        }
        
        return sum/2;
        
      }
    
   else return -1;   

}
    • yinqiankong + 0 comments

      clear solution.