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  1. Practice
  2. Algorithms
  3. Strings
  4. Anagram
  5. Discussions

Anagram

  • DIVYANSHUCEG + 7 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 + 0 comments

      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;
    }
}
    • 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 :)