Counting occurrences of a character in String (consecutive as one)

Question

I am trying to figure out how to count the occurrences of a certain character in String such that the consecutive occurrence is considered as one. Like in string "PPPPAPAAPP" the occurrence of P as 3 and A as 2.

Possible duplicate of [Simple way to count character occurrences in a string](https://stackoverflow.com/questions/6100712/simple-way-to-count-character-occurrences-in-a-string) — vinS, Dec 09 '17 at 06:47
@LalitVerma i want to count unique occurrences as one and consecutive occurrence also one and finally add them. — Dinesh Neupane, Dec 09 '17 at 07:07
@LalitVerma i mean in the string "PPPPAPAAPP" character P is repeated 7 times but I want the consecutive repetition as 1. that means the total count for P is 3. — Dinesh Neupane, Dec 09 '17 at 07:12

score 1 · Answer 1 · answered Dec 09 '17 at 07:46

I think this will help you,

   import java.util.*;

      public class MyClass {
       public static void main(String args[]) {
             Map<String,Integer> countMap = new HashMap<String,Integer>();
             String s = "PPPPPAAPAP";
             String prev = null;
             for(int i=0;i<s.length();i++){
              String c = String.valueOf(s.charAt(i));
              if(prev==null){
                countMap.put(c,new Integer(1));
              }else{
              if(!c.equals(prev)){

                if(countMap.containsKey(c)){

                      Integer count = countMap.get(c);
                      count = count +1;
                      countMap.put(c,count);

                }
               else{

                     countMap.put(c,new Integer(1));

               }
            }
        }
        prev = c;
      }
      for(String keys :countMap.keySet()){
        System.out.println("Keys:"+keys+"\t"+countMap.get(keys));
      }
     }
    }

score 0 · Answer 2 · answered Dec 09 '17 at 07:47

0

This shows your total correctly:

 public static void main (String Args[]){
Scanner in =new Scanner(System.in);
String s =in.next();
//HashSet hd =new HashSet();
int flag=0;
int count=1;
char a=s.charAt(0);
for(int i=1;i<s.length();i++)
{
    //hd.add(s.charAt(i));
    if(a==s.charAt(i))
    {
        continue;
    }
    else
    {
        a=s.charAt(i);
        count++;
    }
}
System.out.println(count);
        }

answered Dec 09 '17 at 07:47

Lalit Verma

782
10
25

i mean total count for a single character – Dinesh Neupane Dec 09 '17 at 08:02
See the another answer than, she has already done that – Lalit Verma Dec 09 '17 at 08:02

score 0 · Answer 3 · answered Dec 09 '17 at 09:07

If you have to cound only letters or e.g. limit range of ASCII, then you could avoid using Map:

public static void main(String... args) {
    int[] letters = calcLetterGroups("PPPPPAAPAP");
    printExistedLetterGroups(letters);
}

public static int[] calcLetterGroups(String str) {
    int[] arr = new int['Z' - 'A' + 1];
    char prv = '\0';

    for (char ch : str.toUpperCase().toCharArray()) {
        if (ch >= 'A' && ch <= 'Z' && prv != ch) {
            arr[ch - 'A']++;
            prv = ch;
        }
    }

    return arr;
}

public static void printExistedLetterGroups(int... letters) {
    for (char ch = 'A'; ch <= 'Z'; ch++)
        if (letters[ch - 'A'] > 0)
            System.out.printf("%s: %d\n", String.valueOf(ch), letters[ch - 'A']);
}

Counting occurrences of a character in String (consecutive as one)

3 Answers3