Printing characters with their count in a String

Total number of Character in ASCII table is 256 (0 to 255). 0 to 31(total 32 character ) is called as ASCII control characters (character code 0-31). 32 to 127 character is called as ASCII printable characters (character code 32-127). 128 to 255 is called as The extended ASCII codes (character code 128-255)

There are multiple solution available with O(n) time and space complexity using Set,Map , O(n2) using Brute force iterating each character element of a String.

Solution 1 : 

But I am representing a solution with O(n) time complexity and In place O(1) space complexity as follow : 

/**


 * @author sachin


 *


 */


public class CountDuplicates {


 public static void charsCount(String str){

  int[] intArr = new int[256];

  char[] charArrInput =  str.toCharArray();

  int charLength = charArrInput.length;

  for (int i = 0; i < charLength; i++) {

   intArr[charArrInput[i]]++;

  }

  int j =0;

  for (int c : intArr) {

   if(c > 0){

    System.out.println( "char=" + (char)j + " and count=" + c );

   }

   j++;

  }

 }

 public static void main(String[] args) {

 String a = "My Name Is Sachin Srivastava and I am a Java Architect passionate "

 + "about writing,blogging & coding , author of tek9g.blogspot.com and vedology.blogspot.com";

 charsCount(a);

 }

}


Output : 

char=  and count=21

char=& and count=1

char=, and count=2

char=. and count=4

char=9 and count=1

char=A and count=1

char=I and count=2

char=J and count=1

char=M and count=1

char=N and count=1

char=S and count=2

char=a and count=15

char=b and count=4

char=c and count=6

char=d and count=4

char=e and count=5

char=f and count=1

char=g and count=9

char=h and count=3

char=i and count=8

char=k and count=1

char=l and count=4

char=m and count=4

char=n and count=7

char=o and count=14

char=p and count=3

char=r and count=4

char=s and count=6

char=t and count=10

char=u and count=2

char=v and count=4

char=w and count=1

char=y and count=2

 

Solution 2 :

Using a HashMap : O(n) time and space complexity

       

  Map<Character, Integer> map = new HashMap<>();

  char[] charArrInput =  str.toCharArray();

  for (char c : charArrInput) {

   if(!map.containsKey(c)){

    map.put(c, 1);

   } else {

    map.put(c, map.get(c)+1);

   }

  }

  for (Entry<Character, Integer> entry : map.entrySet()) {

   System.out.println("char=" + entry.getKey() + " and count=" + entry.getValue());

  }
Output: 

 

char=A and count=1

char=I and count=2

char=J and count=1

char=M and count=1

char=N and count=1

char=S and count=2

char=  and count=21

char=a and count=15

char=b and count=4

char=c and count=6

char=d and count=4

char=e and count=5

char=& and count=1

char=f and count=1

char=g and count=9

char=h and count=3

char=i and count=8

char=k and count=1

char=, and count=2

char=l and count=4

char=m and count=4

char=n and count=7

char=. and count=4

char=o and count=14

char=p and count=3

char=r and count=4

char=s and count=6

char=t and count=10

char=u and count=2

char=v and count=4

char=w and count=1

char=y and count=2

char=9 and count=1
     
 

Solution 3 :   Using brute force in the O(n2)   , basically iterating each  n element and comparing and counting remaining  n-1 element  making it quadratic solution with respect to time.

Solution 3 is not recommended , Solution 1 is the best solution and if one not able to do like solution1 then Solution 2 would be simple to use.

Enjoy tek9g  blog and coding !!!!!!!!!