package array;
public class Character_count {
public static void main(String[] args) {
String s="Gunasekaran";
char[]c=s.toCharArray();
for(int j=0;j<c.length;j++) {
char key=c[j];
int count=1;
if (key == '*') {
continue;
}
for(int i=j+1; i<c.length; i++) {
if(key==c[i]) {
c[i]='*';
count++;
}
}
System.out.println(key+"appears"+ count);
}
}
}
.........................................................................
OUTPUT:
G Appears 1 Times
u Appears 1 Times
n Appears 2 Times
a Appears 3 Times
s Appears 1 Times
e Appears 1 Times
k Appears 1 Times
r Appears 1 Times
FINDING THE CHARACTER WHICH HAS THE ONE COUNT VALUE
package array;
public class Character_count {
public static void main(String[] args) {
String s="Gunasekaran";
char[]c=s.toCharArray();
for(int j=0;j<c.length;j++) {
char key=c[j];
int count=1;
if (key == '*') {
continue;
}
for(int i=j+1; i<c.length; i++) {
if(key==c[i]) {
c[i]='*';
count++;
}
}
if (count == 1) { // using this condition for printing character has one count value
System.out.println(key+ " Appears "+ count+ " Times");
}
}
}
}
.........................................................................
OUTPUT:
G Appears 1 Times
u Appears 1 Times
s Appears 1 Times
e Appears 1 Times
k Appears 1 Times
r Appears 1 Times
Top comments (0)
Subscribe
For further actions, you may consider blocking this person and/or reporting abuse
Top comments (0)