1) Frequency of Each letter in a given String(key != '*')
2) Non-repeated char. in a given String(count==1)
3) Repeated Char. in a given String(count>1)
4) First Repeated Char. in a given String(count>1) use break(First Repeated Char)
5) Last repeated char. in a given String(char last=' ');
ex :
if ((key != '*')&&(count>1))// if condition don't print repeated letter
{
last=key;
}
6) First Non-repeated char. in a given String(count==1)---break;
7) Last non-repeated char. in a given String(char last=' ');
ex :
if ((key != '*')&&(count==1))// if condition don't print repeated letter
{
last=key;
}
8) Most frequent letter in a given String(code at last)
9)finding same letter or numb near by near(letter)
count of letter (or) Frequency of Each letter in a given String
package afterfeb13;
public class findchar {
public static void main(String[] args) {
char[] s = { 'n', 'e', 'e', 'l', 'a', 'k', 'a', 'n', 'd', 'a', 'n' };
for (int j = 0; j < s.length; j++)// selects each character from the array one by one.
{
char key = s[j];// s[j] represent all char because j++
int count = 1;// frist char
for (int i = j+1; i < s.length; i++) //Why j + 1?// Avoid Repetition: you don’t want to compare it with itself,
//and you don’t want to count the same 'n' multiple times.
{
if (key == s[i]) {
s[i] = '*';//mark for repeated letter
count++;// inside each how many time
}
}
if (key != '*')// if condition don't print repeated letter
System.out.println(key + " appears " + count);//print letter not repeated
}
}
}
Output:
n appears 3
e appears 2
l appears 1
a appears 3
k appears 1
d appears 1
convert string to char and find Frequency of Each letter in a given String
package afterfeb13;
public class findchar {
public static void main(String[] args) {
String c = "neelakandan";
char[] s=c.toCharArray();// string to char
for (int j = 0; j < s.length; j++)// selects each character from the array one by one.
{
char key = s[j];// s[j] represent all char because j++
int count = 1;// frist char
for (int i = j+1; i < s.length; i++) //Why j + 1?// Avoid Repetition: you don’t want to compare it with itself,
//and you don’t want to count the same 'n' multiple times.
{
if (key == s[i]) {
s[i] = '*';//mark for repeated letter
count++;// inside each how many time
}
}
if (key != '*')// if condition don't print repeated letter
System.out.println(key + " appears " + count);//print letter not repeated
}
}
}
Output:
n appears 3
e appears 2
l appears 1
a appears 3
k appears 1
d appears 1
change 1 :if ((key != '*')&&(count==1)) char present one time(non duplicate char)
Output:
l appears 1
k appears 1
d appears 1
change 2 :if ((key != '')&&(count==3)) char present three time(duplicate char))(if ((key != '')&&(count>1)))
Output:
n appears 3
e appears 2
a appears 3
change 3 :if ((key != '*')&&(count==1)) {break;}char present one time and first char(non duplicate char) break should apply one char
Output:l appears 1
find last char
package afterfeb13;
public class findchar {
public static void main(String[] args) {
String c = "neelakandan";
char last=' ';
char[] s=c.toCharArray();// string to char
for (int j = 0; j < s.length; j++)// selects each character from the array one by one.
{
char key = s[j];// s[j] represent all char because j++
int count = 1;// frist char
for (int i = j+1; i < s.length; i++) //Why j + 1?// Avoid Repetition: you don’t want to compare it with itself,
//and you don’t want to count the same 'n' multiple times.
{
if (key == s[i]) {
s[i] = '*';//mark for repeated letter
count++;// inside each how many time
}
}
if ((key != '*')&&(count>1))// if condition don't print repeated letter
{
last=key;
}
}System.out.println(last);
}
}
Output:
a
Most frequent letter in a given String
package afterfeb13;
public class findchar {
public static void main(String[] args) {
String c = "neelakandan";
char last=' ';
char max_char=' ';
int max =0;
char[] s=c.toCharArray();// string to char
for (int j = 0; j < s.length; j++)// selects each character from the array one by one.
{
char key = s[j];// s[j] represent all char because j++
int count = 1;// frist char
for (int i = j+1; i < s.length; i++) //Why j + 1?// Avoid Repetition: you don’t want to compare it with itself,
//and you don’t want to count the same 'n' multiple times.
{
if (key == s[i]) {
s[i] = '*';//mark for repeated letter
count++;// inside each how many time
}
}
if ((key != '*')&&(count>1))// if condition don't print repeated letter
{
if(count>max)
{
max=count;// max char
max_char=key;//max char
}
}
}System.out.println(max +" times "+max_char);
}
}
Output:
3 times n
finding same letter or numb near by near(letter)
package afterfeb13;
public class newpartice {
public static void main(String[] args) {
int[] num = { 0, 1, 1, 0 };
for (int j = 0; j < num.length - 1; j++) {
if (num[j] == num[j + 1])
{
System.out.println(num[j]);
}
}
}
}
output:1
String amma,appa,chennai
package afterfeb13;
public class newpartice {
public static void main(String[] args) {
String s = "amma";
char[] c = s.toCharArray();
for (int j = 0; j < c.length - 1; j++) {
if (c[j] == c[j + 1])// c=h//h=e//e=n//n=n//take n==n
{
System.out.println(c[j]);
}
}
}
}
Output:m
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