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Neelakandan R
Neelakandan R

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count of letter

#java #beginners #fullstack #payilagam

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 } } }
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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 } } }
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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); } }
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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); }
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}

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]); } } } }
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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]); } } } }
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Output:m

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