Collections.frequency() method might be useful in this case.

public class test
{
public static void main(String[] args)
{
String str = "ababccbdde";
int i,j,k;
char[] ch = str.toCharArray();
int len=ch.length;
for ( i = 0; i < len; i++) {
int counter = 0;
char c=str.charAt(i);
for ( j = 0; j < len; j++) {
if (c==ch[j]){
counter++;
ch[j]='\u0000';
}
}
if(counter>0)System.out.println(c+"="+counter);

}
}
}

package selfpractice.array;

public class DistinctArray2 {

static String str = "sgsgsgsgs";
static String distinct = "";
static char[] charary = str.toCharArray();
static StringBuilder hello = new StringBuilder("");
static char[] tempcharary = new char[charary.length];
static int i;
static int count;
public static char[] getDistinctLetter()
{
int incr = 0;

for ( i = 0; i < charary.length; i++)
{
if (!(DistinctArray2.isletteroccure(charary[i])))
{

tempcharary[incr] = charary[i];
hello.append( tempcharary[incr]);
++incr;


}
}

char[] distinctresult = hello.toString().toCharArray();

for (int j = 0; j < distinctresult.length; j++)
{
for (int k = 0; k < charary.length; k++)
{
if (distinctresult[j] == charary[k])
{
++count;
}

}

System.out.println("LETTER " + tempcharary[j] + " OCCURE " + count + " TIMES ");
count = 0;

}

return tempcharary;

}



public static Boolean isletteroccure(char ch)
{
for (int j = 0; j < charary.length; j++)
{
if (ch == tempcharary[j])
{
return true;
}
}

return false;
}
public static void main(String[] args) {

DistinctArray2.getDistinctLetter();


}
}

Program to accept the string from command line and display
1. No. of words in string
2. No. of characters in each word
3. No. of characters in string

public class Str
{
public static void main(String args[])
{
int ch=0;
System.out.println("String is = ");
for(int i=0;i<args.length;i++)
{
System.out.println(args[i]);
}
System.out.println("Number of words in the string = "+args.length);
for(int i=0;i<args.length;i++)
{
String s=args[i];
ch=s.length()+ch;
System.out.println("Number of characters in -"+args[i]+"- = "+s.length());
}
System.out.println("No. of characters in string = "+ch);
System.out.println("No. of characters in string with white spaces ="+(ch+(args.length-1)));
}
}

package String;
import java.util.Scanner;
public class Reverse {
@SuppressWarnings("resource")
public static void main(String[] args) {
int charCount = 0;
System.out.println("Enter the string:");

Scanner in = new Scanner(System.in);
String str = in.nextLine();
for(char ch = 'a' ; ch <= 'z' ; ch++ )
{
for(int i =0 ; i<str.length()-1; i++)
{
if(str.charAt(i) == ch)
{
charCount++;
}
}
if(charCount!=0){
System.out.println(ch+"="+charCount);
charCount=0;
}
}
}
}

public class Binary_Octal {

Scanner scan;
int num;

void getVal() {

System.out.println("Binary to Octal");
scan = new Scanner(System.in);

System.out.println("\nEnter the number :");
num = Integer.parseInt(scan.nextLine(), 2);
}

void convert() {

String octal = Integer.toOctalString(num);

System.out.println("Octal Value is : " + octal);
}
}

class MainClass {

public static void main(String args[]) {

Binary_Octal obj = new Binary_Octal();

obj.getVal();
obj.convert();
}
}


Sample Output

Binary to Octal

Enter the number :
1010
Octal Value is : 12

import java.util.Scanner;

public class count {
public static void main(String[] args) {

Scanner scn = new Scanner(System.in);
System.out.println("Enter String");
String str = scn.nextLine();
System.out.println("Enter charcter to know its occurrence");
char c = scn.next().charAt(0);
int count =0;
for(int i=0;i<str.length();i++){
char ch = str.charAt(i);
if(ch==c){
count++;
}
}
System.out.println("count is:"+count);
}
}

Hello Vandana, time complexity of your solution is O(n) as it need to iterate through array, can you better it?

Scanner s = new Scanner(System.in);
System.out.println("Enter some sentence");
String sentence = s.nextLine();
System.out.println("Enter word to to follow ");
char word = s.next().charAt(0);
int counter =0;
char[] chars = sentence.toCharArray();

for (int i = 0; i < chars.length; i++) {
System.out.print("| " + i + " |");
System.out.println(" " + chars[i] + " |");
System.out.println("-------------");
}

for (int i = 0; i < chars.length; i++) {

if(sentence.charAt(i) == word){
counter++;
}
}
System.out.println("Word " + word + " appears in sentence " + counter + " times");


s.close();

count the no of characters in a string

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
String s= " hello srikanth";
int count=0;;
for(int i=0;i<s.length();i++)
{boolean b=Character.isWhitespace(s.charAt(i));


if(b==false)
{
count++;
}
}

System.out.println(count);
}
}

id this fine or not

static int c = 0;

public static void main(String[] args) {
String s = "aabbcfasdgalsgmvxnxbcsasfweasdgkandjkakjgaasdg";
char[] chars = s.toCharArray();
Arrays.sort(chars);
String sortedString = new String(chars);

count(sortedString);
}

public static void count(String str) {
char[] chars = str.toCharArray();
Set set = new HashSet();
for(int i = 0; i < chars.length; i ++) {
set.add(chars[i]);
}
List list = new ArrayList();
for (Character string : set) {
list.add(string);
int occ = StringUtils.countOccurrencesOf(str, string.toString());
System.out.print(string.toString() + occ);
}
}

Use regular expressions

Hello @Matt, can you please share the regular expression to count occurrence of any character in string, that will help better ..

Hello All, I m a beginner in java . I need to write a program for below:
Given a Sentence , i need to color a sentence using minimum no. of Consonants. Each word in sentence can be colored by a single consonant
ex: Hello How are you?
Minimum no. of consonant to color sentence : 2
i.e using o,e ( o occurs in 3 words: Hello,How ,you & e in are )
therefore answer is 2


Please help

str.split(c).length() - 1

String s="helppp";
int c=0;
char o=0;

for (int i = 0; i 0)
{
System.out.println("char is repeated "+o+" at "+c+" times");
}
else
{
System.out.println("no char is repeated");
}

import java.util.HashMap;
import java.util.Map;
import java.util.Set;

public class OccurenceofChar {

public static void main(String[] args) {
// TODO Auto-generated method stub
Occurence occur=new Occurence();
occur.Occurence("Today is Monday", 'o');

}

}
class Occurence{
public static int Occurence(String word, char letter) {
Map map=new HashMap<>();
char c[]=word.toCharArray();
for(Character ch:c) {
if(map.containsKey(ch)) {
map.put(ch,map.get(ch)+1);
}
else {
map.put(ch, 1);
}
}
Set> entryset=map.entrySet();
int occurence=0;
for(Map.Entry entry: entryset) {
if(entry.getKey()==letter) {
occurence=entry.getValue();
}
}
System.out.println(occurence);

return occurence;
}
}

Nice solution, what are the test cases you have thought? Can you explain the Big(O) for space and time?

Wow surprised no one solved this recursively. Here you go with binary search:

public static void main(String[] args) {
int x = stringCounter("abacaaadeaac");
System.out.println(x);
}


public static int stringCounter(String string){

if(string.length() < 2){
if(string.contains("a")){
return 1;
}
return 0;
}

return stringCounter(string.substring(0,string.length()/2)) + stringCounter(string.substring(string.length()/2));


}
}

Recursive solution is good idea but I guess you need one more parameter to count occurence of a given character, right now you are just counting "a"

import java.util.*;
public class solution {
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
String in=s.nextLine();
char c[]=in.toCharArray();
List b=new ArrayList();
for(char addchar:c)
{
b.add(addchar);
}
int g=0;
int j=0;
for(int i=0;ig)
{
j=i;
g=n;
}
}
System.out.println(c[j]);
}
}

function run() {

var value = "add something here with many characters";
var array = value.split("");
var occurrences = {};

array.forEach(function(val) {

if(occurrences[val] == null || occurrences[val] == undefined){
occurrences[val] = 1;
} else {
occurrences[val] += 1;
}
});

console.log(occurrences);
}

public static void main(String[]args){
String s="laxmikanta";
int count=s.length()-s.replace("a","").length();
System.out.println("No of occurrence of a is:"+count);

Good way to solve this problem for all characters.

import java.util.Scanner;
public class HelloWorld{

public static void main(String []args){
// System.out.println("Hello World");
Scanner sc=new Scanner(System.in);
String str=sc.nextLine();
int l=str.length();
int max=0,temp=0,k=0;

for(int i=0;imax)
{
max=temp;
k=i;
}
}

System.out.println("Character "+str.charAt(k)+" with max count "+max);

}
}

Hello @Unknown, blogger at the < sign, I assume its i < max right?

import java.util.*;
class MaximumOccurring{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
String str=sc.nextLine();
char ch[]=str.toCharArray();

char temp;
for(int i=0;ich[j]){
temp=ch[i];
ch[i]=ch[j];
ch[j]=temp;
}
}
}
int count=1;
int arr[]=new int[str.length()];
int k=0;
char result[]=new char[str.length()];
for(int i=0;imax){
max=arr[i];
index=i;
}
}

System.out.println(ch[index]);
}
}

Hello all, String val=banana , a has been occurred 3 times, can any one help an code , i am new to coding.

did you try this code?

using Java 8

Stream charStream = word.chars().mapToObj(obj -> (char) obj);
System.out.println(charStream.filter(c -> c == character).count());

neat example Anshul, thx for sharing with us.

public static void MaxOccChar() {
String s="myy nameeee is Sama".toLowerCase();

Map ch= new TreeMap();
for(int i =0;i> retS=ch.entrySet();
int max=0; char c='a';
for(Entry e:retS) {
if(max<e.getValue()) {
max=e.getValue();
c=e.getKey();}
}
System.out.println("Character "+c+" occurs maximum times & the count is "+max);

}

public class sample4
{
public void vc(String str)
{
int vowels=0,consonants=0;
for(int i=0;i<str.length();i++)
{
if((97<=str.charAt(i) && str.charAt(i)<=122) || (65<=str.charAt(i) && str.charAt(i)<=90))
{
if('a'==str.charAt(i) || 'e'==str.charAt(i) || 'i'==str.charAt(i) || 'o'==str.charAt(i) || 'u'==str.charAt(i))
{
vowels++;
}
else
{
consonants++;
}
}
}
System.out.print("The total vowels are :"+vowels);
System.out.println();
System.out.print("The total consonants are :"+consonants);
}

public static void main(String args[])
{
Scanner sc=new Scanner(System.in);
String asp=sc.next();
sample4 as=new sample4();
as.vc(asp);
}

}

HashMap countM=new HashMap();

for(int i=0;i<str.length();i++) {
//System.out.println("SubString"+String.valueOf(str.charAt(i)));
if(countM.get(String.valueOf(str.charAt(i)))==null) {
countM.put(String.valueOf(str.charAt(i)), 1);
}else {
//System.out.println(String.valueOf(str.charAt(i))+" : "+countM.get(String.valueOf(str.charAt(i))));
countM.put(String.valueOf(str.charAt(i)), countM.get(String.valueOf(str.charAt(i)))+1);
}
}
System.out.println(countM);

public class MaxiOccurChar {

public static void main(String[] args) {
int count;
String s="Java";
int a[]=new int[s.length()];
int max=a[0];
int pos=0;
// for counting char occurence
for (int i = 0; i < s.length(); i++) {
count=0;
for (int j = i; j < s.length(); j++) {
if(s.charAt(i)==s.charAt(j))
{
count++;
}
}
a[i]=count;
System.out.println(a[i]+" Count of chararcter "+s.charAt(i));
// maximum occured char

if(a[i]>max){
max=a[i];
pos=i;
}
}
System.out.println("Maximum occur: "+max+" times for char: "+s.charAt(pos));


}

}

public class Main {
public static void main(String[] args) {
int c1=0,c2=0;
String s="java";
char c=s.charAt(0);
for(int i=0;ic2){
c=s.charAt(i);
c2=c1; }}
c1=0;
}
System.out.println(c);
}
}

public static void main(String[] args) {
String str="nnnaaaavvvsss is navin ssss lllllll tttttt";int result=0;
HashMap hm=new HashMap();
for(int i=0;i<=str.length()-1;i++) {
if(hm.get(str.charAt(i))==null) {
hm.put(str.charAt(i), 1);
}
else {
hm.put(str.charAt(i), hm.get(str.charAt(i))+1);
if(result<hm.get(str.charAt(i)))
result=hm.get(str.charAt(i));
}
}

System.out.println(result);
}

public class MaxOccuringChar {

public static void main(String[] args) {
String s="atrtrayiuytrsaaouhgaaouja";

char[] arr=s.toCharArray();
Map map=new HashMap<>();
Integer count=0;
for (char c : arr) {
if(map.containsKey(c)) {
count= map.get(c) + 1;
map.put(c, count);
}else {
count=count+1;
map.put(c, count);
}
}

int max = map.values().stream().max((a,b)-> a>b?a:b).get();
System.out.println(max);
}
}

public static void findMaxOccurance(String str)
{
int count=0;
int max_occurance=0;
Character max_occ_character= 0;
Map char_map=new HashMap<>();
for(int i=0;i<str.length();i++)
{
count=1;
if(char_map.containsKey(str.charAt(i))) {
count = char_map.get(str.charAt(i)) + 1;
}
char_map.put(str.charAt(i), count);
}
for(Character key: char_map.keySet())
{
if(max_occurance < char_map.get(key)) {
max_occ_character=key;
max_occurance = char_map.get(key);
}
}
System.out.println("max Occurrence character is " + max_occ_character +" Frequency is "+ max_occurance);
}

public static void findMaxOccurance(String str)
{
int count=0;
int max_occurance=0;
Character max_occ_character= 0;
Map char_map=new HashMap<>();
for(int i=0;i<str.length();i++)
{
count=1;
if(char_map.containsKey(str.charAt(i))) {
count = char_map.get(str.charAt(i)) + 1;
}
char_map.put(str.charAt(i), count);
}
for(Character key: char_map.keySet())
{
if(max_occurance < char_map.get(key)) {
max_occ_character=key;
max_occurance = char_map.get(key);
}
}
System.out.println("max Occurrence character is " + max_occ_character +" Frequency is "+ max_occurance);
}

String str = "ababac";
char chr[] = str.toCharArray();
char c =' ';
char f=' ';
int temp=0;
for(int i=0;itemp)
{
temp =count1;
f=chr[i];
}
}
System.out.println("char is = "+f+" count is = "+temp);

public class Sample{

public static void main(String args[]){
String str="sare jhan se aacha";
int temp=0;
int Char=0;

for(int i=0;i<str.length();i++){
int count=0;
char currchar=str.charAt(i);
for(int j=0;j<str.length();j++)
{
if(str.charAt(j)==currchar && str.charAt(j)!=' '){
count++;
}
}
if(temp<count){
temp=count;
Char=i;
}
}
char z=str.charAt(Char);
System.out.println("Highest occurance character is = "+z + "\n" +"occurance count is = "+ temp);

}
}

class Occ
{
public static void main(String[] arg)
{
String s = "This is a kumar sharma";
String[] a = s.split("a");
System.out.println(a.length);
}
}

class Occ
{
public static void main(String[] arg)
{
String s = "This is a kumar sharma";
String[] a = s.split("a");
System.out.println(a.length);
}
}