CareerCup

public class DuplicateFrequency {

	public static void main(String[] args) {
		
		String str = "I am awesomely in Love With ZoomBA!";
		
		char [] strArray = str.toCharArray();
		
		int [] ar = {1,16,2,3,3,4,4,8,6,5,4};
		
		Map<Character, Integer> map = new TreeMap<Character, Integer>();
		
		for (int i=0;  i<strArray.length ; i++) {
			
			if (!map.containsKey(strArray[i]))
			map.put(strArray[i], 1);
			
			else
				map.put(strArray[i],map.get(strArray[i]) +1) ;
		}
		
		map.forEach((k,v) -> System.out.println ("key ="+ k + ", No. of time Repeated "+ v) );
		 
	}

// ZoomBA
s = "I am awesomely in Love With ZoomBA!" 
ms = mset(  s.value )
// produces 
/* 
{ =6, a=2, A=1, !=1, B=1, e=3, h=1, I=1, i=2, l=1, L=1, m=3, n=1, o=4, s=1, t=1, v=1, w=1, W=1, y=1, Z=1} // HashMap
*/

public class DuplicateFrequency {

	public static void main(String[] args) {
		
		String str = "I am awesomely in Love With ZoomBA!";
		
		char [] strArray = str.toCharArray();
		
		int [] ar = {1,16,2,3,3,4,4,8,6,5,4};
		
		Map<Character, Integer> map = new TreeMap<Character, Integer>();
		
		for (int i=0;  i<strArray.length ; i++) {
			
			if (!map.containsKey(strArray[i]))
			map.put(strArray[i], 1);
			
			else
				map.put(strArray[i],map.get(strArray[i]) +1) ;
		}
		
		map.forEach((k,v) -> System.out.println ("key ="+ k + ", No. of time Repeated "+ v) );
		 
	}

public class DuplicateFrequency {

	public static void main(String[] args) {
		
		String str = "I am awesomely in Love With ZoomBA!";
		
		char [] strArray = str.toCharArray();
		
		int [] ar = {1,16,2,3,3,4,4,8,6,5,4};
		
		Map<Character, Integer> map = new TreeMap<Character, Integer>();
		
		for (int i=0;  i<strArray.length ; i++) {
			
			if (!map.containsKey(strArray[i]))
			map.put(strArray[i], 1);
			
			else
				map.put(strArray[i],map.get(strArray[i]) +1) ;
		}
		
		map.forEach((k,v) -> System.out.println ("key ="+ k + ", No. of time Repeated "+ v) );
		 
	}

Pseudo code follows

s = "some string with repeated characters"
map<char, int> m

for each c in some
	m[c]++

for each e in m
	print e.value

pseudo code below

s = "some string with repeated characters"
map<char, int> m

for each c in some
	m[c]++

for each e in m
	print e.value

func printUniqueCharacter(givenString: String) -> (value: Array<Any>, ley: Array<Any>) {

var myCharDict = Dictionary<Character,Int>()

for item in givenString.characters {

if myCharDict[item] != nil {
let value = myCharDict[item]
myCharDict.updateValue(value!+1, forKey: item)
} else {
myCharDict[item] = 1
}
}
let value = Array(myCharDict.values)
let key = Array(myCharDict.keys)

return (value,key)
}

Void stringStats(string& str) {
	if(str.empty()) {
		return;
}
int histo[(‘z’-’a’ + 1)] = {0};
for(size_t i(0); i < str.size(); ++i) {
	histo[str[i]]++;
}
for(size_t i(0); i < (sizeof(histo)/sizeof(histo[0])); ++i ) {
	if(histo[i] == 0) {
		continue;
}
Std::cout << (‘a’+i) << “ : ” << histo[i] << std::endl();
}
}

var listGrouped = myString.GroupBy(x => x,   //the element you want to group by
                                                            (key,    //the element you grouped by
                                                            element  //the new list of strings grouped 
                                                            ) => new
                                                            {
                                                                Key = key,
                                                                Count = element.Count()
                                                            });

            foreach (var item in listGrouped)
            {
                Console.WriteLine("- - - - - - - -");
                Console.WriteLine(item.Key);
                Console.WriteLine(item.Count);
            }

            Console.ReadKey();

It gives this result:

- - - - - - - -
L
1
- - - - - - - -
e
1
- - - - - - - -
o
2
- - - - - - - -
n
1
- - - - - - - -
a
1
- - - - - - - -
r
1
- - - - - - - -
d
1

String text = "text to be tested";
		char[] chars = text.toCharArray();
		Map<Character,Integer> count = new HashMap<>();
		for (char c : chars) {
			if(count.containsKey(c))
			{
				int i = count.get(c);
				count.put(c, ++i);
			}
			else
				count.put(c, 1);
		}
		count.entrySet().stream().forEach(System.out::println  );

/**
   * Javascript Implementation.
   * uniqueCharByCount
   
Given a string, print out all of the unique characters 
and the number of times it appeared in the string

Input:
geeksforgeek
acaaabbbacdddd

Output:
{ g: 2, e: 4, k: 2, s: 1, f: 1, o: 1, r: 1 }
{ a: 5, c: 2, b: 3, d: 4 }
*/


function uniqueCharByCount(str) {
  let charByCountMap = {};
  
  for(let index = 0, length = str.length; index < length; index++) {
    const char = str[index];
    if( !charByCountMap[char] ) {
      charByCountMap[char] = 1;
    } else {
      charByCountMap[char] = charByCountMap[char] + 1;
    }
  }
  console.log(charByCountMap);
}

const str = 'acaaabbbacdddd';
console.log("uniqueCharByCount -> "+str+" \n");
uniqueCharByCount(str);