Develop Reference

I'm solving Leetcode problem: 14 using Javascript

I sorted the elements and comparing the first and last string to check the common prefixes. It works for most of the cases, but not for the input ["dog","racecar","car"]. The expected output is "", but what I'm getting is "c" (The "r" in "car" and "r" in "racecar"). I can tell the code to remove the last char, but this will break the other cases such as ["car", "car", "car"]. Not sure what am I missing. Any insights would help me improve.
Thanks
var longestCommonPrefix = function(strs) {
let count=0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length-1]
for(let i=0; i< firstString.length; i++) {
if(firstString.charAt(i) === lastString.charAt(i)) {
count++
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog","racecar","car"])

You need to break out of the loop as soon as a match is not found. Otherwise, for example, ra and ca match on the second index, the a - which is undesirable.
var longestCommonPrefix = function(strs) {
let count = 0
const sortedString = strs.sort()
const firstString = sortedString[0]
const lastString = sortedString[sortedString.length - 1]
for (let i = 0; i < firstString.length; i++) {
if (firstString.charAt(i) === lastString.charAt(i)) {
count++
} else {
break;
}
}
console.log(firstString.substring(0, count))
};
longestCommonPrefix(
["dog", "racecar", "car"])
or, refactored a bit
const longestCommonPrefix = (strs) => {
strs.sort();
const firstString = strs[0];
const lastString = strs[strs.length - 1];
let prefixSoFar = '';
for (let i = 0; i < firstString.length; i++) {
if (firstString[i] === lastString[i]) {
prefixSoFar += firstString[i];
} else {
return prefixSoFar;
}
}
return prefixSoFar;
};
console.log(longestCommonPrefix(["dog", "racecar", "car"]));

Alibaba interview: print a sentence with min spaces

I saw this interview question and gave a go. I got stuck. The interview question is:
Given a string
var s = "ilikealibaba";
and a dictionary
var d = ["i", "like", "ali", "liba", "baba", "alibaba"];
try to give the s with min space
The output may be
i like alibaba (2 spaces)
i like ali baba (3 spaces)
but pick no.1
I have some code, but got stuck in the printing.
If you have better way to do this question, let me know.
function isStartSub(part, s) {
var condi = s.startsWith(part);
return condi;
}
function getRestStr(part, s) {
var len = part.length;
var len1 = s.length;
var out = s.substring(len, len1);
return out;
}
function recPrint(arr) {
if(arr.length == 0) {
return '';
} else {
var str = arr.pop();
return str + recPrint(arr);
}
}
// NOTE: have trouble to print
// Or if you have better ways to do this interview question, please let me know
function myPrint(arr) {
return recPrint(arr);
}
function getMinArr(arr) {
var min = Number.MAX_SAFE_INTEGER;
var index = 0;
for(var i=0; i<arr.length; i++) {
var sub = arr[i];
if(sub.length < min) {
min = sub.length;
index = i;
} else {
}
}
return arr[index];
}
function rec(s, d, buf) {
// Base
if(s.length == 0) {
return;
} else {
}
for(var i=0; i<d.length; i++) {
var subBuf = [];
// baba
var part = d[i];
var condi = isStartSub(part, s);
if(condi) {
// rest string
var restStr = getRestStr(part, s);
rec(restStr, d, subBuf);
subBuf.unshift(part);
buf.unshift(subBuf);
} else {
}
} // end loop
}
function myfunc(s, d) {
var buf = [];
rec(s, d, buf);
console.log('-- test --');
console.dir(buf, {depth:null});
return myPrint(buf);
}
// Output will be
// 1. i like alibaba (with 2 spaces)
// 2. i like ali baba (with 3 spaces)
// we pick no.1, as it needs less spaces
var s = "ilikealibaba";
var d = ["i", "like", "ali", "liba", "baba", "alibaba"];
var out = myfunc(s, d);
console.log(out);
Basically, my output is, not sure how to print it....
[ [ 'i', [ 'like', [ 'alibaba' ], [ 'ali', [ 'baba' ] ] ] ] ]

This problem is best suited for a dynamic programming approach. The subproblem is, "what is the best way to create a prefix of s". Then, for a given prefix of s, we consider all words that match the end of the prefix, and choose the best one using the results from the earlier prefixes.
Here is an implementation:
var s = "ilikealibaba";
var arr = ["i", "like", "ali", "liba", "baba", "alibaba"];
var dp = []; // dp[i] is the optimal solution for s.substring(0, i)
dp.push("");
for (var i = 1; i <= s.length; i++) {
var best = null; // the best way so far for s.substring(0, i)
for (var j = 0; j < arr.length; j++) {
var word = arr[j];
// consider all words that appear at the end of the prefix
if (!s.substring(0, i).endsWith(word))
continue;
if (word.length == i) {
best = word; // using single word is optimal
break;
}
var prev = dp[i - word.length];
if (prev === null)
continue; // s.substring(i - word.length) can't be made at all
if (best === null || prev.length + word.length + 1 < best.length)
best = prev + " " + word;
}
dp.push(best);
}
console.log(dp[s.length]);

pkpnd's answer is along the right track. But word dictionaries tend to be quite large sets, and iterating over the entire dictionary at every character of the string is going to be inefficient. (Also, saving the entire sequence for each dp cell may consume a large amount of space.) Rather, we can frame the question, as we iterate over the string, as: given all the previous indexes of the string that had dictionary matches extending back (either to the start or to another match), which one is both a dictionary match when we include the current character, and has a smaller length in total. Generally:
f(i) = min(
f(j) + length(i - j) + (1 if j is after the start of the string)
)
for all j < i, where string[j] ended a dictionary match
and string[j+1..i] is in the dictionary
Since we only add another j when there is a match and a new match can only extend back to a previous match or to the start of the string, our data structure could be an array of tuples, (best index this match extends back to, total length up to here). We add another tuple if the current character can extend a dictionary match back to another record we already have. We can also optimize by exiting early from the backwards search once the matched substring would be greater than the longest word in the dictionary, and building the substring to compare against the dictionary as we iterate backwards.
JavaScript code:
function f(str, dict){
let m = [[-1, -1, -1]];
for (let i=0; i<str.length; i++){
let best = [null, null, Infinity];
let substr = '';
let _i = i;
for (let j=m.length-1; j>=0; j--){
let [idx, _j, _total] = m[j];
substr = str.substr(idx + 1, _i - idx) + substr;
_i = idx;
if (dict.has(substr)){
let total = _total + 1 + i - idx;
if (total < best[2])
best = [i, j, total];
}
}
if (best[0] !== null)
m.push(best);
}
return m;
}
var s = "ilikealibaba";
var d = new Set(["i", "like", "ali", "liba", "baba", "alibaba"]);
console.log(JSON.stringify(f(s,d)));
We can track back our result:
[[-1,-1,-1],[0,0,1],[4,1,6],[7,2,10],[11,2,14]]
[11, 2, 14] means a total length of 14,
where the previous index in m is 2 and the right index
of the substr is 11
=> follow it back to m[2] = [4, 1, 6]
this substr ended at index 4 (which means the
first was "alibaba"), and followed m[1]
=> [0, 0, 1], means this substr ended at index 1
so the previous one was "like"
And there you have it: "i like alibaba"

As you're asked to find a shortest answer probably Breadth-First Search would be a possible solution. Or you could look into A* Search.
Here is working example with A* (cause it's less bring to do than BFS :)), basically just copied from Wikipedia article. All the "turning string into a graph" magick happens in the getNeighbors function
https://jsfiddle.net/yLeps4v5/4/
var str = 'ilikealibaba'
var dictionary = ['i', 'like', 'ali', 'baba', 'alibaba']
var START = -1
var FINISH = str.length - 1
// Returns all the positions in the string that we can "jump" to from position i
function getNeighbors(i) {
const matchingWords = dictionary.filter(word => str.slice(i + 1, i + 1 + word.length) == word)
return matchingWords.map(word => i + word.length)
}
function aStar(start, goal) {
// The set of nodes already evaluated
const closedSet = {};
// The set of currently discovered nodes that are not evaluated yet.
// Initially, only the start node is known.
const openSet = [start];
// For each node, which node it can most efficiently be reached from.
// If a node can be reached from many nodes, cameFrom will eventually contain the
// most efficient previous step.
var cameFrom = {};
// For each node, the cost of getting from the start node to that node.
const gScore = dictionary.reduce((acc, word) => { acc[word] = Infinity; return acc }, {})
// The cost of going from start to start is zero.
gScore[start] = 0
while (openSet.length > 0) {
var current = openSet.shift()
if (current == goal) {
return reconstruct_path(cameFrom, current)
}
closedSet[current] = true;
getNeighbors(current).forEach(neighbor => {
if (closedSet[neighbor]) {
return // Ignore the neighbor which is already evaluated.
}
if (openSet.indexOf(neighbor) == -1) { // Discover a new node
openSet.push(neighbor)
}
// The distance from start to a neighbor
var tentative_gScore = gScore[current] + 1
if (tentative_gScore >= gScore[neighbor]) {
return // This is not a better path.
}
// This path is the best until now. Record it!
cameFrom[neighbor] = current
gScore[neighbor] = tentative_gScore
})
}
throw new Error('path not found')
}
function reconstruct_path(cameFrom, current) {
var answer = [];
while (cameFrom[current] || cameFrom[current] == 0) {
answer.push(str.slice(cameFrom[current] + 1, current + 1))
current = cameFrom[current];
}
return answer.reverse()
}
console.log(aStar(START, FINISH));

You could collect all possible combinations of the string by checking the starting string and render then the result.
If more than one result has the minimum length, all results are taken.
It might not work for extrema with string who just contains the same base string, like 'abcabc' and 'abc'. In this case I suggest to use the shortest string and update any part result by iterating for finding longer strings and replace if possible.
function getWords(string, array = []) {
words
.filter(w => string.startsWith(w))
.forEach(s => {
var rest = string.slice(s.length),
temp = array.concat(s);
if (rest) {
getWords(rest, temp);
} else {
result.push(temp);
}
});
}
var string = "ilikealibaba",
words = ["i", "like", "ali", "liba", "baba", "alibaba"],
result = [];
getWords(string);
console.log('all possible combinations:', result);
console.log('result:', result.reduce((r, a) => {
if (!r || r[0].length > a.length) {
return [a];
}
if (r[0].length === a.length) {
r.push(a);
}
return r;
}, undefined))

Use trie data structure
Construct a trie data structure based on the dictionary data
Search the sentence for all possible slices and build a solution tree
Deep traverse the solution tree and sort the final combinations
const sentence = 'ilikealibaba';
const words = ['i', 'like', 'ali', 'liba', 'baba', 'alibaba',];
class TrieNode {
constructor() { }
set(a) {
this[a] = this[a] || new TrieNode();
return this[a];
}
search(word, marks, depth = 1) {
word = Array.isArray(word) ? word : word.split('');
const a = word.shift();
if (this[a]) {
if (this[a]._) {
marks.push(depth);
}
this[a].search(word, marks, depth + 1);
} else {
return 0;
}
}
}
TrieNode.createTree = words => {
const root = new TrieNode();
words.forEach(word => {
let currentNode = root;
for (let i = 0; i < word.length; i++) {
currentNode = currentNode.set(word[i]);
}
currentNode.set('_');
});
return root;
};
const t = TrieNode.createTree(words);
function searchSentence(sentence) {
const marks = [];
t.search(sentence, marks);
const ret = {};
marks.map(mark => {
ret[mark] = searchSentence(sentence.slice(mark));
});
return ret;
}
const solutionTree = searchSentence(sentence);
function deepTraverse(tree, sentence, targetLen = sentence.length) {
const stack = [];
const sum = () => stack.reduce((acc, mark) => acc + mark, 0);
const ret = [];
(function traverse(tree) {
const keys = Object.keys(tree);
keys.forEach(key => {
stack.push(+key);
if (sum() === targetLen) {
const result = [];
let tempStr = sentence;
stack.forEach(mark => {
result.push(tempStr.slice(0, mark));
tempStr = tempStr.slice(mark);
});
ret.push(result);
}
if(tree[key]) {
traverse(tree[key]);
}
stack.pop();
});
})(tree);
return ret;
}
const solutions = deepTraverse(solutionTree, sentence);
solutions.sort((s1, s2) => s1.length - s2.length).forEach((s, i) => {
console.log(`${i + 1}. ${s.join(' ')} (${s.length - 1} spaces)`);
});
console.log('pick no.1');

splitting a string into a multidimensional array

I have a list of strings, I want to check if the string contains a specific word, and if it does split all the words in the string and add it to an associative array.
myString = ['RT #Arsenal: Waiting for the international', 'We’re hungry for revenge #_nachomonreal on Saturday\'s match and aiming for a strong finish']
wordtoFind = ['#Arsenal']
I want to loop through the wordtoFind and if it is in myString, split up myString into individual words and create an object like
newWord = {#Arsenal:[{RT:1},{Waiting:1},{for:1},{the:1},{international:1}]}
for(z=0; z <wordtoFind.length; z++){
for ( i = 0 ; i < myString.length; i++) {
if (myString[i].indexOf(wordtoFind[z].key) > -1){
myString[i].split(" ")
}
}
}

I would say something likes would work, this also counts the amount of occurrences of a word in a sentence. JavaScript does not have associative arrays like PHP for instance. They just have objects or numbered arrays:
var myString = ['RT #Arsenal: Waiting for the international', 'We’re hungry for revenge #_nachomonreal on Saturday\'s match and aiming for a strong finish'];
var wordtoFind = ['#Arsenal'];
var result = {};
for(var i = 0, l = wordtoFind.length; i < l; i++) {
for(var ii = 0, ll = myString.length; ii < ll; ii++) {
if(myString[ii].indexOf(wordtoFind[i]) !== -1) {
var split = myString[ii].split(' ');
var resultpart = {};
for(var iii = 0, lll = split.length; iii < lll; iii++) {
if(split[iii] !== wordtoFind[i]) {
if(!resultpart.hasOwnProperty(split[iii])) {
resultpart[split[iii]] = 0;
}
resultpart[split[iii]]++;
}
}
result[wordtoFind[i]] = resultpart;
}
}
}
console.log(result);
//{"#Arsenal":{"RT":1,"Waiting":1,"for":1,"the":1,"international":1}}

This method makes use of the forEach-function and callbacks.
The containsWord-function was left with a for-loop for now to reduce some callbacks, this can obviously be changed.
var myString = [
'RT #Arsenal: Waiting for the international',
'We’re hungry for revenge #_nachomonreal on Saturday\'s match and aiming for a strong finish',
'#Arsenal: one two three four two four three four three four'
];
var wordtoFind = ['#Arsenal'];
// define the preprocessor that is used before the equality check
function preprocessor(word) {
return word.replace(':', '');
}
function findOccurences(array, search, callback, preprocessor) {
var result = {};
var count = 0;
// calculate the maximum iterations
var max = search.length * array.length;
// iterate the search strings that should be matched
search.forEach(function(needle) {
// iterate the array of strings that should be searched in
array.forEach(function(haystack) {
if (containsWord(haystack, needle, preprocessor)) {
var words = haystack.split(' ');
// iterate every word to count the occurences and write them to the result
words.forEach(function(word) {
countOccurence(result, needle, word);
})
}
count++;
// once every iteration finished, call the callback
if (count == max) {
callback && callback(result);
}
});
});
}
function containsWord(haystack, needle, preprocessor) {
var words = haystack.split(' ');
for (var i = 0; i < words.length; i++) {
var word = words[i];
// preprocess a word before it's compared
if (preprocessor) {
word = preprocessor(word);
}
// if it matches return true
if (word === needle) {
return true;
}
}
return false;
}
function countOccurence(result, key, word) {
// add array to object if it doesn't exist yet
if (!result.hasOwnProperty(key)) {
result[key] = [];
}
var entry = result[key];
// set the count to 0 if it doesn't exist yet
if (!entry.hasOwnProperty(word)) {
entry[word] = 0;
}
entry[word]++;
}
// call our function to find the occurences
findOccurences(myString, wordtoFind, function(result) {
// do something with the result
console.log(result);
}, preprocessor);
// output:
/*
{ '#Arsenal':
[ RT: 1,
'#Arsenal:': 2,
Waiting: 1,
for: 1,
the: 1,
international: 1,
one: 1,
two: 2,
three: 3,
four: 4 ] }
*/
Feel free to ask any questions, if the answer needs clarification.
I hope this fits your needs.

You're on the right track. You just need to store the split string into the associative array variable.
var assocArr = [];
for(z=0; z <wordtoFind.length; z++){
for ( i = 0 ; i < myString.length; i++) {
if (myString[i].indexOf(wordtoFind[z]) > -1){
myString[i].split(" ").forEach(function(word){
assocArr.push(word);
});
}
}
}

I think the key problem that stuck you is the data structure. The optimal structure should be something like this:
{
#Arsenal:[
{RT:1, Waiting:1, for:1, the:1, international:1},
{xxx:1, yyy:1, zzz:3}, //for there are multiple ones in 'myString' that contain the same '#Arsenal'
{slkj:1, sldjfl:2, lsdkjf:1} //maybe more
]
someOtherWord:[
{},
{},
....
]
}
And the code:
var result = {};
//This function will return an object like {RT:1, Waiting:1, for:1, the:1, international:1}.
function calculateCount(string, key) {
var wordCounts = {};
string.split(" ").forEach(function (word) {
if (word !== key) {
if (wordCounts[word] === undefined) wordCounts[word] = 1;
else wordCounts[word]++;
}
});
return wordCounts;
}
//For each 'word to find' and each string that contain the 'word to find', push in that returned object {RT:1, Waiting:1, for:1, the:1, international:1}.
wordToFind.forEach(function (word) {
var current = result[word] = [];
myString.forEach(function (str) {
if (str.indexOf(word) > -1) {
current.push(
calculateCount(str, word)
);
}
}); //Missed the right parenthesis here
});

Checking if the characters in a string are all unique

I am trying to solve this problem using JS by just using an array.
var str = 'abcdefgh';
for (i = 0; i < 255; i++) {
arr[i] = false;
}
function check() {
for (i = 0; i < str.length; i++) {
if (arr[str.charCodeAt(i)] == true) {
return false;
}
arr[str.charCodeAt(i)] = true;
}
return true;
}
I am initializing an array of fixed size 256 to have the boolean value false.
Then i am setting the value for the corresponding ASCII index to true for characters in the string. And if i find the same character again, i am returning false.
While running the program, i am getting false returned even if the string doesn't have any duplicate characters.

Fill a Set with all characters and compare its size to the string's length:
function isUnique(str) {
return new Set(str).size == str.length;
}
console.log(isUnique('abc')); // true
console.log(isUnique('abcabc')); // false

Use object for faster result
function is_unique(str) {
var obj = {};
for (var z = 0; z < str.length; ++z) {
var ch = str[z];
if (obj[ch]) return false;
obj[ch] = true;
}
return true;
}
console.log(is_unique("abcdefgh")); // true
console.log(is_unique("aa")); // false

use .match() function for each of the character. calculate occurrences using length. Guess thats it.
(str.match(/yourChar/g) || []).length

We can also try using indexOf and lastIndexOf method:
function stringIsUnique(input) {
for (i = 0; i < input.length; i++) {
if (input.indexOf(input[i]) !== input.lastIndexOf(input[i])) {
return false;
}
}
return true;
}

You are using arr[str.charCodeAt(i)] which is wrong.
It should be arr[str[i].charCodeAt(0)]
var arr = [];
var str="abcdefgh";
for (i=0;i<255;i++){
arr[i]=false;
}
function check(){
for (i=0;i<str.length;i++){
if (arr[str[i].charCodeAt(0)]==true){
return false;
}
arr[str[i].charCodeAt(0)]=true;
}
console.log(arr);
return true;
}
check();

Time complexity = O(n)
Space complexity = O(n)
const isUnique = (str) => {
let charCount = {};
for(let i = 0; i < str.length; i++) {
if(charCount[str[i]]){
return false;
}
charCount[str[i]] = true;
}
return true;
}
const isUniqueCheekyVersion = (str) => {
return new Set(str).size === str.length;
}
Solution 3:
Transform string to chars array, sort them and then loop through them to check the adjacent elements, if there is a match return false else true
Solution 4:
It's similar to Solution 1 except that we use a Set data structure which is introduced in recent versions of javascript

// no additional Data structure is required. we can use naive solution
// Time Complexity:O(n^2)
function isUnique(str) {
for (let i = 0; i < str.length; i++) {
for (let j = 1 + i; j < str.length; j++) {
if (str[i] === str[j]) {
return false;
}
}
}
return true;
}
// if you can use additional Data structure
// Time Complexity:O(n)
function isUniqueSecondMethos(str) {
let dup_str = new Set();
for (let i = 0; i < str.length; i++) {
if (dup_str.has(str[i])) {
return false;
}
dup_str.add(str[i]);
}
return true;
}
console.log(isUniqueSecondMethos('hello'));

Use an object as a mapper
function uniqueCharacterString(inputString) {
const characterMap = {};
let areCharactersUnique = true;
inputString.trim().split("").map((ch)=>{
if(characterMap[ch]===undefined) {
characterMap[ch] = 1;
} else {
areCharactersUnique = false;
}
})
return areCharactersUnique;
}

Algo
*1. step -first string is ->stack *
*2.step-string covert to CharArray *
3. step - use iteration in array ['s','t','a','c','k']
4. step - if(beginElement !== nextElement){return true}else{return false}
Implement code
function uniqueChars(string){
var charArray = Array.from(string) //convert charArray
for(var i=0;i<charArray.length;i++){
if(charArray[i] !== charArray[i+1]){
return true
}
else{
return false
}
}
}
var string ="stack"
console.log(uniqueChars(string))
Time complexity
O(nlogn)

Algo
Counting frequency of alphabets. e.g. 'Mozilla' will returns Object{ M: 1, o: 1, z: 1, i: 1, l: 2, a: 1 }. Note that, the bitwise NOT operator (~) on -~undefined is 1, -~1 is 2, -~2 is 3 etc.
Return true when all occurrences appear only once.
Implement code
var isUnique = (str) => {
const hash = {};
for (const key of str) {
hash[key] = -~hash[key];
}
return Object.values(hash).every((t) => t === 1);
};
console.log(isUnique('Mozilla'));
console.log(isUnique('Firefox'));
Another alternative could be:
var isUnique = (str) => {
const hash = {};
for (const i in str) {
if (hash[str[i]]) return false;
hash[str[i]] = true;
}
return true;
};
console.log(isUnique('Mozilla'));
console.log(isUnique('Firefox'));

To make efficient one, you can use simple hash map
let isUnique = (s) => {
let ar = [...s];
let check = {};
for (let a of ar) {
if (!check[a]) {
check[a] = 1;
} else {
return false
}
}
return true;
}
alert("isUnique : "+isUnique("kailu"));
Time complexity & Space complexity
using ES6
let isUnique = (s)=>{
return new Set([...s]).size == s.length;
}
console.log("using ES6 : ",isUnique("kailu"));

We can use split method of string:
const checkString = (str) => {
let isUniq = true;
for (let i = 0; i < str.length; i++) {
if (str.split(str[i]).length > 2) {
isUniq = false;
break;
}
}
return isUniq;
};
console.log(checkString("abcdefgh")); //true
console.log(checkString("aa")); //false

String split and count the number of occurrences and also

I have a string
var stringIHave = "Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$";
How to get the count of the number of occurrences of each entry, The occurrence I get, is from a JSON like Java = 8 and etc...

First of all you need to split your srting to array:
var keywordsArr = stringIHave.split( '$$' );
then you need to have an object for example to store counts:
var occur = {};
and then just create simple for loop to count all occurrences:
for( var i = 0; i < keywordsArr.length; i++ ) {
occur[ keywordsArr[ i ] ] = ( occur[ keywordsArr[ i ] ] || 0 ) + 1;
}
now your object occur will have names as keys and count as values.
See jsFiddle demo.
Also as you have at end of your string $$ you maybe will need to remove last item from keywordsArr so just do after split function call:
keywordsArr.pop();
See demo without last element.
So final code will be like:
var stringIHave = "Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$",
keywordsArr = stringIHave.split( '$$' ),
occur = {};
keywordsArr.pop();
for( var i = 0; i < keywordsArr.length; i++ ) {
occur[ keywordsArr[ i ] ] = ( occur[ keywordsArr[ i ] ] || 0 ) + 1;
}
for( var key in occur ) {
document.write( key + ' - ' + occur[key] + '<br/>' );
} ​

I'd suggest the following:
function stringCount(haystack, needle) {
if (!needle || !haystack) {
return false;
}
else {
var words = haystack.split(needle),
count = {};
for (var i = 0, len = words.length; i < len; i++) {
if (count.hasOwnProperty(words[i])) {
count[words[i]] = parseInt(count[words[i]], 10) + 1;
}
else {
count[words[i]] = 1;
}
}
return count;
}
}
console.log(stringCount("Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$", '$$'));
​
JS Fiddle demo.
References:
Object.hasOwnProperty().
parseInt().
String.split().

It's not entirely clear what final objective is. Following creates an object from string that looks like
Object created:
{
"Java": 8,
"jQuery": 4,
"Hibernate": 1,
"Spring": 1,
"Instagram": 1
}
JS:
var str = 'Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$';
var arr = str.split('$$')
var obj = {};
for (i = 0; i < arr.length; i++) {
if (arr[i] != '') {
if (!obj[arr[i]]) {
obj[arr[i]] = 0;
}
obj[arr[i]]++;
}
}
You can loop over the object to get all values or simply look up one value
var jQueryOccurences= obj['jQuery'];
DEMO: http://jsfiddle.net/25hBV/1/

Now a days you can do
const str = "Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$";
var result = str.split("$$").reduce(function(acc, curr) {
curr && (acc[curr] = (acc[curr] + 1) || 1);
return acc
}, {});
console.log(result);

Split the string into an array, and putting the array into an object takes care of duplicates and counts occurences as key/value pairs in the object, see fiddle!
var stringIHave = "Java$$Java$$jQuery$$Java$$jQuery$$Java$$Java$$Java$$Hibernate$$Java$$Java$$Spring$$Instagram$$jQuery$$jQuery$$",
s = stringIHave.split('$$');
obj = {};
for (var i=s.length; i--;) {
obj[s[i]] = (s[i] in obj) ? obj[s[i]]+1 : 1;
}
// obj.Java == 8
FIDDLE

If you want it short and sweet:
// variable declarations
var arParts = stringIHave.match(/\w+/g),
result = {},
i = 0,
item;
// Copy the array to result object
while (item = arParts[i++]) result[item] = (result[item] || 0 ) + 1;
demo