Count number of occurrences for each char in a string

Shorter answer, with reduce:

let s = 'hello';
var result = [...s].reduce((a, e) => { a[e] = a[e] ? a[e] + 1 : 1; return a }, {}); 
console.log(result); // {h: 1, e: 1, l: 2, o: 1}

You can use the maps in ES6 in Javascript. Provides a cleaner and concise code in my opinion. Here is how I would go about

function countChrOccurence ('hello') {
 let charMap = new Map();
 const count = 0;
  for (const key of str) {
   charMap.set(key,count); // initialize every character with 0. this would make charMap to be 'h'=> 0, 'e' => 0, 'l' => 0, 
  }

  for (const key of str) {
    let count = charMap.get(key);
    charMap.set(key, count + 1);
  }
// 'h' => 1, 'e' => 1, 'l' => 2, 'o' => 1

  for (const [key,value] of charMap) {
    console.log(key,value);
  }
// ['h',1],['e',1],['l',2],['o',1]
}  

let str = "atul kumar srivastava";
let obj ={};
for(let s of str)if(!obj[s])obj[s] = 1;else obj[s] = obj[s]  + 1;
console.log(obj)

This is really, really simple in JavaScript (or any other language that supports maps):

// The string
var str = "I want to count the number of occurrences of each char in this string";

// A map (in JavaScript, an object) for the character=>count mappings
var counts = {};

// Misc vars
var ch, index, len, count;

// Loop through the string...
for (index = 0, len = str.length; index < len; ++index) {
    // Get this character
    ch = str.charAt(index); // Not all engines support [] on strings

    // Get the count for it, if we have one; we'll get `undefined` if we
    // don't know this character yet
    count = counts[ch];

    // If we have one, store that count plus one; if not, store one
    // We can rely on `count` being falsey if we haven't seen it before,
    // because we never store falsey numbers in the `counts` object.
    counts[ch] = count ? count + 1 : 1;
}

Now counts has properties for each character; the value of each property is the count. You can output those like this:

for (ch in counts) {
    console.log(ch + " count: " + counts[ch]);
}