small type correction

Author: rohan-paul

Javascript/OOP-Prototypal-Inheritence/print-All-Prototypes-of-Objects.js

@@ -1,40 +1,34 @@
 // Write a function to list Down All Prototype Properties of an Javascript Object
 
-printPrototypeProperties = obj => {
-    var result = [];
+printPrototypeProperties = (obj) => {
+  var result = []
 
-    let objProp = Object.getOwnPropertyNames(obj)
-    console.log(objProp)
+  let objProp = Object.getOwnPropertyNames(obj)
+  console.log(objProp)
 
-    // console.log(Object.keys(obj))
+  // console.log(Object.keys(obj))
 
-    for (let i of objProp) {
-
-      if (result.indexOf(i) === -1 ) {
-        result.push(i);
-      }
+  for (let i of objProp) {
+    if (result.indexOf(i) === -1) {
+      result.push(i)
     }
-    return result;
   }
-
+  return result
+}
 
 let myObj = {
-	"a": 1,
-	"b": 2,
-    "c": 3
+  a: 1,
+  b: 2,
+  c: 3,
 }
 
-console.log(printPrototypeProperties(myObj));
-
+console.log(printPrototypeProperties(myObj))
 
 /* Object.getOwnPropertyNames vs Object.keys
 
 https://stackoverflow.com/questions/22658488/object-getownpropertynames-vs-object-keys
 
 
-203
-down vote
-accepted
 There is a little difference. Object.getOwnPropertyNames(a) returns all own properties of the object a. Object.keys(a) returns all enumerable own properties. It means that if you define your object properties without making some of them enumerable: false these two methods will give you the same result.
 
 It's easy to test:
@@ -53,4 +47,4 @@ If you define a property without providing property attributes descriptor (meani
 a.test = 21;
 then such property becomes an enumerable automatically and both methods produce the same array.
 
-*/
+*/

Javascript/js-basics/Closure/closure-use-case-create-private-variable.js

@@ -26,3 +26,6 @@ Output :
 5
 The counter currently is at 14
  */
+
+
+

Large-Collection-of-Popular-Problems-with-Solutions/General-JS-Problems/raise-To-Power-bit-shifting.js

@@ -1,18 +1,16 @@
 // Regular iterative approach
 
 raiseToPower = (base, exponent) => {
-
-  let result = 1;
+  let result = 1
 
   for (let i = 1; i <= exponent; i++) {
-    result *= base;
+    result *= base
   }
 
-  return result;
+  return result
 }
 
-console.log(raiseToPower(5, 3));
-
+console.log(raiseToPower(5, 3))
 
 /* SOLUTION - 2 as given in - https://js-algorithms.tutorialhorizon.com/2015/10/23/implement-a-power-function/ - "If our exponent is 10,000 then our loop has to run 10,000 times. But we can optimize this logic to use only log2(10,000) ~ 14 times by using the binary form of the exponent."
 
@@ -39,28 +37,25 @@ https://en.wikipedia.org/wiki/Bitwise_operation#AND
 
 */
 
-
 /* Steps - Optimized logic:
 If exponent’s last binary digit is one then multiply the result with base
 Divide the exponent by 2 by doing one right shift. Note from above, a single ritht shift means juse diving the number by 2
 Multiply base with itself
  */
 
 raiseToPower_binary = (base, exponent) => {
-
   if (exponent < 0) {
-    throw new Error('Negative exponent are not accepted')
+    throw new Error("Negative exponent are not accepted")
   }
 
-  let result = 1;
+  let result = 1
 
   while (exponent) {
-
     // Scan the binary representation of exponent on each iteration of the loop. If it finds a 1 then it multiplies the result by base (just like the iterative approach).
     // Finding a 1 means that the current value of exponent is not divisible by 2
 
     if ((exponent & 1) === 1) {
-      result *= base;
+      result *= base
     }
 
     // Divide the exponent by 2 by doing one right shift. Note from above, a single ritht shift means juse diving the number by 2
@@ -70,8 +65,7 @@ raiseToPower_binary = (base, exponent) => {
     base *= base
   }
 
-  return result;
-
+  return result
 }
 
-console.log(raiseToPower_binary(5, 3));
+console.log(raiseToPower_binary(5, 3))