js-challenges

Collection of different JavaScript challenges with my solutions.

1. Valid Palindrome ( solution in Big O (log n) )

   solution for this challenge

   
   

   function palindromeCheck( str ){
   
   	let isPalindrome = false;
   	let strLength = str.length;
   
   	for ( let i=0; i < strLength; i++){
   		console.log('i =', i , "  ", 'strLength - 1 - i =', strLength - 1 - i);
   
   		if (str[i] == str[strLength - 1 - i]){
   			isPalindrome = true;
   		}else{
   			isPalindrome = false;
   			break;
   		}
   
   		/*
   		* Once the left pointer and the right pointer reach the middle 
   		* of the current string we can break the loop. The middle of 
   		* the string can be one character if we have add number 
   		* of characters or for even number of chars if position of 
   		* the left pointer minus position of the right pointer === 1 that 
   		* means those pointers are at adjacent elements and we can break 
   		* the loop because at this point we've checked all characters
   		*/
   		
   		if(i == strLength - 1 - i || strLength - 1 - i - i == 1){
   			console.log('pointers reached the middle of the string');
   			break;
   		}
   	}
   
   	return isPalindrome;
   }
   
   palindromeCheck( 'abcdcba' ) // true
   
   palindromeCheck( 'abcddcba' ) // true
   
   palindromeCheck( 'abcdfcba' ) // false

2. Reverse the string ( solution in Big O (log n) )

   solution for this challenge

   
   

   function reverse (str){
   
   let tempStr1 = "";
   let tempStr2 = "";
   let middleOfStr = Math.floor(str.length / 2) - 1;
   let rightPointer = middleOfStr;
   
   for ( let i = str.length - 1; i > middleOfStr; i--){
       console.log('i =', i , " ", 'rightPointer =', rightPointer);
   
       tempStr1 += str[i];
   
       if (-1 !== rightPointer){
           tempStr2 += str[rightPointer];
           rightPointer--;
       }
   }
   return tempStr1 + tempStr2;
   }
   
   reverse("abcdef");  // 'fedcba'
   
   reverse("javaScript");  // 'tpircSavaj'
   
   reverse("Hello, world!");  // '!dlrow ,olleH'

3. Create a custom function to flatten an array without built-in methods ( solution in Big O (N) )

   solution for this challenge

   
   

   function arrayCustomFlatten(arr){
   
       let tempArr = [];
   
       let iife = (function customFlattenFn (arr) {
       
               arr.forEach( (elm) => {
       
                   if ( !Array.isArray(elm) ){
                       tempArr.push(elm);
                   }else{
                       customFlattenFn(elm);            
                   }
               })
               return tempArr;
           })(arr)
   
       tempArr = null;
       return iife;
   }
   
   arrayCustomFlatten([1,2, [3,4 , [5]], [[6]] ])  // [1, 2, 3, 4, 5, 6]
   
   arrayCustomFlatten([ [[[9]]], [7, [[8], [[9]]]] ])  // [9, 7, 8, 9]

4. Create a custom sleep function which can pause the code execution for the specified amount of time.

   Use case example:

   console.log(Date.now())
   await sleepNow(3000) // sleep for 3 seconds
   console.log(Date.now())

   solution for this challenge

   
   

   sleepNow = function (time){
       return new Promise((resolve) => {
           setTimeout( ()=> {
               resolve();
           }, time)
       })
   }