Ui developer Interview Questions

My answer: The long method finishes first and then short method executes and completes. Expected answer: The short method returns first!

the short method returns first! is the right answer since two methods will run parallel. As javascript is async events handling. You may right if had quested him is the methods run in sequence.

@Anonymous. It looks like you have not read the question well. It says that there is no setTimeout[) involved. If you have read the question, I have some questions for the answer that you provided. 1. Is asynchronous same as multithreading? I remember reading that JavaScript is single-threaded. 2. Suppose the short method depends on some value modified by the long method. As you said, lets suppose that the short method returns first even though it is next to the long method in sequence. In this case, how do you manage your program?

Program

What program they asked

Reverse string

My Answer : var a =[1,2,3,4] for (var i=0; i

Some Correction : It will print 4 times "4". Fix: just use "let"(block scope) rather than "var" It should be : console.log(a[i]) to print (1,2,3,4)

Yes, ofcourse, it'll print 4 times 4 (not 5 times). Another way could be: using closure function. var a = [1,2,3,4] for (var i=0; i

3 divs in same row Ans: <div> <div>C1</div> <div>C2</div> <div>C3</div> </div> .Row { display: table; width: 100%; /*Optional*/ table-layout: fixed; /*Optional*/ border-spacing: 10px; /*Optional*/ } .Column { display: table-cell; background-color: red; /*Optional*/ }

Sample div { display : inline-block; } <div> <div>C1</div> <div>C2</div> <div>C3</div> </div>

Sample div { display : inline-block; } <div> <div>C1</div> <div>C2</div> <div>C3</div> </div>

tail recursive algo

a = "Hi I am John"; ind =[]; for(k=0; k

str1 = "Hi abc tm" len1 = len(str1) ln = len1 space = " " str3 = [] cnt = 0 while len1 !=0: if str1[len1-1]!=" ": len1 = len1-1 cnt = cnt + 1 if str1[len1-1]==" ": len2 = ln-cnt while len2 != ln: if str1[len2] != " ": print(str1[len2],end="") len2 = len2 + 1 else: break print(space,end="") cnt = cnt + 1 len1 = len1-1 len2 = ln-cnt while len2 != ln: if str1[len2] != " ": print(str1[len2],end="") len2 = len2 + 1 else: break

Starting with the basic and easiest solution is that we go floor by floor. in the worst case, the answer would be the 100th floor. Another strategy could be you divide and then find out. for example, if we divide it in a set of 3 (30-30-40). And so.

Worst case 19 10,20,30..........90,100,.....91........99 ==19 efficient Answer is 14...

First floor

<div> Add Row {{x[0]}} {{x[1]}} {{x[2]}} <div> {{y[0]}} </div> </div>

<div> Add Row {{x[0]}} {{x[1]}} {{x[2]}} <div> {{y[0]}} </div> </div> var app = angular.module("myEmpList", []); app.controller("myEmpCtrl", function($scope) { $scope.empHeaders = []; $scope.empList=[]; $scope.addRow = function () { if($scope.empList.length == 0){ $scope.empHeaders.push(["Column Names", "Edit Buttons", "Delete Buttons"]); } $scope.empList.push(['Name', '', '']) ; } $scope.deleteMe = function(y){ $scope.empList.splice(y,1); if($scope.empList.length == 0 ) $scope.empHeaders.length = 0; } $scope.editMe = function(ind){ var rowInd = "row" + ind; var nameObj = document.getElementById(rowInd).getElementsByTagName("td")[0]; nameObj.innerHTML = ''; } }).directive("editBtn",function(){ return { restrict: "E", template: 'Edit' } }).directive("deleteBtn",function(){ return { restrict: "E", template: 'Delete' } });