//for each

let names=['alice','bob','peter','jane'];

names.forEach((name)=>{

console.log(name+" ji");

})

//map

let newnames=names.map((name)=>{

return name;

})

console.log(newnames);

//filter

let condnames=names.filter((name)=>{

return name==='alice'

})

console.log(condnames)

//destructuring

let obj={

name:"marry",

age:[20,30,40],

address:{

karachi:"gulshan",

lahore:"laal qila"

},

country:"Pakistan"

}

let access1=obj.age.at(0)

let access2=obj.address.karachi

console.log(access1);

console.log(access2);

let {karachi,lahore} =obj.address

console.log(lahore);

let [f,s] =obj.age

console.log(s)

//spread

//premetives pass by value

let a=4;

let b=a;

console.log(b+1);

console.log(a);

//nonpremitives pass by ref

let array1=[1,2,3]

// let array2=array1

// array2.pop()

// console.log(array2);

// console.log(array1);

//to solve this prob we use spread op

let array2=[...array1];

array2.pop()

console.log(array2);

console.log(array1);

//same for object

let obj1={

name:"hello",

color:"black"

}

let obj2={...obj1}

obj2.name="mak";

console.log(obj1)

console.log(obj2)

//spread op is only level one operator (que asked in interview)

let list={

add:{

name:"mak"

},

tag:"sad"

}

let list2={...list}

list2.add.name="fak";

console.log(list)

console.log(list2)

//rest op

//bajay ye k ap multiple var banay in fun par we can store val in an array code optimization and memorysaves wrt space

let bilal =function(...val){

console.log(val)

}

bilal(1,2,3,4);

//que 1

//concat two arrays fetched from two diff apis

//concat is just mix or paste two arrays into one

//interview que

let concat1=["raima","maria","andleeb"];

let concat2=["ali","adnan","mubashir"];

let students=[...concat1,...concat2]

console.log(students);

//if i have to merge then I should have a logic

let topstudents=concat1.concat(concat2).sort()//logic is sorting according to alphabetical order

console.log(topstudents)

//Custom Sorting with Comparator Function:

let num1=[3,4,1,3,70]

let sortednum1=num1.sort((a,b)=>a-b)

console.log(sortednum1)

let num2=[20,60,5,3]

let arr=num1.concat(num2).sort((a,b)=>a-b);

console.log(arr);

let arr1=[...num1,...num2].sort((a,b)=>a-b)

console.log(arr1);

//que2

//filter an array of obj to display item based on search query eg filter prod by type

let products=[

{name:"phone",type:"electronic"},

{name:"pen",type:"stationary"},

{name:"computer",type:"electronic"},

{name:"hairbrush",type:"beauty"},

{name:"jacket",type:"clothing"}

]

let search =products.filter((product)=>{

return product.type==='beauty'

}).map((product)=>{

return product.name

})

console.log(search)

//que3

//map array of obj and display on div

let lastname=[

{name:"kamran",type:"boy"},

{name:"bilal",type:"boy"},

{name:"alina",type:"girl"},

{name:"momina",type:"girl"},

]

// lastname.map((name)=>{

// return (

// <div> <h1>{name.type} </h1> </div>

// )

// })

//que4

//

let relatives=[

{name:"phupo",percen:"1%",type:"notimmed"},

{name:"mamoo",percen:"30%",type:"notimmed"},

{name:"khala",percen:"80%",type:"notimmed"},

{name:"father",percen:"0%",type:"immed"},

]

let object={}

relatives.forEach((relative)=>{

if(object[relative.type]){

object[relative.type].push(relative)

}else{

object[relative.type]=[]

object[relative.type].push(relative)

}

})

console.log(object)

//que5

//if you want to remove an obj from array

let countries=[

{name:"Japan",id:1},

{name:"China",id:2},

{name:"Canda",id:3},

{name:"Korea",id:4},

]

let updated_countries=countries.filter((country)=>{

return country.id!==3

})

console.log(updated_countries)