How To find second Largest value in array|Javascript Coding Interview Question #4 |

Javascript Coding Interview Questions - #4 | Javascript Coding Challenge Interview Questions 2022
How To find second Largest value/element in an array in javascript?
Javascript interview questions and answers, javascript techinacl questions, javascript interview coding exercises, javascript interview questions and answers 2020, javascript interview questions and answers 2019, javascript interview, javascript interview prep, javascript in 12 minutes

Javascript interview questions and answers series

Javascript Interview Questions and Answers - Part 1

Javascript Interview Questions and Answers - Part 2

Javascript Interview Questions and Answers - Part 3

Javascript Interview Questions and Answers - Part 4

[KetanUniverseOfficial]

const numbers = [5, 2, 9, 1, 7];

numbers.sort((a, b) => b - a);
const secondLargest = numbers[1];

console.log(secondLargest); // Output: 7

[beluga]

for those who like another way.


const array = [41, 2, 3, 55, 58];

let max = array [0];
let secMax = array [0];
for(let i= 0 ;i< array.length ; i++){
if(array[i]>max){
secMax = max;
max = array[i];

}else if(array[i]>secMax && array[i]!= max){
secMax = array[i];

}
}console.log(secMax);

[brucewayne-]

Here is a simple method :

let numbers =
numbers.sort(function(a, b){return a - b});
numbers.pop();
let secondLargest = numbers[numbers.length -1]
console.log(numbers);
console.log(secondLargest);

[manickamsubramanian]

Store the prev value and replace when a higher value is encountered. This can reduce traversing the array twice

const array = [2, 5, 6, 7, 2, 3, 8, 9, 2, 4, 5, 7, 56, 65, 99, 88, 66, 77, 65]

var return_val=0
var second_largest = 0
const maxFunction =(arr)=>{
return_val = arr.reduce((prev, curr)=>{
if(prev > curr){
second_largest = curr > second_largest? curr:second_largest
return prev
}
second_largest = prev > second_largest? prev:second_largest
return curr
})
console.log(second_largest)
return return_val
}

[Ganesh]

is it necessary to use this complicated code in interview or can we write using sort()

[ashutoshsutar]

function seclargst(data)
{
data.splice( data.indexOf( (Math.max(...data))), 1)
return Math.max(...data)
}
numbers = [5, 2, 9, 1, 7];

[ramyashreemohan]

let arr = [5, 23, 500, 600, 108, 4, 496, 0, -100, 35, 400, 0];

function arrSecondLargest(arrayy)
{
let max = arrayy[0];
let secondMax;
for(let i = 0; i<arrayy.length; i++)
{
if(arrayy[i+1]>max)
{
secondMax = max; //5 23
max = arrayy[i+1]; //23 500
}
else if(arrayy[i+1]>secondMax)
{
secondMax = arrayy[i+1];
}
}
console.log("Max is "+max);
console.log("Second max is "+secondMax);
return secondMax;
}

let result = arrSecondLargest(arr);
console.log("Second largest element in the array is "+result);

[vishvhina]

Really superb bro can you translate english

[hammadhsn]

you are solving a problem but not thinking about the time complexity and other aspects.

[KrishnaDasPC]

short answer:

const nums = [1, 2, 2, 3, 4, 8, 5, 9]
const sortedNum = nums.sort();

[PragyaJain-qcif]

let arr=[1, 0, 3, 43, 45, 6, 76, 876, 4, 6, 2, 132, 600, 1023, 54, -1]
let
let

function secondLargestNum(arr){
if(arr.length>2){
for(let i=0; i<arr.length;i++){
if(arr[i]>largest && arr[i]>secondLargest){
secondLargest= largest;
largest=arr[i];
}else if(arr[i]>secondLargest && largest>secondLargest){
secondLargest=arr[i]
}
}
return secondLargest;
}else{
return null;
}


}

[akshaykumar]

const givenAarray = [5, 6, 66, 6666, 7, 888, 99];
const sortedarray = givenAarray.sort(function(a, b){ return a - b });

// largest


//second largest