Find Second Largest Number From Array | Java Interview Questions and Answers

[@developer00007]

Answer: Yes it will definitely work. I have few approaches to solve this problem: Approach 1: Step 1: Add all the elements in a Set. Step 2: Now add all elements from set into a new ArrayList and Sort. Step 3 : Get size -2 th element Time Complexity O(nlogn) Space Complexity O(n) Approach 2: (Using Heap or PriorityQueue) Step 1: Add All elements into a set to remove duplicates Step 2: Iterate through each elements of the set and add it into the Min Heap. Step 3: Check if the size of Heap >2 then pop the top elements. Step 4: Return Top element Time Complexity: O(nlog2) Space Complexity:O(n) If there were no Duplicates then Space Complexity can Be constant.

[@bheemshankar_pk]

Good Logic Thanks,what if there are null values in array ,then how to find 2nd largest??

[@sandeshkulkarni6941]

I think first we have to remove duplicate elements from array and then we sort, After that there no problem if we give any updated array

[@stayvloging]

public class SecondLargestNumberFromTheArray { public static void main(String[] args) { int arr[] = { 1, 4, 35, 34, 35, 35 }; int largest = arr[0]; int secondLargest = arr[0]; for (int i = 0; i < arr.length; i++) { if (largest < arr[i]) { secondLargest = largest; largest = arr[i]; } else if (secondLargest < arr[i] && largest > arr[i]) { secondLargest = arr[i]; } } System.out.println(secondLargest); } }

[@trendinguniverse6113]

I am thinking about same but it will work if there are duplicate elements in array like that in pprogram

[@stayvloging]

@@trendinguniverse6113 yes it will work

[@sravan77751]

if the first element is largest it will not work { 35, 4, 1, 34, 35, 35 };

[@harikishangrandhe3381]

@@sravan77751I guess it will work

[@prashanthreddy-sd9ju]

In java 8 you would write omething like Arrays.stream(are).sorted(comparator.reverseorder()).limit(2).skip(1). findfirsf().get();

[@unknownplayer0383]

program will continue to work as expected. thank you for the video

[@sathiskumarp]

It will work as expected, since we are checking the highest number logic!!!

[@rmcf3972]

Will this work in an array of 100,000 integers?

[@newanurag]

Just using the builtin functions. where is the brain applied?

[@jeckrazi7691]

Hi @CloudTech, Stream.of(1,4,5,35,34,35) .sorted((n1,n2)-> Integer.compare(n2,n1)) .distinct() .limit(2) .skip(1) .forEach(System.out::println);

[@bhargavmehta97]

What about array of negative numbers?

[@manukarthikeya4560]

Is this code from gfg ??

[@dreamplaying-wg1jn]

Yes bro it will work

[@stayvloging]

public class SecondLargestNumberFromTheArray { public static void main(String[] args) { int arr[] = { 1, 4, 35, 34, 35, 35 }; int largest = arr[0]; int secondLargest = arr[0]; for (int i = 0; i < arr.length; i++) { if (largest < arr[i]) { secondLargest = largest; largest = arr[i]; } else if (secondLargest < arr[i] && largest > arr[i]) { secondLargest = arr[i]; } } System.out.println(secondLargest); } }

[@omkar_gopchade143]

I think it will definitely work

[@akhilraj5268]

I don't think last code is suitable for all testcases

[@cloudtech5260]

Any idea on what test case moght not pass.

[@dipalialone1298]

When all elements are same then it will fail

[@developer00007]

@@dipalialone1298They Have Handle it. Watch again

[@princemohammed921]

For bigger test cases it is not working

[@nitheeshreddy6863]

hi bro i have used this logic can u please check this.. int array[]=new int [] {1,4,5,35,35,35,34}; Arrays.sort(array); //1,2,4,34,35,35; int highest=array[array.length-1]; for(int i=array.length-2;i>=0;i--) { if(highest>array[i]) { System.out.println(array[i]); break; } }

[@iceberg5956]

Awsm

[@venkateshvardhineedi9571]

Super

[@darknight66655]

yes it will work surr**

[@siddumaradi7045]

yes it'll work sir

[@trendinguniverse6113]

Bro pls make remove duplicate elements from array

[@girijasankardas1189]

you can easily remove duplicate element by using HashSet brother.

[@coolraviraj24]

yes it will

[@chaitanaywaskar6008]

at some cases of gfg , this code don't work .

[@gautamnegi6868]

Yes

[@RameshKumar-g8x7f]

If I add second largest element two time Then this logic will give error

[@ChennaiCineCuts]

Better to make it as Hashset to remove duplicates

[@cloudtech5260]

Yes that can be done. 👍

[@shaikhkhizar3167]

This logic won't work, if we take the array elements as a input from user.

[@hariprasadmutalikdesai6504]

But code working with user input with me. and code is import java.util.Arrays; import java.util.Scanner; public class MainClass { public static void main(String[] args) { printsecondlargest(); } public static void printsecondlargest() { int n; Scanner sc = new Scanner(System.in); System.out.println("Enter the No of Elements to store in array : "); n = sc.nextInt(); int[] array = new int[n]; System.out.println("Enter the array Elements : "); for (int i=0; i 0; i--) { if(array[i] != array[size-1]) { System.out.println("Second Largest Element is "+array[i]); return; } } System.out.println("No such Element Present"); } } Output: Enter the No of Elements to store in array : 6 Enter the array Elements : 13 35 35 23 34 13 Second Largest Element is 34

[@bhushantaywadevlogs9552]

Very bad logic...

[@yogeshganpule2695]

This will fail in case you have 2 duplicate elements as smallest . for ex int[] arr = {1, 2, 5, 7, 8, 9, -14, 56, 98, -3, -14, 2};