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.

[@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

[@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!!!

[@bheemshankar_pk]

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

[@omkar_gopchade143]

I think it will definitely work

[@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);
}
}

[@newanurag]

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

[@bhargavmehta97]

What about array of negative numbers?

[@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

[@SaravanaKumar-lx8ue]

int max = 0;
int smax = 0;
if(arr.length < 2)
return -1;
for(int i = 0; i max){
smax = max;
max = arr[i];
} else if ((arr[i]>smax) && (arr[i]!=max)) {
smax = arr[i];
}
}
if(smax == 0)
return -1;
return smax;
smax will hold the value of second largest, this code has only one iteration of the array. Complexity is O(n).

[@rmcf3972]

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

[@hArsh_noob007]

Nope

[@manukarthikeya4560]

Is this code from gfg ??

[@siddumaradi7045]

yes it'll work sir

[@princemohammed921]

For bigger test cases it is not working

[@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);

[@darknight66655]

yes it will work surr**

[@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

[@trendinguniverse6113]

Bro pls make remove duplicate elements from array

[@girijasankardas1189]

you can easily remove duplicate element by using HashSet brother.

[@coolraviraj24]

yes it will

[@gautamnegi6868]

Yes

[@chaitanaywaskar6008]

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

[@Jaanufufjf]

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};