Finding the second highest number in array in Java

Question

I'm having difficulty to understand the logic behind the method to find the second highest number in array. The method used is to find the highest in the array but less than the previous highest (which has already been found). The thing that I still can't figure it out is why || highest_score == second_highest is necessary. For example I input three numbers: 98, 56, 3. Without it, both highest and second highest would be 98. Please explain.

int second highest = score[0];  
if (score[i] > second_highest && score[i] < highest_score || highest_score == second_highest)   
    second_highest = score[i];

Your code snippet does not indicate how highest_score was initialized — Martin OConnor
– Martin OConnor, Commented Apr 11, 2010 at 1:39
Should second highest be second_highest? Is there a loop? — trashgod
– trashgod, Commented Apr 11, 2010 at 1:39
Very similar threads: stackoverflow.com/questions/1811846/… and stackoverflow.com/questions/1412751/… — Alexandros Gezerlis
– Alexandros Gezerlis, Commented Apr 11, 2010 at 5:41

polygenelubricants · Accepted Answer · 2010-04-11 02:00:00Z

42

I'm not convinced that doing what you did fixes the problem; I think it masks yet another problem in your logic. To find the second highest is actually quite simple:

 static int secondHighest(int... nums) {
    int high1 = Integer.MIN_VALUE;
    int high2 = Integer.MIN_VALUE;
    for (int num : nums) {
      if (num > high1) {
        high2 = high1;
        high1 = num;
      } else if (num > high2) {
        high2 = num;
      }
    }
    return high2;
 }

This is O(N) in one pass. If you want to accept ties, then change to if (num >= high1), but as it is, it will return Integer.MIN_VALUE if there aren't at least 2 elements in the array. It will also return Integer.MIN_VALUE if the array contains only the same number.

edited Apr 11, 2010 at 2:00

answered Apr 11, 2010 at 1:49

polygenelubricants

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2 Comments

loc Over a year ago

To the array [2,2,1] it gives back 2. Isn't it wrong?

Jhabar Over a year ago

the 'else if (num > high2)' check should be 'else if (num != high1 && num > high2)'

Scott Smith · Accepted Answer · 2010-04-11 02:12:29Z

16

// Initialize these to the smallest value possible
int highest = Integer.MIN_VALUE;
int secondHighest = Integer.MIN_VALUE;

// Loop over the array
for (int i = 0; i < array.Length; i++) {

    // If we've found a new highest number...
    if (array[i] > highest) {

        // ...shift the current highest number to second highest
        secondHighest = highest;

        // ...and set the new highest.
        highest = array[i];
    } else if (array[i] > secondHighest)
        // Just replace the second highest
        secondHighest = array[i];
    }
}

// After exiting the loop, secondHighest now represents the second
// largest value in the array

Edit:

Whoops. Thanks for pointing out my mistake, guys. Fixed now.

edited Apr 11, 2010 at 2:12

answered Apr 11, 2010 at 1:41

Scott Smith

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4 Comments

Anthony Pegram Over a year ago

This code will not handle the condition of the current element exceeding the second highest value but not the first highest value.

polygenelubricants Over a year ago

This fails on input like {1, 0}.

hitesh141 Over a year ago

super... i love this logic. can you tell me what about the third max??

Scott Smith Over a year ago

@hitesh - is that sarcasm? OP didn't ask about the third max. If s/he wanted the nth max, the algorithm would obviously have been different.

Nathan S. · Accepted Answer · 2010-04-11 01:37:18Z

5

If the first element which second_highest is set to initially is already the highest element, then it should be reassigned to a new element when the next element is found. That is, it's being initialized to 98, and should be set to 56. But, 56 isn't higher than 98, so it won't be set unless you do the check.

If the highest number appears twice, this will result in the second highest value as opposed to the second element that you would find if you sorted the array.

answered Apr 11, 2010 at 1:37

Nathan S.

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1 Comment

Arkku Over a year ago

+1 for pointing out that this returns the second highest value instead of the element.

Keshav Gera · Accepted Answer · 2021-07-02 11:33:11Z

4

let array = [0,12,74,26,82,176,189,8,55,3,189]; 
let highest=0 ;
let secondHighest = 0;
for (let i = 0; i < array.length; i++) { 
if (array[i] > highest) { 
    // ...shift the current highest number to second highest
    secondHighest = highest; 
    // ...and set the new highest.
    highest = array[i]; 
} else if (highest > array[i] > secondHighest) { 
   // Just replace the second highest
   secondHighest = array[i]; 
  }
}
console.log(secondHighest);

edited Jul 2, 2021 at 11:33

answered Jul 6, 2016 at 11:19

Keshav Gera

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1 Comment

Frits Over a year ago

Please try to avoid just dumping a code as an answer and try to explain what it does and why. Your code might not be obvious for people who do not have the relevant coding experience.

subhashis · Accepted Answer · 2015-02-03 06:51:51Z

3

The answers I saw wont work if there are two same largest numbers like the below example.

        int[] randomIntegers = { 1, 5, 4, 2, 8, 1, 8, 9,9 };
        SortedSet<Integer> set = new TreeSet<Integer>();
        for (int i: randomIntegers) {
            set.add(i);
        }
        // Remove the maximum value; print the largest remaining item
        set.remove(set.last());
        System.out.println(set.last());

I have removed it from the Set not from the Array

answered Feb 3, 2015 at 6:51

subhashis

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Comments

Boris · Accepted Answer · 2017-11-30 12:07:28Z

2

My idea is that you assume that first and second members of the array are your first max and second max. Then you take each new member of an array and compare it with the 2nd max. Don't forget to compare 2nd max with the 1st one. If it's bigger, just swap them.

   public static int getMax22(int[] arr){
    int max1 = arr[0];
    int max2 = arr[1];
    for (int i = 2; i < arr.length; i++){
        if (arr[i] > max2)
        {
            max2 = arr[i];
        }

        if (max2 > max1)
        {
            int temp = max1;
            max1 = max2;
            max2 = temp;
        }
    }
     return max2;
}

answered Nov 30, 2017 at 12:07

Boris

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Comments

Chander Shivdasani · Accepted Answer · 2012-09-24 21:20:55Z

1

 public static int secondLargest(int[] input) {
            int largest,secondLargest;

            if(input[0] > input[1]) {
                largest = input[0];
                secondLargest = input[1];
            }
            else {
                largest = input[1];
                secondLargest = input[0];
            }

            for(int i = 2; i < input.length; i++) {
                if((input[i] <= largest) && input[i] > secondLargest) {
                    secondLargest = input[i];
                }

                if(input[i] > largest) {
                    secondLargest = largest;
                    largest = input[i];
                }
            }

            return secondLargest;
        }

answered Sep 24, 2012 at 21:20

Chander Shivdasani

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Comments

Neeraj Gahlawat · Accepted Answer · 2017-08-22 04:57:52Z

import java.util.Scanner;

public class SecondHighestFromArrayTest {

public static void main(String[] args) {
    Scanner scan = new Scanner(System.in);
    System.out.println("Enter size of Array");
    int size = scan.nextInt();
    int[] arr = new int[size];
    for (int i = 0; i < size; i++) {
        arr[i] = scan.nextInt();
    }
    System.out.println("second highest element " + getSecondHighest(arr));
}

public static int getSecondHighest(int arr[]) {
    int firstHighest = arr[0];
    int secondHighest = arr[0];
    for (int i = 0; i < arr.length; i++) {
        if (arr[i] > firstHighest) {
            secondHighest = firstHighest;
            firstHighest = arr[i];
        } else if (arr[i] > secondHighest) {
            secondHighest = arr[i];
        }
    }
    return secondHighest;
}

}

Soudipta Dutta · Accepted Answer · 2018-06-17 01:20:06Z

1

The second largest element in the array : IN Java:

class test2{
    public static void main(String[] args) {

int a[] = {1,2,3,9,5,7,6,4,8};
Arrays.sort(a);
int aa = a[a.length -2 ];
System.out.println(aa);


    }//main

}//end

In Python :

a = [1, 2, 3, 9, 5, 7, 6, 4, 8]

aa = sorted(list(a))
print(aa)
aaa = aa[-2]
print(aaa)

answered Jun 17, 2018 at 1:20

Soudipta Dutta

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Comments

Stranger · Accepted Answer · 2013-02-14 17:37:15Z

0

If time complexity is not an issue, then You can run bubble sort and within two iterations, you will get your second highest number because in the first iteration of the loop, the largest number will be moved to the last. In the second iteration, the second largest number will be moved next to last.

answered Feb 14, 2013 at 17:37

Stranger

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1 Comment

Jeff Foster Over a year ago

You mean selection sort, not bubble sort. Bubble sort only swaps adjacent elements.

Juliano Alves · Accepted Answer · 2013-07-19 18:11:06Z

If you want to 2nd highest and highest number index in array then....

public class Scoller_student {

    public static void main(String[] args) {
        System.out.println("\t\t\tEnter No. of Student\n");
        Scanner scan = new Scanner(System.in);
        int student_no = scan.nextInt();

        // Marks Array.........
        int[] marks;
        marks = new int[student_no];

        // Student name array.....
        String[] names;
        names = new String[student_no];
        int max = 0;
        int sec = max;
        for (int i = 0; i < student_no; i++) {
            System.out.println("\t\t\tEnter Student Name of id = " + i + ".");

            names[i] = scan.next();
            System.out.println("\t\t\tEnter Student Score of id = " + i + ".\n");

            marks[i] = scan.nextInt();
            if (marks[max] < marks[i]) {
                sec = max;
                max = i;
            } else if (marks[sec] < marks[i] && marks[max] != marks[i]) {
                sec = i;
            }
        }

        if (max == sec) {
            sec = 1;
            for (int i = 1; i < student_no; i++) {
                if (marks[sec] < marks[i]) {
                    sec = i;
                }
            }
        }

        System.out.println("\t\t\tHigherst score id = \"" + max + "\" Name : \""
            + names[max] + "\" Max mark : \"" + marks[max] + "\".\n");
        System.out.println("\t\t\tSecond Higherst score id = \"" + sec + "\" Name : \""
            + names[sec] + "\" Max mark : \"" + marks[sec] + "\".\n");

    }
}

Deep · Accepted Answer · 2014-07-02 10:41:41Z

public class secondLargestElement 
{
    public static void main(String[] args) 
    {
        int []a1={1,0};
        secondHigh(a1);
    }

    public static void secondHigh(int[] arr)
    {
        try
        {
            int highest,sec_high;
            highest=arr[0];
            sec_high=arr[1];

                for(int i=1;i<arr.length;i++)
                {
                    if(arr[i]>highest)
                    {           
                        sec_high=highest;
                        highest=arr[i]; 
                    }
                    else 
                    // The first condition before the || is to make sure that second highest is not actually same as the highest , think 
                        // about {5,4,5}, you don't want the last  5 to be reported as the sec_high
                        // The other half after || says if the first two elements are same then also replace the sec_high with incoming integer
                        // Think about {5,5,4}
                    if(arr[i]>sec_high && arr[i]<highest || highest==sec_high)
                        sec_high=arr[i];
                }
            //System.out.println("high="+highest +"sec"+sec_high);
            if(highest==sec_high)
                System.out.println("All the elements in the input array are same");
             else
                 System.out.println("The second highest element in the array is:"+ sec_high);

         }

        catch(ArrayIndexOutOfBoundsException e)
        {
        System.out.println("Not enough elements in the array");
        //e.printStackTrace();
        }
    }
}

Prakash H · Accepted Answer · 2015-01-08 13:46:49Z

You can find Largest and Third largest number of unsorted array as well.

 public class ThirdLargestNumber {
        public static void main(String[] args) {
            int arr[] = { 220, 200, 100, 100, 300, 600, 50, 5000, 125, 785 };
            int first = 0, second = 0, third = 0, firstTemp = 0, secondTemp = 0;
            for (int i = 0; i <= 9 /*
                                     * Length of array-1. You can use here length
                                     * property of java array instead of hard coded
                                     * value
                                     */; i++) {
                if (arr[i] == first) {
                    continue;
                }
                if (arr[i] > first) {
                    firstTemp = first;
                    secondTemp = second;
                    first = arr[i];
                    second = firstTemp;
                    if (secondTemp > third) {
                        third = secondTemp;
                    }
                } else {
                    if ((arr[i] == second) || (arr[i]) == first) {
                        continue;
                    }
                    if ((arr[i] > second) && (arr[i]) < first) {
                        secondTemp = second;
                        second = arr[i];
                        if (secondTemp > third) {
                            third = secondTemp;
                        }
                    } else {
                        if (arr[i] > third) {
                            third = arr[i];
                        }
                    }
                }
            }
            // System.out.println("Third largest number: " + third);
            System.out.println("Second largest number: " + second);
            // System.out.println("Largest number: " + first);
        }
    }

Ashish Singh · Accepted Answer · 2015-07-31 12:52:48Z

0

I think for finding the second Highest no we require these lines,if we can use inbuilt function

int[] randomIntegers = {1, 5, 4, 2, 8, 1, 1, 6, 7, 8, 9};
    Arrays.sort(randomIntegers);
    System.out.println(randomIntegers[randomIntegers.length-2]);

answered Jul 31, 2015 at 12:52

Ashish Singh

11

Comments

Praveen M P · Accepted Answer · 2015-11-04 14:53:34Z

I am giving solution that's not in JAVA program (written in JavaScript), but it takes o(n/2) iteration to find the highest and second highest number.
Working fiddler link Fiddler link

 var num=[1020215,2000,35,2,54546,456,2,2345,24,545,132,5469,25653,0,2315648978523];
var j=num.length-1;
var firstHighest=0,seoncdHighest=0;
num[0] >num[num.length-1]?(firstHighest=num[0],seoncdHighest=num[num.length-1]):(firstHighest=num[num.length-1],   seoncdHighest=num[0]);
j--;
for(var i=1;i<=num.length/2;i++,j--)
{
   if(num[i] < num[j] )
   {
          if(firstHighest < num[j]){
          seoncdHighest=firstHighest;
           firstHighest= num[j];
          }
           else if(seoncdHighest < num[j] ) {
               seoncdHighest= num[j];

           }
   }
   else {
       if(firstHighest < num[i])
       {
           seoncdHighest=firstHighest;
           firstHighest= num[i];

       }
       else if(seoncdHighest < num[i] ) {
            seoncdHighest= num[i];

       }
   }

}

Manash Ranjan Dakua · Accepted Answer · 2015-11-18 16:29:02Z

public class SecondandThirdHighestElement {
    public static void main(String[] args) {
        int[] arr = {1,1,2,3,8,1,2,3,3,3,2,3,101,6,6,7,8,8,1001,99,1,0};
        // create three temp variable and store arr of first element in that temp variable so that it will compare with other element
        int firsttemp = arr[0];
        int secondtemp = arr[0];
        int thirdtemp = arr[0];
        //check and find first highest value from array by comparing with other elements if found than save in the first temp variable 
        for (int i = 0; i < arr.length; i++) {
            if(firsttemp <arr[i]){
                firsttemp =  arr[i];
            }//if

        }//for
        //check and find the second highest variable by comparing with other elements in an array and find the element and that element should be smaller than first element array
        for (int i = 0; i < arr.length; i++) {
            if(secondtemp < arr[i] && firsttemp>arr[i]){
                secondtemp = arr[i];
            }//if
        }//for
        //check and find the third highest variable by comparing with other elements in an array and find the element and that element should be smaller than second element array

        for (int i = 0; i < arr.length; i++) {
            if(thirdtemp < arr[i] && secondtemp>arr[i]){
                thirdtemp = arr[i];
            }//if
        }//for

        System.out.println("First Highest Value:"+firsttemp);
        System.out.println("Second Highest Value:"+secondtemp);
        System.out.println("Third Highest  Value:"+thirdtemp);

    }//main
}//class

James Johney · Accepted Answer · 2016-01-04 22:33:49Z

If this question is from the interviewer then please DONT USE SORTING Technique or Don't use any built in methods like Arrays.sort or Collection.sort. The purpose of this questions is how optimal your solution is in terms of performance so the best option would be just implement with your own logic with O(n-1) implementation. The below code is strictly for beginners and not for experienced guys.

  public void printLargest(){


    int num[] ={ 900,90,6,7,5000,4,60000,20,3};

    int largest = num[0];

    int secondLargest = num[1];

    for (int i=1; i<num.length; i++)
    {
        if(largest < num[i])
        {
            secondLargest = largest;
            largest = num[i];


        }
        else if(secondLargest < num[i]){
            secondLargest = num[i];
        }
    }
    System.out.println("Largest : " +largest);
    System.out.println("Second Largest : "+secondLargest);
}

SagaciousLearner · Accepted Answer · 2016-02-12 03:05:53Z

Problem: The problem is to get the second largest array element.

Observation: Second largest number is defined as the number that has the minimum difference when subtracted from the maximum element in the array.

Solution: This is a two pass solution. First pass is to find the maximum number. Second pass is to find the element that has minimum difference with the maximum element as compared to other array elements. Example: In the array [2, 3, 6, 6, 5] maximum = 6 and second maximum = 5 , since it has the minimum difference to the maximum element 6 - 5 = 1 the solution for second largest = 5

function printSecondMax(myArray) {
  var x, max = myArray[0];
  // Find maximum element 
  for(x in myArray){
     if(max < myArray[x]){
        max = myArray[x];
     }
  }
  var secondMax = myArray[0], diff = max - secondMax;
  // Find second max, an element that has min diff with the max
  for(x in myArray){
    if(diff != 0 && (max - myArray[x]) != 0 && diff > (max - myArray[x])){
        secondMax = myArray[x];
        diff = max - secondMax;
    }
  }
  console.log(secondMax);
}

Complexity : O(n), This is the simplest way to do this.

For finding maximum element even more efficiently one can look into max heap, a call to max-heapify will take O(log n) time to find the max and then pop-ing the top element gives maximum. To get the second maximum, max-heapify after pop-ing the top and keep pop-ing till you get a number that is less than maximum. That will be the second maximum. This solution has O(n log n) complexity.

Pang · Accepted Answer · 2016-02-12 03:13:34Z

0

private static int SecondBiggest(int[] vector)
{
    if (vector == null)
    {
        throw new ArgumentNullException("vector");
    }
    if (vector.Length < 2)
    {
        return int.MinValue;
    }

    int max1 = vector[0];
    int max2 = vector[1];
    for (int i = 2; i < vector.Length; ++i)
    {
        if (max1 > max2 && max1 != vector[i])
        {
            max2 = Math.Max(max2, vector[i]);
        }
        else if (max2 != vector[i])
        {
            max1 = Math.Max(max1, vector[i]);
        }
    }
    return Math.Min(max1, max2);
}

This treats duplicates as the same number. You can change the condition checks if you want to all the biggest and the second biggest to be duplicates.

edited Feb 12, 2016 at 3:13

Pang

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answered Feb 14, 2013 at 2:31

csaam

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Comments

Farhan Ghaffar · Accepted Answer · 2016-03-03 20:41:23Z

0

Please try this one: Using this method, You can fined second largest number in array even array contain random number. The first loop is used to solve the problem if largest number come first index of array.

public class secondLargestnum {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] array = new int[6];
        array[0] = 10;
        array[1] = 80;
        array[2] = 5;
        array[3] = 6;
        array[4] = 50;
        array[5] = 60;
        int tem = 0;
        for (int i = 0; i < array.length; i++) {
            if (array[0]>array[i]) {
                tem = array[0];
            array[0] = array[array.length-1];
            array[array.length-1] = tem;
            }
        }
        Integer largest = array[0];
        Integer second_largest = array[0];

        for (int i = 0; i < array.length; i++) {

            if (largest<array[i]) {
                second_large = largest;
                largest = array[i];
            }
            else if (second_large<array[i]) {
                second_large = array[i];

            }

        }
System.out.println("largest number "+largest+" and second largest number "+second_largest);

    }

}

edited Mar 3, 2016 at 20:41

answered Mar 2, 2016 at 16:32

Farhan Ghaffar

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1 Comment

Mogsdad Over a year ago

Please edit with more information. Code-only and "try this" answers are discouraged, because they contain no searchable content, and don't explain why someone should "try this". We make an effort here to be a resource for knowledge.

SanA · Accepted Answer · 2016-03-20 12:59:00Z

0

public class SecondHighInIntArray {

public static void main(String[] args) {
    int[] intArray=new int[]{2,2,1};
            //{2,2,1,12,3,7,9,-1,-5,7};
    int secHigh=findSecHigh(intArray);
    System.out.println(secHigh);
}

private static int findSecHigh(int[] intArray) {

    int highest=Integer.MIN_VALUE;
    int sechighest=Integer.MIN_VALUE;
    int len=intArray.length;
    for(int i=0;i<len;i++)
    {
        if(intArray[i]>highest)
        {
            sechighest=highest;
            highest=intArray[i];
            continue;
        }

        if(intArray[i]<highest && intArray[i]>sechighest)
        {
            sechighest=intArray[i];
            continue;
        }


    }
    return sechighest;
}

}

edited Mar 20, 2016 at 12:59

answered Mar 20, 2016 at 12:18

SanA

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Comments

dur · Accepted Answer · 2016-03-20 13:28:24Z

0

Second Largest in O(n/2)

public class SecMaxNum {

    // second Largest number with O(n/2)
    /**
     * @author Rohan Kamat
     * @Date Feb 04, 2016
     */
    public static void main(String[] args) {
        int[] input = { 1, 5, 10, 11, 11, 4, 2, 8, 1, 8, 9, 8 };
        int large = 0, second = 0;

        for (int i = 0; i < input.length - 1; i = i + 2) {
            // System.out.println(i);
            int fist = input[i];
            int sec = input[i + 1];
            if (sec >= fist) {
                int temp = fist;
                fist = sec;
                sec = temp;
            }
            if (fist >= second) {
                if (fist >= large) {
                    large = fist;
                } else {
                    second = fist;
                }

            }

            if (sec >= second) {
                if (sec >= large) {
                    large = sec;
                } else {
                    second = sec;
                }
            }
        }
    }
}

edited Mar 20, 2016 at 13:28

dur

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answered Feb 4, 2016 at 11:09

rohan kamat

5815 silver badges12 bronze badges

1 Comment

very hit Over a year ago

Hi, your solution is not O(n/2), because the times of comparison determines the compute complexity for this question. Even if you times of iteration is n/2, but every iteration you would compare more than 2 times so that the whole times of comparison is still more than n times. BTW, when we talk about compute complexity, we usually ignore the coefficient, O(0.5n), O(2n), and O(4n), all of them can be seen as the same class.

dur · Accepted Answer · 2016-03-20 21:44:37Z

public class SecondHighest {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        /*
         * Find the second largest int item in an unsorted array.
         * This solution assumes we have atleast two elements in the array
         * SOLVED! - Order N. 
         * Other possible solution is to solve with Array.sort and get n-2 element.
         * However, Big(O) time NlgN 
         */

        int[] nums = new int[]{1,2,4,3,5,8,55,76,90,34,91};
        int highest,cur, secondHighest = -1;

        int arrayLength = nums.length;
        highest = nums[1] > nums[0] ? nums[1] : nums[0];
        secondHighest = nums[1] < nums[0] ? nums[1] : nums[0];

        if (arrayLength == 2) {
            System.out.println(secondHighest);

        } else {

            for (int x = 0; x < nums.length; x++) {

                cur = nums[x];
                int tmp;

                if (cur < highest && cur > secondHighest)
                    secondHighest = cur;

                else if (cur > secondHighest && cur > highest) {
                    tmp = highest;
                    highest = cur;
                    secondHighest = tmp;
                }

            }   

            System.out.println(secondHighest);

        }   
    }
}

devasghar · Accepted Answer · 2016-08-14 12:53:15Z

0

Use following function
`

public static int secHigh(int arr[]){
            int firstHigh = 0,secHigh = 0;
            for(int x: arr){
                if(x > firstHigh){
                    secHigh = firstHigh;
                    firstHigh = x;
                }else if(x > secHigh){
                    secHigh = x;
                }
            }
            return secHigh;
        }

Function Call

int secondHigh = secHigh(arr);

answered Aug 14, 2016 at 12:53

devasghar

613 bronze badges

Comments

Keith OYS · Accepted Answer · 2016-12-30 04:25:08Z

0

import java.util.Scanner;

public class SecondLargest {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Enter size of array : ");
        int n = sc.nextInt();
        int ar[] = new int[n];
        for(int i=0;i<n;i++)
        {
            System.out.print("Enter value for array : ");
            ar[i] = sc.nextInt();
        }
        int m=ar[0],m2=ar[0];
        for(int i=0;i<n;i++)
        {
            if(ar[i]>m)
                m=ar[i];
        }
        for(int i=0;i<n;i++)
        {
            if(ar[i]>m2 && ar[i]<m)
                m2=ar[i];
        }
        System.out.println("Second largest : "+m2);
        sc.close();
    }
}

edited Dec 30, 2016 at 4:25

Keith OYS

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answered Dec 30, 2016 at 3:38

Asish

12 bronze badges

Comments

Rahul Chauhan · Accepted Answer · 2017-05-30 12:01:36Z

0

public void findMax(int a[]) {
    int large = Integer.MIN_VALUE;
    int secondLarge = Integer.MIN_VALUE;
    for (int i = 0; i < a.length; i++) {
        if (large < a[i]) {
            secondLarge = large;
            large = a[i];
        } else if (a[i] > secondLarge) {
            if (a[i] != large) {
                secondLarge = a[i];
            }
        }
    }
    System.out.println("Large number " + large + " Second Large  number " + secondLarge);
}

The above code has been tested with integer arrays having duplicate entries, negative values. Largest number and second largest number are retrived in one pass. This code only fails if array only contains multiple copy of same number like {8,8,8,8} or having only one number.

edited May 30, 2017 at 12:01

answered May 30, 2017 at 11:56

Rahul Chauhan

264 bronze badges

Comments

Dheeraj Devadiga · Accepted Answer · 2017-06-16 12:57:05Z

0

public class SecondLargestNumber
{
  public static void main(String[] args)
  {
    int[] var={-11,-11,-11,-11,115,-11,-9};
    int largest = 0;
    int secLargest = 0;
    if(var.length == 1)
    {
      largest = var[0];
      secLargest = var[0];
    }
    else if(var.length > 1)
    {
      largest= var[0];
      secLargest = var[1];
      for(int i=1;i<var.length;i++)
      {
        if(secLargest!=largest)
        {
          if(var[i]>largest)
          { 
            secLargest = largest;
            largest = var[i];
          }
          else if(var[i]>secLargest && var[i] != largest)
          {
            secLargest= var[i];
          }
        }
        else
        {
          if(var[i]>largest)
          {
           secLargest = largest;
           largest = var[i];
          }
          else
          {
           secLargest = var[i];
          }
       }
    }
  }

    System.out.println("Largest: "+largest+" Second Largest: "+secLargest);
  }
}

answered Jun 16, 2017 at 12:57

Dheeraj Devadiga

111 bronze badge

1 Comment

Gary99 Over a year ago

It may help if you explain your answer some. What was the problem with the OP's code? How does this fix it?

Venkata · Accepted Answer · 2017-07-16 05:20:17Z

   /* Function to print the second largest elements */
    void print2largest(int arr[], int arr_size)
    {
   int i, first, second;

   /* There should be atleast two elements */
   if (arr_size < 2)
   {
    printf(" Invalid Input ");
    return;
    }

   first = second = INT_MIN;
   for (i = 0; i < arr_size ; i ++)
   {
    /* If current element is smaller than first
       then update both first and second */
    if (arr[i] > first)
    {
        second = first;
        first = arr[i];
    }

    /* If arr[i] is in between first and 
       second then update second  */
    else if (arr[i] > second && arr[i] != first)
        second = arr[i];
   }
   if (second == INT_MIN)
    printf("There is no second largest elementn");
    else
    printf("The second largest element is %dn", second);
    }

RKP · Accepted Answer · 2017-07-21 20:02:04Z

0

I have got the simplest logic to find the second largest number may be, it's not. The logic find sum of two number in the array which has the highest value and then check which is greater among two simple............

int ar[]={611,4,556,107,5,55,811};
int sum=ar[0]+ar[1];
int temp=0;
int m=ar[0];
int n=ar[1];
for(int i=0;i<ar.length;i++){
    for(int j=i;j<ar.length;j++){
        if(i!=j){
        temp=ar[i]+ar[j];
        if(temp>sum){
            sum=temp;
            m=ar[i];
            n=ar[j];
        }
        temp=0;

    }
    }
}
if(m>n){
    System.out.println(n);

}
else{
    System.out.println(m);
}

answered Jul 21, 2017 at 20:02

RKP

496 bronze badges

Comments

Avinash Kadam · Accepted Answer · 2018-04-07 10:06:35Z

0

The simplest way is -

public class SecondLargest {
    public static void main(String[] args) {
        int[] arr = { 1, 2, 5, 6, 3 };

        int first = Integer.MIN_VALUE;
        int second = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) {
            // If current element is smaller than first then update both first
            // and second
            if (arr[i] > first) {
                second = first;
                first = arr[i];
            }
            // If arr[i] is in between first and second then update second
            else if (arr[i] > second && arr[i] != first) {
                second = arr[i];
            }
        }
    }
}

answered Apr 7, 2018 at 10:06

Avinash Kadam

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