finding the second largest value in an array using java

Question

my question is that I need to find the second largest value from my array but I am getting the same value which is equal to the first value. please help

int[] nums = { 50, 6, 60, 70, 80, 90, 9, 150, 2, 35 };
int max = 0;
int secmax = 0;

for (int x = 0; x < nums.length; x++) {
    if (nums[x] > max)
        max = nums[x];
    if (nums[x] > secmax && secmax != max)
        secmax = nums[x];
}

System.out.println("1st H value: " + max);
System.out.println("2nd H Value: " + secmax);

Kamil Jarosz · Accepted Answer · 2021-07-18 19:20:46Z

7

your mistake is the conditions in the loop use this code:

public class Main{
    public static void main(String[] args){
        int[] nums = { 6, 9, 11, 1, 10 };

        int max = Integer.MIN_VALUE;
        int secmax = Integer.MIN_VALUE;

        for(int x=0; x<nums.length; x++) {
            if(nums[x]>max ) {
                secmax = max;
                max=nums[x];
            }else if(nums[x]>secmax){
                secmax=nums[x];
            }
         }
        System.out.println("1st H value: " + max);
        System.out.println("2nd H Value: " + secmax);
    }
}

edited Jul 18, 2021 at 19:20

Kamil Jarosz

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answered Feb 16, 2019 at 8:13

saeedata

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1 Comment

Haagenti Over a year ago

I think you should also take into consideration the value of 'secmax' variable in the above snippet. For eg: The above solution would simply fail for the test values 6 9 11 1 10

Ruslan · Accepted Answer · 2019-02-16 11:55:49Z

1

You should use if.. else if structure inside your for loop:

for (int item : nums) {
    if (item > max) {
        secmax = max;
        max = item;
    } else if (item > secmax) {
        secmax = item;
    }
}

Run time of this algorithm is O(n). There is also quite concise solution in case the array is sorted:

Arrays.sort(nums);     // Dual-Pivot Quicksort O(nlogn)
System.out.println(nums[nums.length - 1]);    // largest item
System.out.println(nums[nums.length - 2]);    // second largest item
...

You can get any n-largest item, but in this case the run time will be O(nlogn)

answered Feb 16, 2019 at 11:55

Ruslan

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Comments

SujJi · Accepted Answer · 2021-06-19 18:51:01Z

1

This could be the simplest one

public static void main(String[] args) {
    int[] array = { 1, 2, 3, -1, -2, 4 };
    Arrays.sort(array);
    System.out.println(array[array.length-2]);
    
}

answered Jun 19, 2021 at 18:51

SujJi

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Comments

Freeman · Accepted Answer · 2019-02-16 08:09:48Z

Step 1:

Iterate the given array

Step 2 (first if condition arr[i] > largest):

If current array value is greater than largest value then

Move the largest value to secondLargest and make

current value as largest

Step 3 (second if condition arr[i] > secondLargest )

If the current value is smaller than largest and greater than secondLargest then the current value becomes secondLargest

public class SecondLargest {

        public static void main(String[] args) {

            int arr[] = {50,06,60,70,80,90,9,150,2,35};
            int largest = arr[0];
            int secondLargest = arr[0];

            System.out.println("The given array is:" );
            for (int i = 0; i < arr.length; i++) {
                System.out.print(arr[i]+"\t");
            }
            for (int i = 0; i < arr.length; i++) {

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

                } else if (arr[i] > secondLargest) {
                    secondLargest = arr[i];

                }
            }

            System.out.println("\nSecond largest number is:" + secondLargest);

        }
    }

output :

The given array is:
50  6   60  70  80  90  9   150 2   35  
Second largest number is:90

Ruslan · Accepted Answer · 2019-02-16 11:32:49Z

0

int max=nums[0],secMax=nums[0]      
for (int x = 0; x < nums.length; x++) {
    if (nums[x] > max) {
        secMax=max;
        max = nums[x];
    }
}

secMax wil have the second largest

edited Feb 16, 2019 at 11:32

Ruslan

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answered Feb 16, 2019 at 8:21

Akash Pal

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Comments

Elyzer · Accepted Answer · 2019-03-28 17:41:17Z

0

Initialize max and scmax(second max) with 2 first value of array
Then I compare the rest with max and possibly scmax. at each comparison I adjust.

int arr[] = {-50,10,80,78,67,86,34,276,8};
        int max, scmax;
        if(arr[0]>=arr[1]){
            max = arr[0]; scmax=arr[1];}
        else{max = arr[1]; scmax=arr[0];}

        for (int i = 2; i < arr.length; i++) {
            if (max <= arr[i]) {
                scmax = max;
                max = arr[i];
            } else if(scmax<=arr[i]){ scmax = arr[i];}
        }
            System.out.println("max"+max);
         System.out.println("scmax"+scmax);

edited Mar 28, 2019 at 17:41

answered Mar 28, 2019 at 15:48

Elyzer

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4 Comments

Antony Hutchison Over a year ago

Some explanation of the answer would be helpful together with the logic/reason why this answers the question.

Elyzer Over a year ago

I have a little trouble with the interface, i'm just starting on this platform(sorry). the idea is that i initialize max and scmax with the first two values of the array. then when if i find a value greater than max, it becomes max and the old max becomes scmax

double-beep Over a year ago

While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply.

Elyzer Over a year ago

Thank you for notice! i edited and change the code.

Divyansh Chaudhary · Accepted Answer · 2020-02-06 14:56:14Z

0

public class Secondlargest {
public static void main(String[] args) {
    int arr[]= {5,3,6,8,9,11,5,74};
    List <Integer> array = new ArrayList<Integer> ();

    for(int i=0;i<arr.length;i++){
        if (array.isEmpty()){
            array.add(arr[i]);
        }
        else if(array.contains(arr[i])) {
        }
        else{
            array.add(arr[i]);
        }
    }

    Collections.sort(array);
    System.out.println("Second Largest "+ array.get(array.size()-2));
}

}

answered Feb 6, 2020 at 14:56

Divyansh Chaudhary

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Comments

Edward Abattam · Accepted Answer · 2022-01-09 15:44:09Z

0

Very easy answer using java streams

public static int secondLargestNumberInArray(Integer[] numbers) {
    return Arrays.stream(numbers).sorted(Collections.reverseOrder()).skip(1).findFirst().get();
}

In case you want to use primitive method parameter

public static int secondLargestNumberInArray(int[] numbers) {
    Integer[] arr = new Integer[numbers.length];
    Arrays.setAll(arr, i -> numbers[i]);
    return Arrays.stream(arr).sorted(Collections.reverseOrder()).skip(1).findFirst().get();
}

answered Jan 9, 2022 at 15:44

Edward Abattam

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