CODEWITHSUNDEEP
java
Mohamed
Mohamed

Reputation: 31

finding the second largest value in an array using java

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);

Upvotes: 1

Views: 10939

Answers (8)

Edward Abattam
Edward Abattam

Reputation: 91

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();
}

Upvotes: 0

saeedata
saeedata

Reputation: 947

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);
    }
}

Upvotes: 7

SujJi
SujJi

Reputation: 9

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]);
    
}

Upvotes: 1

Divyansh Chaudhary
Divyansh Chaudhary

Reputation: 31

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));
}

}

Upvotes: 0

Elyzer
Elyzer

Reputation: 1

  1. Initialize max and scmax(second max) with 2 first value of array
  2. 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);

Upvotes: 0

Ruslan
Ruslan

Reputation: 6290

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)

Upvotes: 1

Akash Pal
Akash Pal

Reputation: 1115

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

Upvotes: 0

Freeman
Freeman

Reputation: 12708

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

Upvotes: 0

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