Finding the second smallest integer in array

Question

We are required in our assignment to find the second smallest integer in one array recursively. However, for the sake of understanding the subject more, I want to do it iter

Answer 1

Here is TimeComlexity Linear O(N):

  public static int secondSmallest(int[] arr) {
    if(arr==null || arr.length < 2) {
        throw new IllegalArgumentException("Input array too small");
    }

    //implement
    int firstSmall = -1;
    int secondSmall = -1;

    //traverse to find 1st small integer on array
    for (int i = 0; i<arr.length;i++)
        if (firstSmall == -1 || arr[firstSmall]>arr[i])
            firstSmall = i;

     //traverse to array find 2 integer, and skip first small
    for (int i = 0;i<arr.length;i++) {
        if (i != firstSmall && (secondSmall == -1 || arr[secondSmall] > arr[i]))
            secondSmall = i;
    }

    return arr[secondSmall];
}

Answer 2

public class SecondSmallestNumberInArray 
        {
            public static void main(String[] args) 
            {
                int arr[] = { 99, 76, 47, 85, 929, 52, 48, 36, 66, 81, 9 };
                int smallest = arr[0];
                int secondSmallest = arr[0];

                System.out.println("The given array is:");
                boolean find = false;
                boolean flag = true;

                for (int i = 0; i < arr.length; i++) 
                {
                    System.out.print(arr[i] + "  ");
                }

                System.out.println("");

                while (flag) 
                {
                    for (int i = 0; i < arr.length; i++) 
                    {
                        if (arr[i] < smallest) 
                        {   
                            find = true;
                            secondSmallest = smallest;
                            smallest = arr[i];
                        } else if (arr[i] < secondSmallest) {   
                            find = true;
                            secondSmallest = arr[i];
                        }
                    }
                    if (find) {
                        System.out.println("\nSecond Smallest number is Array : ->  " + secondSmallest);
                        flag = false;
                    } else {
                        smallest = arr[1];
                        secondSmallest = arr[1];
                    }
                }
            }
        }

    **Output is**

    D:\Java>java SecondSmallestNumberInArray

    The given array is:

    99  76  47  85  929  52  48  36  66  81  9

    Second Smallest number is Array : ->  36

    D:\Java>

Answer 3

Here's a Swift version that runs in linear time. Basically, find the smallest number. Then assign the 2nd minimum number as the largest value. Then loop through through the array and find a number greater than the smallest one but also smaller than the 2nd smallest found so far.

func findSecondMinimumElementLinear(in nums: [Int]) -> Int? {
    // If the size is less than 2, then returl nil.
    guard nums.count > 1 else { return nil }

    // First, convert it into a set to reduce duplicates.
    let uniqueNums = Array(Set(nums))

    // There is no point in sorting if all the elements were the same since it will only leave 1 element
    // after the set removed duplicates.
    if uniqueNums.count == 1 { return nil }

    let min: Int = uniqueNums.min() ?? 0 // O(n)
    var secondMinNum: Int = uniqueNums.max() ?? 0 // O(n)
    // O(n)
    for num in uniqueNums {
        if num > min && num < secondMinNum {
            secondMinNum = num
        }
    }

    return secondMinNum
}

Answer 4

Try this one.

    public static void main(String args[]){
        int[] array = new int[]{10, 30, 15, 8, 20, 4};

        int min, secondMin;

        if (array[0] > array[1]){
            min = array[1];
            secondMin = array[0];
        }
        else{
            min = array[0];
            secondMin = array[1];
        }

        for (int i=2; i<array.length; i++){
            if (array[i] < min){
                secondMin = min;
                min = array[i];
            }
            else if ((array[i] > min) && (array[i] < secondMin)){
                secondMin = array[i];
            }
        }
       System.out.println(secondMin);
  }

Answer 5

Find the second minimum element of an array in Python, short and simple

def second_minimum(arr):
    second = arr[1]
    first = arr[0]

    for n in arr:
        if n < first:
            first = n
        if n > first and n < second  :
            second = n

    return second

print(second_minimum([-2, 4, 5, -1, 2, 3, 0, -4, 1, 99, -6, -5, -19]))

Answer 6

public static void main(String[] args)  {
    int arr[] = {6,1,37,-4,12,46,5,64,21,2,-4,-3};
    int lowest =arr[0];
    int sec_lowest =arr[0];
    for(int n : arr){
        if (lowest >  n)
        {
            sec_lowest = lowest;
            lowest = n;

        }
        else if (sec_lowest > n && lowest != n)
            sec_lowest = n;
    }

    System.out.println(lowest+"   "+sec_lowest);

    }