Python – Find second smallest number

Yes, except that code relies on a small quirk (that raises an exception in Python 3): the fact that None compares as smaller than a number.

Another value that works is float("-inf"), which is a number that is smaller than any other number.

If you use that instead of None, and just change -inf to +inf and > to <, there’s no reason it wouldn’t work.

Edit: another possibility would be to simply write -x in all the comparisons on x, e.g. do if -x <= m1: et cetera.

Here we want to keep an invariant while we scan the list of numbers, for every sublist it must be

m1<=m2<={all other elements}

the minimum length of a list for which the question (2nd smallest) is sensible is 2, so we establish the invariant examining the first and the second element of the list (no need for magic numbers), next we iterate on all the remaining numbers, maintaining our invariant.

def second_smaller(numbers):
    # if len(numbers)<2: return None or otherwise raise an exception

    m1, m2 = numbers[:2]
    if m2<m1: m1, m2 = m2, m1

    for x in numbers[2:]:
        if x <= m1:
            m1, m2 = x, m1
        elif x < m2:
            m2 = x
    return m2

Addendum

BTW, the same reasoning should be applied to the second_largest function mentioned by the OP

Or just use heapq:

import heapq
def second_smallest(numbers):
    return heapq.nsmallest(2, numbers)[-1]

second_smallest([1, 2, 3, 4])
# Output: 2

l = [41,9000,123,1337]

# second smallest
sorted(l)[1]
123


# second biggest
sorted(l)[-2]
1337

As per the Python in-built function sorted

sorted(my_list)[0]

gives back the smallest number, and sorted(my_list)[1] does accordingly for the second smallest, and so on and so forth.

I am writing the code which is using recursion to find the second smallest element in a list.

def small(l):
 small.counter+=1;
 min=l[0];

 emp=[]

 for i in range(len(l)):
    if l[i]<min:
        min=l[i]

 for i in range(len(l)):
    if min==l[i]:
     emp.append(i)

 if small.counter==2:
    print "The Second smallest element is:"+str(min)
 else:
   for j in range(0,len(emp)):

     l.remove(min)

   small(l)
small.counter = 0

list=[-1-1-1-1-1-1-1-1-1,1,1,1,1,1]
small(list)

You can test it with various input integers.

a = [6,5,4,4,2,1,10,1,2,48]
s = set(a) # used to convert any of the list/tuple to the distinct element and sorted sequence of elements
# Note: above statement will convert list into sets 
print sorted(s)[1] 

There is a easy way to do . First sort the list and get the second item from the list.

def solution(a_list):

    a_list.sort()
    print a_list[1]

solution([1, 2, -8, -2, -10])

You can use in built function ‘sorted‘

def second_smallest(numbers):

count = 0
l = []
for i in numbers:
    if(i not in l):
        l.append(i)
        count+=1
    if(count==2):
        break

return max(l)

def SecondSmallest(x):
    lowest=min(x[0],x[1])
    lowest2 = max(x[0],x[1])
    for item in x:
         if item < lowest:
            lowest2 = lowest
            lowest = item
         elif lowest2 > item and item > lowest:
            lowest2 = item
    return lowest2

SecondSmallest([10,1,-1,2,3,4,5])

To find second smallest in the list, use can use following approach which will work if two or more elements are repeated.

def second_smallest(numbers):
     s = sorted(set(numbers))
     return s[1]

Here is:

def find_second_smallest(a: list) -> int:
    first, second = float('inf')
    for i in range(len(a)):
        if a[i] < first:
            first, second = a[i], first
        elif a[i] < second and a[i] != first:
            second = a[i]
    return second

input: [1, 1, 1, 2]
output: 2

My favourite way of finding the second smallest number is by eliminating the smallest number from the list and then printing the minimum from the list would return me the second smallest element of the list. The code for the task is as below:

mylist=[1,2,3,4]
mylist=[x for x in mylist if x!=min(mylist)]  #deletes the min element from the list
print(min(mylist))

Solution that returns second unique number in list with no sort:

def sec_smallest(numbers):
    smallest = float('+inf')
    small = float('+inf')
    for i in numbers:
        if i < smallest:
            small = smallest
            smallest = i
        elif i < small and i != smallest:
            small = i
    return small

print('Sec_smallest:', sec_smallest([1, 2, -8, -8, -2, 0]))

 mi= min(input_list)
    second_min = float('inf')
    for i in input_list:
        if i != mi:
            if i<second_min:
                second_min=i
    if second_min == float('inf'):
        print('not present')
    else:
        print(second_min)
             

##input_list = [6,6,6,6,6]
#input_list = [3, 1, 4, 4, 5, 5, 5, 0, 2, 2]
#input_list = [7, 2, 0, 9, -1, 8]
# Even if there is same number in the list then Python will not get confused.

This code is also works fine, To find the second smallest number in list.
For this code first we have to sort the values in list. after that we have to initialize the variable as second index.

l1 = [12,32,4,34,64,3,43]
for i in range(0,len(l1)):
       for j in range(0,i+1):
          if l1[i]<l1[j]:
              l1[i],l1[j]=l1[j],l1[i]
min_val = l1[1]
for k in l1:
   if min_val>k:
       break
print(min_val)

I’d like to add another, more general approach:
Here’s a recursive way of finding the i-th minimums of a given list of numbers

def find_i_minimums(numbers,i):
    minimum = float('inf')
    if i==0:
        return []
    less_than_i_minimums = find_i_minimums(numbers,i-1)
    for element in numbers:
        if element not in less_than_i_minimums and element < minimum:
            minimum = element
    return less_than_i_minimums + [minimum]

For example,

>>> find_i_minimums([0,7,4,5,21,2,6,1],3) # finding 3 minimial values for the given list
[0, 1, 2]

( And if you want only the i-th minimum number you’d extract the final value of the list )

The time-complexity of the above algorithm is bad though, it is O(N*i^2) ( Since the recursion depth is i , and at each recursive call we go over all values in ‘numbers’ list whose length is N and we check if the minimum element we’re searching for isn’t in a list of length i-1, thus the total complexity can be described by a geometric sum that will give the above mentioned complexity ).

Here’s a similar but alternative-implementation whose time-complexity is O(N*i) on average. It uses python’s built-in ‘set’ data-structure:

def find_i_minimums(numbers,i):
    minimum = float('inf')
    if i==0:
        return set()
    less_than_i_minimums = find_i_minimums(numbers,i-1)
    for element in numbers:
        if element not in less_than_i_minimums and element < minimum:
            minimum = element
    return less_than_i_minimums.union(set({minimum}))

If your ‘i’ is small, you can use the implementations above and then extract how many minimums you want ( or if you want the second minimum, then in your case run the code for i=2 and just extract the last element from the output data-structure ).
But if ‘i’ is for example greater than log(N) , I’d recommend sorting the list of numbers itself ( for example, using mergesort whose complexity is O(N*log(N)) at worst case ) and then taking the i-th element. Why so? because as stated, the run-time of the algorithm above is not great for larger values of ‘i’.

You might find this code easy and understandable

def secsmall(numbers):
small = max(numbers)
for i in range(len(numbers)):
    if numbers[i]>min(numbers):
        if numbers[i]<small:
            small = numbers[i]
return small

I am assuming "numbers" is a list name.