# Python: Find the second smallest number in a list

## Python List: Exercise - 27 with Solution

Write a Python program to find the second smallest number in a list.

**Example - 1 :**

**Example - 2 :**

**Example - 3 :**

**Example - 4 :**

**Sample Solution:**

**Python Code:**

```
# Define a function named 'second_smallest' that takes a list of numbers 'numbers' as a parameter
def second_smallest(numbers):
# Check if the length of the 'numbers' list is less than 2; return None in this case
if len(numbers) < 2:
return
# Check if there are only two elements in the 'numbers' list, and they are the same; return None in this case
if len(numbers) == 2 and numbers[0] == numbers[1]:
return
# Create an empty set 'dup_items' to store duplicate items and an empty list 'uniq_items' to store unique items
dup_items = set()
uniq_items = []
# Iterate through the elements in the 'numbers' list
for x in numbers:
# Check if 'x' is not in 'dup_items'; if not, add it to 'uniq_items' and 'dup_items'
if x not in dup_items:
uniq_items.append(x)
dup_items.add(x)
# Sort the 'uniq_items' list in ascending order
uniq_items.sort()
# Return the second smallest item from the sorted 'uniq_items' list, which is at index 1
return uniq_items[1]
# Call the 'second_smallest' function with different lists and print the results
print(second_smallest([1, 2, -8, -2, 0, -2]))
print(second_smallest([1, 1, 0, 0, 2, -2, -2]))
print(second_smallest([1, 1, 1, 0, 0, 0, 2, -2, -2]))
print(second_smallest([2, 2])) # Edge case with two identical elements, returns None
print(second_smallest([2])) # Edge case with a single element, returns None
```

Sample Output:

-2 0 0 None None

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