Python – First Even Number in List
Sometimes, while working with lists, we can have a problem in which we need to extract certain numbers occurring first time. This can also be even number. This kind of problem is common in day-day and competitive programming. Lets discuss certain ways in which this task can be performed.
Method #1 : Using loop This is brute force way in which this task can be performed. In this, we loop through the list and when 1st time even number occurs, we store and break the loop.
Python3
test_list = [ 43 , 9 , 6 , 72 , 8 , 11 ]
print ("The original list is : " + str (test_list))
res = None
for ele in test_list:
if not ele % 2 :
res = ele
break
print ("The first even element in list is : " + str (res))
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Output :
The original list is : [43, 9, 6, 72, 8, 11]
The first even element in list is : 6
Time Complexity: O(n) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”.
Method #2 : Using binary search (Sorted list) The binary search can also be employed on list when we have sorted list. In this, we keep looking for smaller even term when we find one and move to rear end in case we don’t find one.
Python3
test_list = [ 3 , 7 , 8 , 9 , 10 , 15 ]
print ("The original list is : " + str (test_list))
res = len (test_list)
low = 0
high = len (test_list) - 1
while low < = high:
mid = (low + high) / / 2
if test_list[mid] % 2 :
low = mid + 1
else :
res = test_list[mid]
high = mid - 1
print ("The first even element in list is : " + str (res))
|
Output :
The original list is : [3, 7, 8, 9, 10, 15]
The first even element in list is : 8
Time Complexity: O(logn), where n is the length of the list test_list
Auxiliary Space: O(1) constant additional space is required
Using Recursion
Python3
def findfirsteven(lst,i):
if lst[i] % 2 = = 0 :
return lst[i]
else :
return findfirsteven(lst,i + 1 )
return - 1
test_list = [ 43 , 9 , 6 , 72 , 8 , 11 ]
print ( "The original list is : " + str (test_list))
print ( "The first even element in list is : " + str ( findfirsteven(test_list, 0 )))
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Output
The original list is : [43, 9, 6, 72, 8, 11]
The first even element in list is : 6
Time Complexity:O(n)
Auxiliary Space: O(n)
Using filter function:
Python3
test_list = [ 43 , 9 , 6 , 72 , 8 , 11 ]
print ( "The original list is : " + str (test_list))
res = list ( filter ( lambda x: (x % 2 = = 0 ), test_list))[ 0 ]
print ( "The first even element in list is : " + str (res))
|
Output
The original list is : [43, 9, 6, 72, 8, 11]
The first even element in list is : 6
Time Complexity : O(n)
Auxiliary Space : O(n)
Method 5 : using a list comprehension.
Python3
test_list = [ 43 , 9 , 6 , 72 , 8 , 11 ]
print ( "The original list is : " + str (test_list))
even_nums = [num for num in test_list if num % 2 = = 0 ]
if even_nums:
res = even_nums[ 0 ]
else :
res = None
if res:
print ( "The first even element in list is :" , res)
else :
print ( "There are no even elements in the list." )
|
Output
The original list is : [43, 9, 6, 72, 8, 11]
The first even element in list is : 6
The time complexity of this method is O(n) as it involves iterating over the entire list.
The auxiliary space used by this method is O(n) as a new list even_nums is created to store the filtered elements.
Last Updated :
06 May, 2023
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