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
# Python3 code to demonstrate # First Even Number in List# using loop# Initializing listtest_list = [43, 9, 6, 72, 8, 11]# printing original listprint("The original list is : " + str(test_list))# First Even Number in List# using loopres = Nonefor ele in test_list: if not ele % 2 : res = ele break # printing result print ("The first even element in list is : " + str(res)) |
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
# Python3 code to demonstrate # First Even Number in List# using binary search# Initializing listtest_list = [3, 7, 8, 9, 10, 15]# printing original listprint("The original list is : " + str(test_list))# First Even Number in List# using binary searchres = len(test_list)low = 0high = len(test_list) - 1while low <= high: mid = (low + high) // 2 if test_list[mid] % 2: low = mid + 1 else: res = test_list[mid] high = mid - 1 # printing result print ("The first even element in list is : " + str(res)) |
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
# Python3 code to demonstrate# First Even Number in List# using recursiondef findfirsteven(lst,i): if lst[i]%2==0: return lst[i] else: return findfirsteven(lst,i+1) return -1# Initializing listtest_list = [43, 9, 6, 72, 8, 11]# printing original listprint("The original list is : " + str(test_list))print ("The first even element in list is : " +str( findfirsteven(test_list,0))) #This code is contributed by Vinay Pinjala |
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
# Python3 code to demonstrate# First Even Number in List# using filter function # Initializing listtest_list = [43, 9, 6, 72, 8, 11] # printing original listprint("The original list is : " + str(test_list)) # First Even Number in List# using filter functionres = list(filter(lambda x: (x % 2 == 0), test_list))[0] # printing resultprint ("The first even element in list is : " + str(res)) |
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
# Initializing listtest_list = [43, 9, 6, 72, 8, 11] # printing original listprint("The original list is : " + str(test_list)) # Finding the first even number in listeven_nums = [num for num in test_list if num % 2 == 0]if even_nums: res = even_nums[0]else: res = None# Printing the resultif res: print("The first even element in list is :", res)else: print("There are no even elements in the list.") |
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.
