Auxiliary Given a list. The task is to print the largest even and largest odd number in a list.
Examples:
Input: 1 3 5 8 6 10 Output: Largest even number is 10 Largest odd number is 5 Input: 123 234 236 694 809 Output: Largest odd number is 809 Largest even number is 694
The first approach uses two methods , one for computing largest even number and the other for computing the largest odd number in a list of numbers, input by the user. Each of the methods prints the largest even and odd number respectively. We maintain one counter for each method, for current largest even or odd and check if the number is divisible by two or not . Accordingly, we print the largest values.
Python3
class LargestOddAndEven: # find largest even number of # the list def largestEven(self, list): # counter for current largest # even number curr = -1 for num in list: # converting number to integer # explicitly num = int(num) # even number is divisible by 2 and # if larger than current largest if(num % 2 == 0 and num > curr): # replace current largest even curr = num print("Largest even number is ", curr) # find largest odd number of the list def largestOdd(self, list): # current largest odd number currO = -1 for num in list: # converting number to integer # explicitly num = int(num) # even number is divisible by 2 and # if larger than current largest if(num % 2 == 1 and num > currO): # replace current largest even currO = num print("Largest odd number is ", currO)list_num = [1, 3, 5, 8, 6, 10]# creating an object of classobj = LargestOddAndEven()# calling method for largest even numberobj.largestEven(list_num)# calling method for largest odd numberobj.largestOdd(list_num) |
Output:
Largest even number is 10 Largest odd number is 5
Time Complexity: O(n)
Auxiliary Space: O(1)
The second approach, uses an optimised version of first approach, where in we compute both the largest values in one method itself. We still maintain two counters, but the for loop that iterates over the list runs only once.
Python3
class LargestOddAndEven: # find largest even number of the list def largestEvenandOdd(self, list): # counter for current largest even # number curr = -1 # counter for current largest odd # number currO = -1 for num in list: # converting number to integer # explicitly num = int(num) # even number is divisible by 2 and # if larger than current largest if(num % 2 == 0 and num > curr): # replace current largest even curr = num elif(num % 2 == 1 and num > currO): # replace current largest even currO = num print("Largest odd number is ", currO) print("Largest even number is ", curr)# input a list of numberslist_num = [123, 234, 236, 694, 809]# creating an object of classobj = LargestOddAndEven()# calling method for largest even and odd numberobj.largestEvenandOdd(list_num) |
Output:
Largest odd number is 809 Largest even number is 694
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 3: Using list Comprehension and max function in python:
- Store even and odd numbers in separate lists using list comprehension.
- print max() of corresponding lists.
Below is the implementation of above approach:
Python3
# Python program for the above approachdef printmax(lis): # Using list comprehension storing # even and odd numbers as separate lists even = [x for x in lis if x % 2 == 0] odd = [x for x in lis if x % 2 == 1] # printing max numbers in corresponding lists print("Largest odd number is ", max(odd)) print("Largest even number is ", max(even))# Input a list of numberslis = [123, 234, 236, 694, 809]printmax(lis)# This code is contributed by vikkycirus |
Output:
Largest odd number is 809 Largest even number is 694
Time Complexity: O(n)
Auxiliary Space: O(1)
Method: Using the Lambda function
Python3
lis = [123, 234, 236, 694, 809]print("largest even number",max(filter(lambda x: x%2==0,lis)))print("largest odd number",max(filter(lambda x: x%2!=0,lis))) |
largest even number 694 largest odd number 809
Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.
Method: Using enumerate function
Python3
lis = [123, 234, 236, 694, 809]even = [x for i,x in enumerate(lis) if x % 2 == 0]odd = [x for i,x in enumerate(lis) if x % 2 == 1] print("large even num",max(even)) print("large odd num",max(odd)) |
large even num 694 large odd num 809
Time Complexity: O(n)
Auxiliary Space: O(n), where n is length of list.
Method: Using recursion
Python3
def largest_even(lis): if len(lis) == 1: if lis[0] % 2 == 0: return lis[0] else: return None else: first = lis[0] rest = lis[1:] rest_largest_even = largest_even(rest) if rest_largest_even is None: if first % 2 == 0: return first else: return None else: if first % 2 == 0: return max(first, rest_largest_even) else: return rest_largest_evendef largest_odd(lis): if len(lis) == 1: if lis[0] % 2 != 0: return lis[0] else: return None else: first = lis[0] rest = lis[1:] rest_largest_odd = largest_odd(rest) if rest_largest_odd is None: if first % 2 != 0: return first else: return None else: if first % 2 != 0: return max(first, rest_largest_odd) else: return rest_largest_oddlis = [123, 234, 236, 694, 809]print("largest even number",largest_even(lis))print("largest odd number",largest_odd(lis))#this code is contributed by Vinay Pinjala. |
largest even number 694 largest odd number 809
Time Complexity: O(n), because n recursive calls are made.
Auxiliary Space: O(n) , because n recursive calls are made and each recursive call pushed into stack.
Method: heapq.nlargest() function:
Python3
import heapqlis = [123, 234, 236, 694, 809]even_numbers = heapq.nlargest(1, (x for x in lis if x % 2 == 0))odd_numbers = heapq.nlargest(1, (x for x in lis if x % 2 != 0))print("largest even number", even_numbers[0])print("largest odd number", odd_numbers[0])#this code is contributed by tvsk. |
largest even number 694 largest odd number 809
Time Complexity: O(n log k), where n is the length of the input list and k is the number of elements returned by the function.
Space Complexity: O(k), where k is the number of elements returned by the function.
Method: using itertools:
Python3
import itertoolsdef printmax(lis): even = [x for x in lis if x % 2 == 0] odd = [x for x in lis if x % 2 == 1] if even: max_even = max(even) else: max_even = None if odd: max_odd = max(odd) else: max_odd = None print("Largest odd number is ", max_odd) print("Largest even number is ", max_even)lis = [123, 234, 236, 694, 809]printmax(lis)#This code is contributed by Jyothi pinjala. |
Largest odd number is 809 Largest even number is 694
Time Complexity: O(n)
Auxiliary Space: O(n)
Method: Using numpy:
- Initialize a list lis with the given values [123, 234, 236, 694, 809].
- Use numpy to create a boolean index array that is True for each element in lis that is even, and False for each
- element that is odd. This is done with the following line of code: np.array(lis)[np.array(lis)%2==0]. The resulting array contains only the even numbers from the original list lis.
- Find the maximum value in the even array using the max() function. This is done with the following line of code: max(even).
- Find the maximum value in the odd array using the max() function. This is done with the following line of code: max(odd).
- Print the maximum even number and the maximum odd number using the print() function. These values were found in steps 4 and 5.
Python3
import numpy as np# Input a list of numberslis = [123, 234, 236, 694, 809]even = np.array(lis)[np.array(lis)%2==0]odd = np.array(lis)[np.array(lis)%2==1]# printing max numbers print("large even num",max(even))print("large odd num",max(odd))#This code is contributed by Rayudu |
Output:
large even num 694
large odd num 809
The time complexity : O(n), where n is the length of the input list. This is because the numpy boolean indexing operation and the max() function each take O(n) time to complete.
The auxiliary space : O(n), where n is the length of the input list. This is because two new arrays are created, one for even numbers and one for odd numbers, each with a maximum size of n.
