Python Program to Print Largest Even and Largest Odd Number in a List
Last Updated :
04 Apr, 2023
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 class
obj = LargestOddAndEven()
# calling method for largest even number
obj.largestEven(list_num)
# calling method for largest odd number
obj.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 numbers
list_num = [123, 234, 236, 694, 809]
# creating an object of class
obj = LargestOddAndEven()
# calling method for largest even and odd number
obj.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:
Below is the implementation of above approach:
Python3
# Python program for the above approach
def 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 numbers
lis = [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)))
Outputlargest 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))
Outputlarge 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_even
def 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_odd
lis = [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.
Outputlargest 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 heapq
lis = [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.
Outputlargest 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 itertools
def 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.
OutputLargest 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 numbers
lis = [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.
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