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Python Program to Print Largest Even and Largest Odd Number in a List

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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)))


Output

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))


Output

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_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.


Output

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 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.


Output

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 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.


Output

Largest odd number is  809
Largest even number is  694

Time Complexity: O(n)
Auxiliary Space: O(n) 

Method: Using numpy:

  1. Initialize a list lis with the given values [123, 234, 236, 694, 809].
  2. Use numpy to create a boolean index array that is True for each element in lis that is even, and False for each
  3. 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.
  4. Find the maximum value in the even array using the max() function. This is done with the following line of code: max(even).
  5. Find the maximum value in the odd array using the max() function. This is done with the following line of code: max(odd).
  6. 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.



Last Updated : 04 Apr, 2023
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