Python - Maximum of K element in other list
Last Updated :
02 May, 2023
Given two lists, extract maximum of elements with similar K in corresponding list.
Input : test_list1 = [4, 3, 6, 2, 8], test_list2 = [3, 6, 3, 4, 3], K = 3
Output : 8
Explanation : Elements corresponding to 3 are, 4, 6, and 8, Max. is 8.
Input : test_list1 = [10, 3, 6, 2, 8], test_list2 = [5, 6, 5, 4, 5], K = 5
Output : 10
Explanation : Elements corresponding to 5 are, 10, 6, and 8, Max. is 10.
Method #1 : Using loop + max()
In this, we extract all elements from list 1 which are equal to K in list 2, and then perform max() to get maximum of them.
Python3
# Python3 code to demonstrate working of
# Maximum of K element in other list
# Using loop + max()
# initializing lists
test_list1 = [4, 3, 6, 2, 9]
test_list2 = [3, 6, 3, 4, 3]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# initializing K
K = 3
res = []
for idx in range(len(test_list1)):
# checking for K in 2nd list
if test_list2[idx] == K :
res.append(test_list1[idx])
# getting Maximum element
res = max(res)
# printing result
print("Extracted Maximum element : " + str(res))
OutputThe original list 1 is : [4, 3, 6, 2, 9]
The original list 2 is : [3, 6, 3, 4, 3]
Extracted Maximum element : 9
Method #2 : list comprehension + max() + zip()
In this, we perform task of pairing elements using zip() and is one-liner solution provided using list comprehension.
Python3
# Python3 code to demonstrate working of
# Maximum of K element in other list
# Using list comprehension + max() + zip()
# initializing lists
test_list1 = [4, 3, 6, 2, 9]
test_list2 = [3, 6, 3, 4, 3]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# initializing K
K = 3
# one liner to solve this problem
res = max([sub1 for sub1, sub2 in zip(test_list1, test_list2) if sub2 == K])
# printing result
print("Extracted Maximum element : " + str(res))
OutputThe original list 1 is : [4, 3, 6, 2, 9]
The original list 2 is : [3, 6, 3, 4, 3]
Extracted Maximum element : 9
Method #3 : Using heapq
Approach
Using heapq.nlargest() to find the max number
Algorithm
1. Create a list of tuples where each tuple contains elements from both test_list1 and test_list2 at corresponding indices.
2. Create a new list using list comprehension to filter out tuples that have the element K in test_list2.
3. Use heapq.nlargest() function to return the K largest elements from the filtered list.
4. Return the first element from the resulting list.
Python3
import heapq
def max_k_element_3(test_list1, test_list2, K):
zipped_list = list(zip(test_list1, test_list2))
filtered_list = [x[0] for x in zipped_list if x[1] == K]
return heapq.nlargest(K, filtered_list)[0]
test_list1 = [4, 3, 6, 2, 9]
test_list2 = [3, 6, 3, 4, 3]
K=3
print(max_k_element_3(test_list1, test_list2, K))
Time complexity: O(n log K) - due to using heapq.nlargest() function
Space complexity: O(n) - for creating a list of tuples
Method #4: Using numpy
Step-by-step approach:
- Import the numpy library.
- Convert both lists into numpy arrays using the np.array() method.
- Use the np.where() method to find the indices of elements in test_list2 that are equal to K.
- Use the resulting indices to slice the corresponding elements in test_list1 and store them in a new numpy array.
- Use the np.amax() method to find the maximum value in the new array.
- Convert the result back to a regular Python integer using the int() method.
- Print the final result.
Python3
import numpy as np
# initializing lists
test_list1 = [4, 3, 6, 2, 9]
test_list2 = [3, 6, 3, 4, 3]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# initializing K
K = 3
# convert lists to numpy arrays
arr1 = np.array(test_list1)
arr2 = np.array(test_list2)
# find indices where arr2 == K
indices = np.where(arr2 == K)[0]
# slice arr1 using indices and store in new array
new_arr = arr1[indices]
# find maximum value in new array
max_val = np.amax(new_arr)
# convert result to integer
res = int(max_val)
# printing result
print("Extracted Maximum element : " + str(res))
OUTPUT :
The original list 1 is : [4, 3, 6, 2, 9]
The original list 2 is : [3, 6, 3, 4, 3]
Extracted Maximum element : 9
Time complexity: O(n), where n is the length of the input lists.
Auxiliary space: O(n), for the numpy arrays and other variables created in the process.
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