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 liststest_list1 = [4, 3, 6, 2, 9]test_list2 = [3, 6, 3, 4, 3]# printing original listsprint("The original list 1 is : " + str(test_list1))print("The original list 2 is : " + str(test_list2))# initializing K K = 3res = []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 elementres = max(res)# printing result print("Extracted Maximum element : " + str(res)) |
The 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 liststest_list1 = [4, 3, 6, 2, 9]test_list2 = [3, 6, 3, 4, 3]# printing original listsprint("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 problemres = max([sub1 for sub1, sub2 in zip(test_list1, test_list2) if sub2 == K])# printing result print("Extracted Maximum element : " + str(res)) |
The 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 heapqdef 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=3print(max_k_element_3(test_list1, test_list2, K)) |
9
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 liststest_list1 = [4, 3, 6, 2, 9]test_list2 = [3, 6, 3, 4, 3]# printing original listsprint("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 arraysarr1 = np.array(test_list1)arr2 = np.array(test_list2)# find indices where arr2 == Kindices = np.where(arr2 == K)[0]# slice arr1 using indices and store in new arraynew_arr = arr1[indices]# find maximum value in new arraymax_val = np.amax(new_arr)# convert result to integerres = 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.
