Python – Check alternate peak elements in List
Given a list, the task is to write a Python program to test if it’s alternate, i.e next and previous elements are either both smaller or larger across the whole list.
Input : test_list = [2, 4, 1, 6, 4, 8, 0]
Output : True
Explanation : 4, 6, 8 are alternate and peaks (2 1).Input : test_list = [2, 4, 1, 6, 4, 1, 0]
Output : False
Explanation : 1 is not peak ( 4 < 1 < 0).
Method #1: Using loop.
In this, we check for each element for next and previous elements using conditional if statement to be either smaller or larger, if any variation is found, the result is flagged false and loop is exited.
Python3
# Python3 code to demonstrate working of# Test for alternate peak List# Using loop# initializing listtest_list = [2, 4, 1, 6, 4, 8, 0] # printing original listprint("The original list is : " + str(test_list))res = Truefor idx in range(1, len(test_list) - 1): # breaking if not alternate peaks if not ((test_list[idx - 1] < test_list[idx] and test_list[idx + 1] < test_list[idx]) or (test_list[idx - 1] > test_list[idx] and test_list[idx + 1] > test_list[idx])): res = False # printing resultprint("Is list forming alternate peaks ? : " + str(res)) |
Output:
The original list is : [2, 4, 1, 6, 4, 8, 0] Is list forming alternate peaks ? : True
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2 : Using all() + generator expression.
In this, we perform the task of checking for all the elements for getting alternate peaks using all(), and generator expression is used to iterate through the entire list.
Python3
# Python3 code to demonstrate working of# Test for alternate peak List# Using all() + generator expression# initializing listtest_list = [2, 4, 1, 6, 4, 8, 0] # printing original listprint("The original list is : " + str(test_list))# checking for all the elements for alternate peaks # one liner solution to problemres = all(((test_list[idx - 1] < test_list[idx] and test_list[idx + 1] < test_list[idx]) or (test_list[idx - 1] > test_list[idx] and test_list[idx + 1] > test_list[idx])) for idx in range(1, len(test_list) - 1)) # printing resultprint("Is list forming alternate peaks ? : " + str(res)) |
Output:
The original list is : [2, 4, 1, 6, 4, 8, 0] Is list forming alternate peaks ? : True
Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1), no extra space is required
Method #3 : Using NumPy
Python3
import numpy as np# initializing listtest_list = [2, 4, 1, 6, 4, 8, 0] # printing original listprint("The original list is : " + str(test_list))# converting list to numpy arrayarr = np.array(test_list)# checking if list has alternate peaksres = np.all(np.diff(np.sign(np.diff(arr))) != 0)# printing resultprint("Is list forming alternate peaks ? : " + str(res))#this code was contributed by Asif_Shaik |
output:
The original list is : [2, 4, 1, 6, 4, 8, 0] Is list forming alternate peaks ? : True
Time Complexity: O(n)
Auxiliary Space: O(n)
