Python - Summation of consecutive elements power

Last Updated : 03 May, 2023

Given a List, the task is to write a Python program to compute summation of power of consecutive elements occurrences frequency.

Examples:

Input : test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Output : 110
Explanation : 2^3 + 3^4 + 4^2 + 5 = 110

Input : test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4]
Output : 105
Explanation : 2^3 + 3^4 + 4^2 = 105

Method #1 : Using loop

In this, we iterate for each element and test for next element, if found to be different, perform summation of power of consecutive elements, else add counter to get frequency to raise number to.

Python3

# Python3 code to demonstrate working of
# Summation of consecutive elements power
# Using loop

# initializing list
test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]

# printing original lists
print("The original list is : " + str(test_list))

freq = 1
res = 0
for idx in range(0, len(test_list) - 1):

    # adding powers
    if test_list[idx] != test_list[idx + 1]:
        res = res + test_list[idx] ** freq
        freq = 1
    else:
        freq += 1

# catering for last element
res = res + test_list[len(test_list) - 1] ** freq

# printing result
print("Computed summation of powers : " + str(res))

Output

The original list is : [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Computed summation of powers : 110

Time Complexity: O(n)
Auxiliary Space: O(1)

Method #2 : Using defaultdict() + loop + sum()

In this we capture frequency using defaultdict using key value pairs and loop is used to iterate through each element. Next, summation is performed using sum().

Python3

# Python3 code to demonstrate working of
# Summation of consecutive elements power
# Using defaultdict() + loop + sum()
from collections import defaultdict

# initializing list
test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]

# printing original lists
print("The original list is : " + str(test_list))

# getting frequency
temp = defaultdict(int)
for ele in test_list:
    temp[ele] += 1

temp = dict(temp)

# computing summation
res = sum([key ** temp[key] for key in temp])

# printing result
print("Computed summation of powers : " + str(res))

Output

The original list is : [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Computed summation of powers : 110

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

Method #3: Using Counter object + list comprehension

Import Counter object from the collections module.
Initialize the list.
Get the frequency of each element using the Counter() method and store it in a dictionary.
Compute the summation using a list comprehension that takes the power of the keys of the dictionary raised to the frequency of each element.
Print the result.

Python3

from collections import Counter

# initializing list
test_list = [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]

# printing original lists
print("The original list is : " + str(test_list))

# getting frequency
temp = Counter(test_list)

# computing summation
res = sum([key ** temp[key] for key in temp])

# printing result
print("Computed summation of powers : " + str(res))

Output

The original list is : [2, 2, 2, 3, 3, 3, 3, 4, 4, 5]
Computed summation of powers : 110

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

Python | Retain K consecutive elements

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Python - Summation of consecutive elements power

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