Python program for sum of consecutive numbers with overlapping in lists

Given a List, perform summation of consecutive elements, by overlapping.

Input : test_list = [4, 7, 3, 2] 
Output : [11, 10, 5, 6] 
Explanation : 4 + 7 = 11, 7 + 3 = 10, 3 + 2 = 5, and 2 + 4 = 6. 

Input : test_list = [4, 7, 3] 
Output : [11, 10, 7] 
Explanation : 4+7=11, 7+3=10, 3+4=7. 

Method 1 : Using list comprehension + zip()

In this, we zip list, with consecutive elements’ using zip() and then perform summation using + operator. The iteration part is done using list comprehension.

Python3

# initializing list
test_list = [4, 7, 3, 2, 9, 2, 1]

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

# using zip() to get pairing.
# last element is joined with first for pairing
res = [a + b for a, b in zip(test_list, test_list[1:] + [test_list[0]])]

# printing result
print("The Consecutive overlapping Summation : " + str(res))

Output

The original list is : [4, 7, 3, 2, 9, 2, 1]
The Consecutive overlapping Summation : [11, 10, 5, 11, 11, 3, 5]

Time Complexity: O(n) where n is the number of elements in the list “test_list”. 
Auxiliary Space: O(1) additional space is not needed. 

Method #2 : Using sum() + list comprehension + zip()

In this, we perform the task of getting summation using sum() and rest all functionalities kept similar to the above method.

Python3

# Python3 code to demonstrate working of
# Consecutive overlapping Summation
# Using sum() + list comprehension + zip()

# initializing list
test_list = [4, 7, 3, 2, 9, 2, 1]

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

# using sum() to compute elements sum
# last element is joined with first for pairing
res = [sum(sub) for sub in zip(test_list, test_list[1:] + [test_list[0]])]

# printing result
print("The Consecutive overlapping Summation : " + str(res))

Output

The original list is : [4, 7, 3, 2, 9, 2, 1]
The Consecutive overlapping Summation : [11, 10, 5, 11, 11, 3, 5]

Method 3 :  using a for loop 

Python3

# Python3 code to demonstrate working of
# Consecutive overlapping Summation
# Using for loop

# initializing list
test_list = [4, 7, 3, 2, 9, 2, 1]

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

# using for loop to compute consecutive overlapping summation
n = len(test_list)
res = [test_list[i] + test_list[(i+1) % n] for i in range(n)]

# printing result
print("The Consecutive overlapping Summation : " + str(res))

Output

The original list is : [4, 7, 3, 2, 9, 2, 1]
The Consecutive overlapping Summation : [11, 10, 5, 11, 11, 3, 5]

The time complexity of this approach is O(n), where n is the length of the list, since we iterate over the list once. 
The auxiliary space complexity is also O(n), since we need to store the result list.