Sometimes, we need to find the specific problem of getting the pair which yields the minimum sum, this can be computed by getting initial two elements after sorting. But in some case, we don’t with to change the ordering of list and perform some operation in the similar list without using extra space. Let’s discuss certain ways in which this can be performed.
Method #1 : Using list comprehension + min() + combination() + lambda This particular task can be performed using the combination of above functions in which we use list comprehension to bind all the functionalities and min function to get the minimum sum, combination function finds all sums internally and lambda function is used to compute the sum.
Python3
# Python3 code to demonstrate# Minimum Pair Sum # using list comprehension + min() + combinations() + lambdafrom itertools import combinations# initializing listtest_list = [3, 4, 1, 7, 9, 1]# printing original listprint("The original list : " + str(test_list))# using list comprehension + min() + combinations() + lambda# Minimum Pair Sum res = min(combinations(test_list, 2), key = lambda sub: abs(sub[0]-sub[1]))# print resultprint("The minimum sum pair is : " + str(res)) |
The original list : [3, 4, 1, 7, 9, 1] The minimum sum pair is : (1, 1)
Time Complexity: O(n*n), where n is the length of the input list. This is because we’re using list comprehension + min() + combination() + lambda which has a time complexity of O(n*n) in the worst case.
Auxiliary Space: O(n), as we’re using additional space res other than the input list itself with the same size of input list.
Method #2 : Using list comprehension + nsmallest() + combination() + lambda This method has potential of not only finding a single minimum but also k minimum sum pairs if required and uses nsmallest function instead of min function to achieve this functionality.
Python3
# Python3 code to demonstrate# Minimum Pair Sum # using list comprehension + nsmallest() + combinations() + lambdafrom itertools import combinationsfrom heapq import nsmallest# initializing listtest_list = [3, 4, 1, 7, 9, 8]# printing original listprint("The original list : " + str(test_list))# using list comprehension + nsmallest() + combinations() + lambda# Minimum Pair Sum # computes 2 min sum pairres = nsmallest(2, combinations(test_list, 2), key = lambda sub: abs(sub[0] + sub[1]))# print resultprint("The minimum sum pair is : " + str(res)) |
The original list : [3, 4, 1, 7, 9, 8] The minimum sum pair is : [(3, 1), (4, 1)]
Time Complexity: O(n*n) where n is the number of elements in the string list. The list comprehension + nsmallest() + combination() + lambda is used to perform the task and it takes O(n*n) time.
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res test_list.
Method #3 : Using for loop
We start by initializing the minimum pair sum to the maximum possible value, using float(‘inf’). We loop through all pairs of distinct elements in the list, using two nested loops. Note that we start the second loop at i+1, since we don’t want to count pairs with the same element twice. For each pair, we calculate the sum of the two elements. If the current pair sum is less than the current minimum pair sum, we update the minimum pair sum and the minimum pair. After we’ve looped through all pairs, we print the minimum sum pair.
Python3
lst = [3, 4, 1, 7, 9, 1]# Initialize the minimum pair sum to the maximum possible valuemin_pair_sum = float('inf')# Loop through all pairs of distinct elements in the listfor i in range(len(lst)): for j in range(i+1, len(lst)): # Calculate the sum of the current pair pair_sum = lst[i] + lst[j] # If the current pair sum is less than the current minimum pair sum, # update the minimum pair sum and the minimum pair if pair_sum < min_pair_sum: min_pair_sum = pair_sum min_pair = (lst[i], lst[j])print("The minimum sum pair is :", min_pair) |
The minimum sum pair is : (1, 1)
Time complexity: O(n^2)
Auxiliary Space: O(1)
