Python – Extract Dictionary Items with Summation Greater than K

Given dictionary with value lists, extract all those items with values summing over K.

Input : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “neveropen” : [1, 2, 3, 4]}, K = 10 
Output : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “neveropen” : [1, 2, 3, 4]} 
Explanation : All elements are greater than 10.

Input : {“Gfg” : [6, 3, 4], “is” : [8, 10, 12], “Best” : [10, 16, 14], “for” : [7, 4, 3], “neveropen” : [1, 2, 3, 4]}, K = 50 
Output : {} 
Explanation : No elements are greater than 50.

Method #1 :  Using dictionary comprehension + sum()

This is one of the ways in which this problem can be solved. In this one-liner, we iterate through the keys and append the dictionary item only if its value’s total crosses K computed using sum().

The original dictionary is : {'Gfg': [6, 3, 4], 'is': [8, 10, 12], 'Best': [10, 16, 14], 'for': [7, 4, 3], 'neveropen': [1, 2, 3, 4]}
The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}

Time Complexity: O(n), where n is the length of the list test_dict
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list

Method #2 : Using filter() + lambda() + sum() + dict()

This is another way in which this task can be performed. In this, computation part is done using filter() and lambda, summation using sum() and result converted to dictionary using dict().

The original dictionary is : {'Gfg': [6, 3, 4], 'is': [8, 10, 12], 'Best': [10, 16, 14], 'for': [7, 4, 3], 'neveropen': [1, 2, 3, 4]}
The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}

Method #3: Using for loop and if statement

• Initialize the dictionary test_dict with the given key-value pairs.
• Print the original dictionary.
• Initialize the value of K.
• Create an empty dictionary res to store the computed dictionary.
• Loop through the key-value pairs in test_dict.
• For each key-value pair, compute the sum of the values using sum(value).
• Check if the sum is greater than K.
• If the sum is greater than K, add the key-value pair to res.
• Print the computed dictionary.

The original dictionary is : {'Gfg': [6, 3, 4], 'is': [8, 10, 12], 'Best': [10, 16, 14], 'for': [7, 4, 3], 'neveropen': [1, 2, 3, 4]}
The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}

Time complexity: O(n), where n is the number of key-value pairs in the dictionary
Auxiliary space: O(k), where k is the number of key-value pairs in the result dictionary (in the worst case, all key-value pairs in the original dictionary satisfy the condition and are included in the result dictionary)

Method #4: Using the reduce() function from the functools module

Step-by-step approach:

Import the functools module.
Initialize the dictionary and K value as given in the problem statement.
Use the reduce() function to iterate through the dictionary and check if the sum of the values is greater than K. If yes, add the key-value pair to the resulting dictionary.
Print the resulting dictionary.

The computed dictionary : {'is': [8, 10, 12], 'Best': [10, 16, 14]}

Time complexity: O(n), where n is the number of key-value pairs in the dictionary.

Auxiliary space: O(1) for the variables used in the code, and O(n) for the resulting dictionary