Python | Least Value test in Dictionary

Last Updated : 21 Apr, 2023

While working with dictionary, we might come to a problem in which we require to ensure that all the values are atleast K in dictionary. This kind of problem can occur while checking status of start or checking for a bug/action that could have occurred. Let’s discuss certain ways in which this task can be performed.

Method #1 : Using all() + dictionary comprehension The combination of above functions can be used to perform the following task. The all function checks for each key and dictionary comprehension checks for the atleast K value.

Python3

# Python3 code to demonstrate working of # Least Value test in Dictionary # Using all() + dictionary comprehension  # Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary  print("The original dictionary is : " + str(test_dict))  # Initialize K  K = 8  # using all() + dictionary comprehension # Least Value test in Dictionary res = all(x >= K for x in test_dict.values())  # printing result  print("Does all keys have atleast K value ? : " + str(res))

Output :

The original dictionary is : {'gfg': 8, 'best': 11, 'is': 10} Does all keys have atleast K value ? : True

Time complexity: O(n), where n is the number of values in the dictionary.
Auxiliary space: O(1).

Method #2 : Using loop This problem can be solved using brute force strategy using loop and we compare each key with K and return True if all elements are atleast K.

Python3

# Python3 code to demonstrate working of # Least Value test in Dictionary # Using loop  # Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary  print("The original dictionary is : " + str(test_dict))  # Initialize K  K = 8  # using loop # Least Value test in Dictionary res = True for key in test_dict:     if test_dict[key] < K:         res = False  # printing result  print("Does all keys have atleast K value ? : " + str(res))

Output :

The original dictionary is : {'gfg': 8, 'best': 11, 'is': 10} Does all keys have atleast K value ? : True

Time complexity: O(n), where n is the number of elements in the dictionary.
Auxiliary space: O(1), as the space used by the variables does not depend on the size of the dictionary.

Method #3 : Using min()
This problem can also be solved by using the min() function which takes dictionary values as argument and returns the least value.

Python3

# Python3 code to demonstrate working of # Least Value test in Dictionary # Using min()    # Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}    # Printing original dictionary  print("The original dictionary is : " + str(test_dict))    # Initialize K  K = 8    # using min() # Least Value test in Dictionary res = min(test_dict.values()) >= K    # printing result  print("Does all keys have atleast K value ? : " + str(res))  #This code is contributed by Edula Vinay Kumar Reddy

Output

The original dictionary is : {'gfg': 8, 'is': 10, 'best': 11} Does all keys have atleast K value ? : True

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

Method 4: Using filter() and len()

Use the filter() function to create a new dictionary containing only the key-value pairs where the value is less than K. This filtered dictionary is then checked to see if its length is zero, which would indicate that all the values in the original dictionary are greater than or equal to K.

Python3

# Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary  print("The original dictionary is : " + str(test_dict))  # Initialize K  K = 8  # using filter() and len() # Least Value test in Dictionary filtered_dict = dict(filter(lambda elem: elem[1] < K, test_dict.items())) res = len(filtered_dict) == 0  # printing result  print("Does all keys have atleast K value ? : " + str(res))

Output

The original dictionary is : {'gfg': 8, 'is': 10, 'best': 11} Does all keys have atleast K value ? : True

Time Complexity: O(n), where n is the number of key-value pairs in the dictionary.
Auxiliary Space: O(m), where m is the number of key-value pairs in the filtered dictionary.

Method #5: Using set() and comparison

Convert the values of the dictionary into a set.
Check if K is less than or equal to the minimum value in the set.
If it is, then all keys have at least K value

Python3

# Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary  print("The original dictionary is : " + str(test_dict))  # Initialize K  K = 8  # using set() and comparison # Least Value test in Dictionary value_set = set(test_dict.values()) res = K <= min(value_set)  # printing result  print("Does all keys have atleast K value ? : " + str(res))

Output

The original dictionary is : {'gfg': 8, 'is': 10, 'best': 11} Does all keys have atleast K value ? : True

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

Method #6: Using Numpy"

Step-by-step approach:

First, the numpy library is imported with the alias "np". This library is used later in the program to perform an array operation.
The program initializes a dictionary called "test_dict" with three key-value pairs. The keys are strings and the values are integers.
The original dictionary is printed using the "print()" function.
The program initializes a variable called "K" with the value 8.
Next, the program creates a numpy array called "values_array" containing the values from the dictionary. It does this by using the "values()" function to get a list of the dictionary's values, then passing that list to the "np.array()" function to create a numpy array.
The program then uses numpy's "all()" function to test if all the values in the "values_array" are greater than or equal to the value of "K". If all values are greater than or equal to "K", then the function returns True. Otherwise, it returns False.
Finally, the program prints the result of the test as a string using the "print()" function.
So, the program's overall goal is to check if all the values in the dictionary

Below is the implementation of the above approach:

Python3

import numpy as np  # Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary print("The original dictionary is: " + str(test_dict))  # Initialize K K = 8  # using numpy library # Least Value test in Dictionary values_array = np.array(list(test_dict.values())) res = np.all(values_array >= K)  # printing result print("Does all keys have atleast K value? : " + str(res))

Output

The original dictionary is : {'gfg': 8, 'is': 10, 'best': 11} Does all keys have atleast K value ? : True

Time complexity: O(n) where n is the number of values in the dictionary.
Auxiliary space: O(n) for the numpy array.

Method #7: Using reduce():

Algorithm:

Initialize the dictionary and K value.
Apply filter and len functions on the dictionary items to filter out the dictionary keys with values less than K.
Check if the length of the resulting dictionary is equal to zero or not.
Print the result.

Python3

from functools import reduce  # Initialize dictionary test_dict = {'gfg' : 8, 'is' : 10, 'best' : 11}  # Printing original dictionary print("The original dictionary is : " + str(test_dict))  # Initialize K K = 8  # using reduce() and lambda function # Least Value test in Dictionary res = reduce(lambda x, y: x and y, map(lambda val: val >= K, test_dict.values()))  # printing result print("Does all keys have atleast K value ? : " + str(res)) #This code is contributed by Vinay pinjala.

Output

The original dictionary is : {'gfg': 8, 'is': 10, 'best': 11} Does all keys have atleast K value ? : True

Time Complexity: O(n), where n is the number of key-value pairs in the dictionary. This is because filter and len functions take O(n) time to execute.

Space Complexity: O(1). The space used in this code is constant as no extra data structure is used.