Given a dictionary, count instances where keys are equal to values.
Input : test_dict = {5:5, 8:9, 7:8, 1:2, 10:10, 4:8}
Output : 2
Explanation : At 2 instances, keys are equal to values.
Input : test_dict = {5:4, 8:9, 7:8, 1:2, 10:10, 4:8}
Output : 1
Explanation : At 1 instance, key is equal to value.
Method #1: Using loop
In this, we count instances where keys are equal to values and increment the counter accordingly.
Python3
# Python3 code to demonstrate working of# Keys Values equal frequency# Using loop# initializing dictionarytest_dict = {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))res = 0for key in test_dict: # checking for equality and incrementing counter if key == test_dict[key]: res += 1# printing resultprint("The required frequency : " + str(res)) |
The original dictionary is : {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}
The required frequency : 3
Time complexity: O(n), where n is the number of values in the dictionary.
Auxiliary Space: O(1), constant extra space is required
Method #2 : Using sum() + list comprehension
In this, the task of counting is performed using sum(), when equal key values are found, 1 is appended to the list, and then at the end, each value is summed.
Python3
# Python3 code to demonstrate working of# Keys Values equal frequency# Using sum() + list comprehension# Initializing dictionarytest_dict = {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}# Printing original dictionaryprint("The original dictionary is : " + str(test_dict))# Computing summation to get frequencyres = sum([1 for key in test_dict if key == test_dict[key]])# Printing resultprint("The required frequency : " + str(res)) |
The original dictionary is : {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}
The required frequency : 3
Method #3: Using filter() + lambda function
Approach:
- Initialize the dictionary.
- Use filter() function along with lambda function to filter out the key-value pairs where key and value are equal.
- Find the length of the filtered list to get the required frequency.
- Print the result.
Below is the implementation of the above approach:
Python3
# Python3 code to demonstrate working of# Keys Values equal frequency# using filter() + lambda# Initializing dictionarytest_dict = {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}# Printing original dictionaryprint("The original dictionary is : " + str(test_dict))# Getting key-value pairs where key and value are equal# using filter() and lambdafiltered_list = list(filter(lambda x: x[0] == x[1], test_dict.items()))# Getting the length of filtered list to get the required frequencyres = len(filtered_list)# Printing resultprint("The required frequency : " + str(res)) |
The original dictionary is : {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}
The required frequency : 3
Time complexity: O(n), where n is the number of items in the dictionary.
Auxiliary space: O(k), where k is the number of key-value pairs where key and value are equal.
Method #4: Using dictionary comprehension
Stepwise approach:
- Initialize the dictionary test_dict.
- Print the original dictionary.
- Use dictionary comprehension to iterate over the items of test_dict and filter the keys that have the same value as their key. Here, we create a new dictionary with the same keys and values as test_dict but only keep the items where the key is equal to the value.
- Take the length of the resulting dictionary to get the frequency of keys with the same value as key.
- Print the result.
Python3
# Python3 code to demonstrate working of# Keys Values equal frequency# using dictionary comprehension# Initializing dictionarytest_dict = {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}# Printing original dictionaryprint("The original dictionary is : " + str(test_dict))# Filter keys with same value as key# using dictionary comprehensionres = len({k: v for k, v in test_dict.items() if k == v})# Printing resultprint("The required frequency : " + str(res)) |
The original dictionary is : {5: 5, 8: 9, 7: 7, 1: 2, 10: 10, 4: 8}
The required frequency : 3
Time complexity: O(n), where n is the number of items in the dictionary.
Auxiliary space: O(n), where n is the number of items in the dictionary (to store the filtered dictionary).
