Python – Count if dictionary position equals key or value

Given a dictionary, count instances where dictionary item position equals key or value. Valid for Py >= 3.6 [ Introduction of dictionary ordering ].

Input : test_dict = {5:3, 2:3, 10:4, 7:3, 8:1, 9:5} 
Output : 2 
Explanation : At 3 and 5th position, values are 3 and 5.

Input : test_dict = {5:3, 2:3, 10:4, 8:1, 9:5} 
Output : 1 
Explanation : At 5th position, value is 5. 
 

Method #1 : Using loop

In this we iterate for each dictionary item and test for each item to check if any position is equal to key or value of dictionary, if found, we iterate the counter.

The original dictionary is : {5: 3, 1: 3, 10: 4, 7: 3, 8: 1, 9: 5}
The required frequency : 3

Method #2 : Using sum() + list comprehension

In this, we assign 1 to each case in which dictionary index is found equal to any of its items, then perform list summation using sum().

The original dictionary is : {5: 3, 1: 3, 10: 4, 7: 3, 8: 1, 9: 5}
The required frequency : 3

Method #3: Using filter() and lambda function

Approach:

1. Initialize a dictionary test_dict.
2. Initialize a variable res to 0.
3. Use a list comprehension to iterate over the enumerated items of the dictionary test_dict.items().
4. For each enumerated key-value pair, check if the index i is present in the key-value pair.
5. If the index i is present in the key-value pair, add 1 to the res variable.
6. Print the resulting frequency of indices that match a key or a value in test_dict.

The original dictionary is : {5: 3, 1: 3, 10: 4, 7: 3, 8: 1, 9: 5}
The required frequency : 3

Time complexity: O(n), where n is the number of items in the dictionary.
Auxiliary Space: O(1), as we use only a constant amount of memory.