Given a dictionary in python, write a program to find the XOR of all the key-value pairs in the dictionary and return in the form of an array.
Note: All the keys and values in the dictionary are integers.
Examples:
Input : dic={1:3, 4:5, 6:7, 3 :8}
Output : [2, 1, 1, 11]
Explanation: XOR of all the key-value pairs in the dictionary are [1^3, 4^5, 6^7, 3^8]
Thus, [2, 1, 1, 11]
Note: Order may change as the dictionary is unordered.
Method 1:
- Start traversing the dictionary.
- Maintain an array for storing the XOR of each key-value pair.
- For each key in the dictionary, find key^value and append the value to the array.
- Finally, return the array.
Below is the implementation of the above approach
Python3
def xorOfDictionary(dic): # Array to store XOR values arr = [] # Traversing the dictionary for i in dic: # Finding XOR of Key value. cur = i ^ dic[i] arr.append(cur) return arrdic = {5: 8, 10: 9, 11: 12, 1: 14}print(xorOfDictionary(dic)) |
[13, 3, 7, 15]
Time complexity: O(n)
Auxiliary space : O(n) for storing XOR values.
Method 2:Using items() function
Python3
def xorOfDictionary(dic): # Array to store XOR values arr = [] # Traversing the dictionary for key, value in dic.items(): # Finding XOR of Key value. cur = key ^ value arr.append(cur) return arrdic = {5: 8, 10: 9, 11: 12, 1: 14}print(xorOfDictionary(dic)) |
[13, 3, 7, 15]
Time complexity: where n is the number of items in the dictionary, because it has to traverse the dictionary once to calculate the XOR of each key-value pair.
Auxiliary space: O(n), because it needs to store all the XOR values in an array.
Method #3 : Using keys() and values() methods
Python3
dic = {5: 8, 10: 9, 11: 12, 1: 14}arr = []x=list(dic.keys())y=list(dic.values())for i in range(0,len(x)): arr.append(x[i]^y[i])print(arr) |
[13, 3, 7, 15]
Time complexity: O(n)
Auxiliary space: O(n) for storing XOR values.
Method 4: Using a list comprehension
Python3
def xorOfDictionary(dic): # List comprehension to compute XOR values return [i ^ dic[i] for i in dic]dic = {5: 8, 10: 9, 11: 12, 1: 14}print(xorOfDictionary(dic)) |
[13, 3, 7, 15]
Time complexity: O(n)
Auxiliary space: O(n)
Method 5:Using the map() function and a lambda function
- Traverse the dictionary using the .items() method.
- For each key-value pair, apply the lambda function, which takes the XOR of the key and value.
- The result of each XOR operation is stored in a list using the map() function.
- Return the list of XOR values.
Code uses the map() function with a lambda function to perform the XOR operation on each key-value pair in the dictionary and store the result in a list.
Python3
dict = {5: 8, 10: 9, 11: 12, 1: 14}xor_res = list(map(lambda kv: kv[0] ^ kv[1], dict.items()))# Resultprint(xor_res) |
[13, 3, 7, 15]
The items() method is used by the map() function to extract the key-value pairs for each item in the dictionary. The list() method is used to turn iterable that the map() function returns into a list. The list of XOR results is printed last.
Time complexity: O(N) as the map() function and the lambda function are both applied to each key-value pair once, which takes O(1) time and there are a total of N key-value pairs. So time complexity is O(N).
Auxiliary space: O(n) as the list is of the same size as the number of key-value pairs in the dictionary. Therefore, the space complexity is O(N).
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