Python – Rearrange dictionary for consecutive value-keys

Sometimes, while working with Python dictionaries, we can have a problem in which we need to rearrange dictionary keys so as a value is followed by same key in dictionary. This problem can have application in competitive programming algorithms and Graph problems. Let’s discuss certain ways in which this task can be performed.

Input : test_dict = {1 : 2, 3 : 2, 2 : 3} 
Output : {1: 2, 2: 3, 3: 2}
Input : test_dict = {1 : 2} 
Output : {1 : 2} 

Method #1 : Using loop + keys()
The combination of above functions can be used to solve this problem. In this, we extract the dictionary keys using keys() and iterate till we find value succeeded by equal key.

Python3

# Python3 code to demonstrate working of
# Rearrange dictionary for consecutive value-keys
# Using loop + keys()

# initializing dictionary
test_dict = {1 : 3, 4 : 5, 3 : 4, 5 : 6}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Rearrange dictionary for consecutive value-keys
# Using loop + keys()
temp = list(test_dict.keys())[0]
res = {}
while len(test_dict) > len(res):
    res[temp] = temp = test_dict[temp]

# printing result
print("The rearranged dictionary : " + str(res))

Output : 

The original dictionary : {1: 3, 4: 5, 3: 4, 5: 6}
The rearranged dictionary : {1: 3, 3: 4, 4: 5, 5: 6}

Method #2 : Using dictionary comprehension + accumulate()
The combination of above functions can be used to solve this problem. In this, we perform the task of pairing using accumulate and rearrange new dictionary using dictionary comprehension.

Python3

# Python3 code to demonstrate working of
# Rearrange dictionary for consecutive value-keys
# Using dictionary comprehension + accumulate()
from itertools import accumulate

# initializing dictionary
test_dict = {1 : 3, 4 : 5, 3 : 4, 5 : 6}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Rearrange dictionary for consecutive value-keys
# Using dictionary comprehension + accumulate()
res = {key : test_dict[key] for key in accumulate(test_dict,
                              lambda key, x :test_dict[key])}

# printing result
print("The rearranged dictionary : " + str(res))

Output : 

The original dictionary : {1: 3, 4: 5, 3: 4, 5: 6}
The rearranged dictionary : {1: 3, 3: 4, 4: 5, 5: 6}

Method #3: Using sorted() and list comprehension

Step-by-step algorithm:

1. Initialize the dictionary and the keys.
2. Sort the keys of the dictionary.
3. Retrieve the values for each key in the sorted keys list and store them in a separate list.
4. Create a new dictionary by zipping the sorted keys and values list.
5. Print the rearranged dictionary.

Python3

test_dict = {1 : 3, 4 : 5, 3 : 4, 5 : 6}

# printing original dictionary
print("The original dictionary : " + str(test_dict))
#sorting the keys of dictionary in ascending order
keys = sorted(test_dict.keys())

#creating list of values by retrieving value corresponding to each key in keys list
values = [test_dict[k] for k in keys]

#creating new dictionary using zip function with keys and values list
res = dict(zip(keys, values))

#printing result
print("The rearranged dictionary : " + str(res))

Output

The original dictionary : {1: 3, 4: 5, 3: 4, 5: 6}
The rearranged dictionary : {1: 3, 3: 4, 4: 5, 5: 6}

Time complexity: O(n log n) where n represents the number of key-value pairs in the input dictionary test_dict. 
Space complexity: O(n)  where n represents the number of key-value pairs in the input dictionary test_dict.