Python – Key with all Characters in String

Sometimes, while working with Python Strings, we can have problem in which we need to extract all the keys that have all the characters of given some random key. This kind of problem has application has many domains such as day-day programming. Let’s discuss a way in which this problem can be solved. 

Method 1: Using all() + dictionary comprehension The combination of above functionalities can be used to solve this problem. In this, we use all() to perform check in whole of dictionary and extraction of items using items(). 

Python3

# Python3 code to demonstrate working of
# Key with all Characters in String
# Using all() + dictionary comprehension

# initializing dictionary
test_dict = { 'gfg' : ['a', 'b', 'c', 'd', 'g'],
              'is' : ['b', 'f', 'e'],
              'best' : ['c', 'd', 'g'],
              'for' : ['n', 'z'],
              'CS' : ['g', 'd'] }

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

# initializing keys
test_str = 'gd'

# Key with all Characters in String
# Using all() + dictionary comprehension
res = list({key for key, val in test_dict.items()
            if all(chr in val for chr in test_str)})

# printing result
print("The keys list : " + str(res))

Output : 

The original dictionary is : {‘is’: [‘b’, ‘f’, ‘e’], ‘best’: [‘c’, ‘d’, ‘g’], ‘for’: [‘n’, ‘z’], ‘CS’: [‘g’, ‘d’], ‘gfg’: [‘a’, ‘b’, ‘c’, ‘d’, ‘g’]} The keys list : [‘best’, ‘CS’, ‘gfg’]

Time Complexity: O(n), where n is the length of the list test_list 
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list

Method 2:  Using nested for loops

Step-by-step approach:

• Used nested for loops to extract keys that contain characters of given string
• Append those keys to output list
• Display output list

Python3

# Python3 code to demonstrate working of
# Key with all Characters in String

# initializing dictionary
test_dict = { 'gfg' : ['a', 'b', 'c', 'd', 'g'],
            'is' : ['b', 'f', 'e'],
            'best' : ['c', 'd', 'g'],
            'for' : ['n', 'z'],
            'CS' : ['g', 'd'] }

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

# initializing keys
test_str = 'gd'

# Key with all Characters in String
x=list(test_dict.keys())
y=list(test_dict.values())
res=[]
for j in range(0,len(y)):
    c=0
    for i in test_str:
        if i in y[j]:
            c+=1
    if(c==len(test_str)):
        res.append(x[j])
# printing result
print("The keys list : " + str(res))

Output

The original dictionary is : {'gfg': ['a', 'b', 'c', 'd', 'g'], 'is': ['b', 'f', 'e'], 'best': ['c', 'd', 'g'], 'for': ['n', 'z'], 'CS': ['g', 'd']}
The keys list : ['gfg', 'best', 'CS']

Time Complexity: O(M*N) M -length of value list N -length of each list in value list
Auxiliary Space: O(N) N- number of keys with characters of string