Python program to multiply all the items in a dictionary

Python program to illustrate multiplying all the items in a dictionary could be done by creating a dictionary that will store all the key-value pairs, multiplying the value of all the keys, and storing it in a variable.

Example:

Input: dict = {‘value1’:5, ‘value2’:4, ‘value3’:3, ‘value4’:2, ‘value5’:1} 
Output: ans = 120 

Input: dict = {‘v1’:10, ‘v2’:7, ‘v3’:2} 
Output: ans = 140

Approach 1:

• Create a dictionary d and store key-value pairs in it.
• Create a variable answer initialized to 1.
• Run a loop to traverse through the dictionary d
• Multiply each value of key with answer and store the result in answer itself.
• Print answer.

Below are the examples of above approach.

Example 1:

Python3

# create a dictionary
d = {
    'value1': 5,
    'value2': 4,
    'value3': 3,
    'value4': 2,
    'value5': 1,
}

# create a variable to store result
answer = 1

# run a loop
for i in d:
    answer = answer*d[i]

# print answer
print(answer)

Output

120

Time complexity: O(n), where n is the number of elements in the dictionary.
Auxiliary space: O(1), as only a single variable is used to store the result.

Example 2:

Python3

# create a dictionary
d = {
    'a': 10,
    'b': 7,
    'c': 2,
}

# create a variable to store result
answer = 1

# run a loop
for i in d:
    answer = answer*d[i]

# print answer
print(answer)

Output

140

Approach 2: Using values() and list() methods

Python3

# create a dictionary
d = {
    'value1': 5,
    'value2': 4,
    'value3': 3,
    'value4': 2,
    'value5': 1,
}
# create a variable to store result
answer = 1
val=list(d.values())

# run a loop
for i in val:
    answer = answer*i
# print answer
print(answer)

Output

120

Time Complexity: O(n), where n is the number of key-value pairs in the dictionary. 

Auxiliary Space Complexity: O(1), as the amount of additional space used is constant regardless of the size of the input. 

Approach 3: Using functools.reduce(), operator.mul

Python3

# create a dictionary
d = {
    'value1': 5,
    'value2': 4,
    'value3': 3,
    'value4': 2,
    'value5': 1,
}
# create a variable to store result
answer = 1
val=list(d.values())
from functools import reduce
import operator

# print answer
print(reduce(operator.mul,val, 1))

Output

120

Time complexity: O(n), where n is the number of key-value pairs in the dictionary.

Auxiliary space: O(n), where n is the number of key-value pairs in the dictionary.