# Python: n^{th} Hamming number

## Python Itertools: Exercise-26 with Solution

Write a Python program to find the nth Hamming number. Use the itertools module.

Hamming numbers are numbers of the form

H = 2^{i} x 3^{j} x 5^{k}
Where i, j, k ≥ 0

The sequence of Hamming numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27. . . consists of all numbers of the form 2i.3j.5k where i, j and k are non-negative integers.

**Sample Solution:**

**Python Code:**

```
import itertools
from heapq import merge
def nth_hamming_number(n):
def num_recur():
last = 1
yield last
x, y, z = itertools.tee(num_recur(), 3)
for n in merge((2 * i for i in x), (3 * i for i in y), (5 * i for i in z)):
if n != last:
yield n
last = n
result = itertools.islice(num_recur(), n)
return list(result)[-1]
print(nth_hamming_number(8))
print(nth_hamming_number(14))
print(nth_hamming_number(17))
```

Sample Output:

9 20 27

**Python Code Editor:**

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**Previous:** Write a Python program to find the first two elements of a given list whose sum is equal to a given value. Use itertools module to solve the problem.

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## Python: Tips of the Day

**Inverts a dictionary with non-unique hashable values:**

Example:

def tips_collect_dictionary(obj): inv_obj = {} for key, value in obj.items(): inv_obj.setdefault(value, list()).append(key) return inv_obj ages = { "Owen": 25, "Jhon": 25, "Pepe": 15, } print(tips_collect_dictionary(ages))

Output:

{25: ['Owen', 'Jhon'], 15: ['Pepe']}

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