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Python: Find the number of combinations

Python Basic - 1: Exercise-62 with Solution

Write a Python program to find the number of combinations that satisfy p + q + r + s = n where n is a given number <= 4000 and p, q, r, s in the range of 0 to 1000.

Sample Solution:

Python Code:

from collections import Counter
print("Input a positive integer: (ctrl+d to exit)") 
pair_dict = Counter()
for i in range(2001):
  pair_dict[i] = min(i, 2000 - i) + 1 
 
while True:
  try:
    n = int(input())
    ans = 0
    for i in range(n + 1):
      ans += pair_dict[i] * pair_dict[n - i]
    print("Number of combinations of a,b,c,d:",ans) 
  except EOFError:
    break

Sample Output:

Input a positive integer: (ctrl+d to exit)
 252
Number of combinations of a,b,c,d: 2731135

Flowchart:

Flowchart: Python - Find the number of combinations

Python Code Editor:

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Previous: Write a Python program that compute the maximum value of the sum of the passing integers.
Next: Write a Python program which adds up columns and rows of given table as shown in the specified figure.

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Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

flat_list = [item for sublist in l for item in sublist]

which means:

flat_list = []
for sublist in l:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR