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Dynamic Programming

bookProblem A. Binomial Coefficient

Let's use the Memoization principle here. Let dp[i][j] be a Binomial Coefficient C(i,j). First, dp initialized with None.

Given dp[n][k]:

  • if it is None, calculate it as c(n-1, k-1) + c(n-1, k)
  • if it is a base case: k==0 or k==n, then dp[n][k] = 1
  • else return dp[n][k]

Note that structure dp depends on n, and you must use it for defined n.


              
123456789101112131415

            
n = 200
dp = [[None for _ in range(n+1)] for _ in range(n+1)]
def c(n, k):
  if k==0 or k==n:
    dp[n][k] = 1
        
  if dp[n][k] == None:
    dp[n][k] = c(n-1, k)+c(n-1, k-1)
        
  return dp[n][k]

print(c(3, 2))
print(c(10, 4))
print(c(11, 5))
print(c(144, 7))
copy

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SectionΒ 3. ChapterΒ 1
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bookProblem A. Binomial Coefficient

Swipe to show menu

Let's use the Memoization principle here. Let dp[i][j] be a Binomial Coefficient C(i,j). First, dp initialized with None.

Given dp[n][k]:

  • if it is None, calculate it as c(n-1, k-1) + c(n-1, k)
  • if it is a base case: k==0 or k==n, then dp[n][k] = 1
  • else return dp[n][k]

Note that structure dp depends on n, and you must use it for defined n.


              
123456789101112131415

            
n = 200
dp = [[None for _ in range(n+1)] for _ in range(n+1)]
def c(n, k):
  if k==0 or k==n:
    dp[n][k] = 1
        
  if dp[n][k] == None:
    dp[n][k] = c(n-1, k)+c(n-1, k-1)
        
  return dp[n][k]

print(c(3, 2))
print(c(10, 4))
print(c(11, 5))
print(c(144, 7))
copy

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Everything was clear?

How can we improve it?

Thanks for your feedback!

SectionΒ 3. ChapterΒ 1
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