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Dynamic Programming
course content

Course Content

Dynamic Programming

Dynamic Programming

1. Intro to Dynamic Programming
What is DPOverlapping Subproblems Property: MemoizationOverlapping Subproblems Property: TabulationExample: Step by Step Solution
2. Problems
Problem A. Binomial CoefficientProblem B. Minimum pathProblem C. Minimum Path in TriangleProblem D. Coin Change
3. Solutions
Problem A. Binomial CoefficientProblem B. Minimum pathProblem C. Minimum Path in TriangleProblem D. Coin Change

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))
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Section 3. Chapter 1
toggle bottom row

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

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Everything was clear?

How can we improve it?

Thanks for your feedback!

Section 3. Chapter 1
toggle bottom row

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

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Everything was clear?

How can we improve it?

Thanks for your feedback!

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

Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
Section 3. Chapter 1
Switch to desktopSwitch to desktop for real-world practiceContinue from where you are using one of the options below
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