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Fibonacci Number using Dynamic Programming:

CODE:

calculated = {}

def fib(n):
if n == 0: # base case 1
return 0
if n == 1: # base case 2
return 1
elif n in calculated:
return calculated[n]
else: # recursive step
calculated[n] = fib(n-1) + fib(n-2)
return calculated[n]

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Time complexity in above Picture

Fibonacci Number using Dynamic Programming:

CODE:

calculated = {}

def fib(n):
if n == 0: # base case 1
return 0
if n == 1: # base case 2
return 1
elif n in calculated:
return calculated[n]
else: # recursive step
calculated[n] = fib(n-1) + fib(n-2)
return calculated[n]

Share and Support
@Python_Codes

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