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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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8.0K viewsedited May 24, 2022 at 04:40

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Create: 2022-05-24
Last Update: 2025-06-29 13:41:03

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
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