Update complexity notation in graph.md and add dynamic programming ex…

…ample in 05-dynamic-programming.md

04-graph.md

@@ -98,6 +98,6 @@ b --> c
 5. Finally, return the maximum distance found.
 
 **Complexity**:
-- Time: O(E) where E is the number of edges in the graph.
-- Space: O(N) where N is the number of nodes in the graph.
+- Time: $O(E)$ where $E$ is the number of edges in the graph.
+- Space: $O(N)$ where $N$ is the number of nodes in the graph.
 

05-dynamic-programming.md

@@ -1,2 +1,36 @@
 ## Dynamic Programming
 
+### Fibonacci sequence problem
+
+**Statement:** Given a number ```n``` $\in\N$, find the its Fibonacci number.
+
+$$
+\begin{align*}
+F_0 &=  0, \\
+F_1 &=  1, \\
+F_n &=  F_{n-1} + F_{n-2}
+\end{align*}
+$$
+
+**Approach**:
+It is possible to solve this problem with brute force recursion.
+
+```python
+def fib(n):
+    if n == 0:
+        return 0
+    if n == 1:
+        return 1
+    return fib(n-1) + fib(n-2)
+```
+
+Nevertheless this leads to repeated work. For example, the call to ```fib(3)``` will call ```fib(2)``` and ```fib(1)```, and the call to ```fib(2)``` will call ```fib(1)``` and ```fib(0)```. This means that the call to ```fib(1)``` is repeated.
+Leading to a complexity of $O(2^n)$.
+
+A better approach is to use **dynamic programming** and store the results of the subproblems.
+
+**Complexity**:
+- Time: $O(n)$
+- Space: $O(n)$
+
+

dynamic_programming/fib.py

@@ -0,0 +1,32 @@
+"""
+fib
+Write a function fib that takes in a number argument, n, and returns 
+the n-th number of the Fibonacci sequence.
+
+The 0-th number of the sequence is 0.
+
+The 1-st number of the sequence is 1.
+
+To generate further numbers of the sequence, calculate the sum of 
+previous two numbers.
+
+Solve this recursively.
+"""
+
+def fib(n):
+    calculated = dict()
+    calculated[0] = 0
+    calculated[1] = 1
+    return _fib(n, calculated)
+
+
+def _fib(n, calculated):
+
+    if n in calculated:
+        return calculated[n]
+
+    calculated[n] = _fib(n-1, calculated) + _fib(n-2, calculated)
+    return calculated[n]
+
+
+print(fib(35)) # -> 9227465

dynamic_programming/sum_possible.py

@@ -0,0 +1,70 @@
+"""
+sum possible
+Write a function sum_possible that takes in an amount and a list of 
+positive numbers. The function should return a boolean indicating whether 
+or not it is possible to create the amount by summing numbers of the list.
+You may reuse numbers of the list as many times as necessary.
+
+You may assume that the target amount is non-negative.
+"""
+
+def sum_possible(amount, numbers):
+    for n in numbers:
+        possible = recur(amount, n, numbers)
+        if possible:
+            return True 
+
+    return False 
+
+def recur(remainder, subtrahend, numbers):
+
+    if subtrahend > remainder:
+        return
+
+    remainder = remainder - subtrahend
+
+    if remainder == 0:
+        return True
+
+
+
+    return False 
+
+
+
+
+
+
+print(sum_possible(4, [1, 2, 3,])) # -> True, 4 + 4
+
+###############################
+# solution with n! complexity #
+###############################
+
+#def sum_possible(amount, numbers):
+#    for j in range(len(numbers)):
+#        possible =_mod(amount, numbers.copy(), j)
+#        if possible:
+#            return True
+#    return False
+#
+#def _mod(amount, numbers, i):
+#    # i need a way to keep track of the numbers i already used for that
+#    # branch of the recursive call
+#    
+#    # i need to try modding simultaneously with all values in the array
+#    if i == len(numbers):
+#        return 
+#
+#    #n = numbers[i]
+#    n = numbers.pop(i)
+#
+#    if n > amount:
+#        return
+#
+#    remainder = amount % n 
+#    if remainder == 0:
+#        return True
+#    
+#    return _mod(remainder, numbers, 0)
+

dynamic_programming/tribonacci.py

@@ -0,0 +1,33 @@
+"""
+tribonacci
+Write a function tribonacci that takes in a number argument, n, and 
+returns the n-th number of the Tribonacci sequence.
+
+The 0-th and 1-st numbers of the sequence are both 0.
+
+The 2-nd number of the sequence is 1.
+
+To generate further numbers of the sequence, calculate the sum of 
+previous three numbers.
+
+Solve this recursively.
+"""
+
+def tribonacci(n):
+    memo = dict()
+    return _trib(n, memo)
+
+
+def _trib(n, memo):
+    if n == 0 or n==1:
+        return 0
+    if n ==2:
+        return 1
+
+    if n in memo:
+        return memo[n]
+
+    memo[n] = _trib(n-1, memo) + _trib(n-2, memo) + _trib(n-3, memo)
+    return memo[n]
+
+print(tribonacci(20)) # -> 35890