Create sol2.py (#1876)

* Create sol2.py

* updating DIRECTORY.md

* Update DIRECTORY.md

* updating DIRECTORY.md

* Update sol2.py

* Update DIRECTORY.md

* updating DIRECTORY.md

* Improve docstrings

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: vinayak <itssvinayak@gmail.com>
Co-authored-by: John Law <johnlaw@linux.com>

DIRECTORY.md

@@ -509,6 +509,7 @@
     * [Soln](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_30/soln.py)
   * Problem 31
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_31/sol1.py)
+    * [Sol2](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_31/sol2.py)
   * Problem 32
     * [Sol32](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_32/sol32.py)
   * Problem 33

project_euler/problem_31/sol2.py

@@ -0,0 +1,56 @@
+"""Coin sums
+
+In England the currency is made up of pound, £, and pence, p, and there are
+eight coins in general circulation:
+
+1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
+It is possible to make £2 in the following way:
+
+1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
+How many different ways can £2 be made using any number of coins?
+
+Hint:
+    > There are 100 pence in a pound (£1 = 100p)
+    > There are coins(in pence) are available: 1, 2, 5, 10, 20, 50, 100 and 200.
+    > how many different ways you can combine these values to create 200 pence.
+
+Example:
+    to make 6p there are 5 ways
+      1,1,1,1,1,1
+      1,1,1,1,2
+      1,1,2,2
+      2,2,2
+      1,5
+    to make 5p there are 4 ways
+      1,1,1,1,1
+      1,1,1,2
+      1,2,2
+      5
+"""
+
+
+def solution(pence: int) -> int:
+    """Returns the number of different ways to make X pence using any number of coins. 
+    The solution is based on dynamic programming paradigm in a bottom-up fashion.
+
+    >>> solution(500)
+    6295434
+    >>> solution(200)
+    73682
+    >>> solution(50)
+    451
+    >>> solution(10)
+    11
+    """
+    coins = [1, 2, 5, 10, 20, 50, 100, 200]
+    number_of_ways = [0] * (pence + 1)
+    number_of_ways[0] = 1  # base case: 1 way to make 0 pence
+
+    for coin in coins:
+        for i in range(coin, pence + 1, 1):
+            number_of_ways[i] += number_of_ways[i - coin]
+    return number_of_ways[pence]
+
+
+if __name__ == "__main__":
+    assert solution(200) == 73682