Create sol2.py

DIRECTORY.md

@@ -499,6 +499,7 @@
     * [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 +1,38 @@
+"""
+Coin sums
+Problem 31
+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?
+"""
+
+
+def solution(pence):
+    """Returns the number of different ways to make X pence using any number of coins. solution is
+    based on dynamic programming paradigm in 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 i in range(len(coins)):
+        for j in range(coins[i], pence + 1, 1):
+            number_of_ways[j] += number_of_ways[j - coins[i]]
+    return number_of_ways[pence]
+
+
+if __name__ == "__main__":
+    assert solution(200) == 73682