Rename Project Euler directories and other dependent changes (#3300)

* Rename all Project Euler directories:

Reason:
The change was done to maintain consistency throughout the directory
and to keep all directories in sorted order.

Due to the above change, some config files had to be modified:
'problem_22` -> `problem_022`

* Update scripts to pad zeroes in PE directories

project_euler/problem_031/sol1.py

@@ -0,0 +1,65 @@
+"""
+Coin sums
+Problem 31: https://projecteuler.net/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 one_pence() -> int:
+    return 1
+
+
+def two_pence(x: int) -> int:
+    return 0 if x < 0 else two_pence(x - 2) + one_pence()
+
+
+def five_pence(x: int) -> int:
+    return 0 if x < 0 else five_pence(x - 5) + two_pence(x)
+
+
+def ten_pence(x: int) -> int:
+    return 0 if x < 0 else ten_pence(x - 10) + five_pence(x)
+
+
+def twenty_pence(x: int) -> int:
+    return 0 if x < 0 else twenty_pence(x - 20) + ten_pence(x)
+
+
+def fifty_pence(x: int) -> int:
+    return 0 if x < 0 else fifty_pence(x - 50) + twenty_pence(x)
+
+
+def one_pound(x: int) -> int:
+    return 0 if x < 0 else one_pound(x - 100) + fifty_pence(x)
+
+
+def two_pound(x: int) -> int:
+    return 0 if x < 0 else two_pound(x - 200) + one_pound(x)
+
+
+def solution(n: int = 200) -> int:
+    """Returns the number of different ways can n pence be made using any number of
+    coins?
+
+    >>> solution(500)
+    6295434
+    >>> solution(200)
+    73682
+    >>> solution(50)
+    451
+    >>> solution(10)
+    11
+    """
+    return two_pound(n)
+
+
+if __name__ == "__main__":
+    print(solution(int(input().strip())))

project_euler/problem_031/sol2.py

@@ -0,0 +1,59 @@
+"""
+Problem 31: https://projecteuler.net/problem=31
+
+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 = 200) -> 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