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_009/sol1.py

@@ -0,0 +1,69 @@
+"""
+Problem 9: https://projecteuler.net/problem=9
+
+A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+    a^2 + b^2 = c^2
+For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+Find the product abc.
+"""
+
+
+def solution() -> int:
+    """
+     Returns the product of a,b,c which are Pythagorean Triplet that satisfies
+     the following:
+     1. a < b < c
+     2. a**2 + b**2 = c**2
+     3. a + b + c = 1000
+    # The code below has been commented due to slow execution affecting Travis.
+    # >>> solution()
+    # 31875000
+    """
+    for a in range(300):
+        for b in range(400):
+            for c in range(500):
+                if a < b < c:
+                    if (a ** 2) + (b ** 2) == (c ** 2):
+                        if (a + b + c) == 1000:
+                            return a * b * c
+
+
+def solution_fast() -> int:
+    """
+     Returns the product of a,b,c which are Pythagorean Triplet that satisfies
+     the following:
+     1. a < b < c
+     2. a**2 + b**2 = c**2
+     3. a + b + c = 1000
+
+    # The code below has been commented due to slow execution affecting Travis.
+    # >>> solution_fast()
+    # 31875000
+    """
+    for a in range(300):
+        for b in range(400):
+            c = 1000 - a - b
+            if a < b < c and (a ** 2) + (b ** 2) == (c ** 2):
+                return a * b * c
+
+
+def benchmark() -> None:
+    """
+    Benchmark code comparing two different version function.
+    """
+    import timeit
+
+    print(
+        timeit.timeit("solution()", setup="from __main__ import solution", number=1000)
+    )
+    print(
+        timeit.timeit(
+            "solution_fast()", setup="from __main__ import solution_fast", number=1000
+        )
+    )
+
+
+if __name__ == "__main__":
+    benchmark()

project_euler/problem_009/sol2.py

@@ -0,0 +1,38 @@
+"""
+Problem 9: https://projecteuler.net/problem=9
+
+A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+    a^2 + b^2 = c^2
+For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+Find the product abc.
+"""
+
+
+def solution(n: int = 1000) -> int:
+    """
+     Return the product of a,b,c which are Pythagorean Triplet that satisfies
+     the following:
+     1. a < b < c
+     2. a**2 + b**2 = c**2
+     3. a + b + c = n
+
+    >>> solution(1000)
+    31875000
+    """
+    product = -1
+    candidate = 0
+    for a in range(1, n // 3):
+        """Solving the two equations a**2+b**2=c**2 and a+b+c=N eliminating c"""
+        b = (n * n - 2 * a * n) // (2 * n - 2 * a)
+        c = n - a - b
+        if c * c == (a * a + b * b):
+            candidate = a * b * c
+            if candidate >= product:
+                product = candidate
+    return product
+
+
+if __name__ == "__main__":
+    print(solution(int(input().strip())))

project_euler/problem_009/sol3.py

@@ -0,0 +1,36 @@
+"""
+Problem 9: https://projecteuler.net/problem=9
+
+A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
+
+    a^2 + b^2 = c^2
+
+For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
+
+There exists exactly one Pythagorean triplet for which a + b + c = 1000.
+Find the product abc.
+"""
+
+
+def solution() -> int:
+    """
+     Returns the product of a,b,c which are Pythagorean Triplet that satisfies
+     the following:
+
+     1. a**2 + b**2 = c**2
+     2. a + b + c = 1000
+
+    # The code below has been commented due to slow execution affecting Travis.
+    # >>> solution()
+    # 31875000
+    """
+    return [
+        a * b * (1000 - a - b)
+        for a in range(1, 999)
+        for b in range(a, 999)
+        if (a * a + b * b == (1000 - a - b) ** 2)
+    ][0]
+
+
+if __name__ == "__main__":
+    print(solution())