Really fast solution, not bruteforce

24/main.py

@@ -0,0 +1,44 @@
+# Get the millionth lexicographic permutation of the digits
+# 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
+
+import math
+
+def fac(n):
+	if n == 0:
+		return 1
+	else:
+		return n * fac(n-1)
+
+def getPermutation(n, s):
+	p = sorted(s)
+	l = len(s)
+	x = n - 1
+	seq = []
+
+	assert n > 0
+	assert l > 0
+	assert n <= fac(l)
+
+	f = fac(l)
+	while l > 0:
+		part = math.floor(f / l)
+		f /= l
+		y = math.floor(x / part)
+		x -= y * part
+		seq.append(y)
+		l -= 1
+
+	res = []
+	for se in seq:
+		res.append(p.pop(se))
+
+	return res
+
+def main():
+	print("Hello, this is Patrick")
+
+	r = getPermutation(1000000, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
+	print(r)
+
+if __name__ == "__main__":
+	main()