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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-238/#TASK2
+#
+# Task 2: Persistence Sort
+# ========================
+#
+# You are given an array of positive integers.
+#
+# Write a script to sort the given array in increasing order with respect to
+# the count of steps required to obtain a single-digit number by multiplying
+# its digits recursively for each array element. If any two numbers have the
+# same count of steps, then print the smaller number first.
+#
+## Example 1
+##
+## Input: @int = (15, 99, 1, 34)
+## Output: (1, 15, 34, 99)
+##
+## 15 => 1 x 5 => 5 (1 step)
+## 99 => 9 x 9 => 81 => 8 x 1 => 8 (2 steps)
+## 1  => 0 step
+## 34 => 3 x 4 => 12 => 1 x 2 => 2 (2 steps)
+#
+## Example 2
+##
+## Input: @int = (50, 25, 33, 22)
+## Output: (22, 33, 50, 25)
+##
+## 50 => 5 x 0 => 0 (1 step)
+## 25 => 2 x 5 => 10 => 1 x 0 => 0 (2 steps)
+## 33 => 3 x 3 => 6 (1 step)
+## 22 => 2 x 2 => 4 (1 step)
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Create a temporary list that contains a list for each
+# element. The list consists of the int and the persistence
+# Then we just need to sort by persistence, then by the number
+
+from operator import itemgetter
+
+def persistence(number: int) -> int:
+    digits = [int(d) for d in str(number)]
+    if len(digits) <= 1:
+        return 0
+    prod = 1
+    for d in digits:
+        prod *= d
+    return 1 + persistence(prod)
+
+
+def persistence_sort(ints: list):
+    print("Input: (" , ", ".join(str(x) for x in ints), ")")
+    tmp = []
+    for x in ints:
+        tmp.append( (x, persistence(x)) )
+    s = sorted(tmp, key=itemgetter(1))
+    s = sorted(s, key=itemgetter(0))
+    #s = sorted(s)
+    print("Output: (" , ", ".join(str(x[0]) for x in s), ")")
+
+persistence_sort([15, 99, 1, 34])
+persistence_sort([50, 25, 33, 22])