Add solution 253.

Signed-off-by: Thomas Köhler <jean-luc@picard.franken.de>

1 files changed, 91 insertions, 0 deletions

diff --git a/challenge-253/jeanluc2020/python/ch-2.py b/challenge-253/jeanluc2020/python/ch-2.py
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+++ b/challenge-253/jeanluc2020/python/ch-2.py
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-253/#TASK2
+#
+# Task 2: Weakest Row
+# ===================
+#
+# You are given an m x n binary matrix i.e. only 0 and 1 where 1 always appear before 0.
+#
+# A row i is weaker than a row j if one of the following is true:
+#
+## a) The number of 1s in row i is less than the number of 1s in row j.
+## b) Both rows have the same number of 1 and i < j.
+#
+# Write a script to return the order of rows from weakest to strongest.
+#
+## Example 1
+##
+## Input: $matrix = [
+##                    [1, 1, 0, 0, 0],
+##                    [1, 1, 1, 1, 0],
+##                    [1, 0, 0, 0, 0],
+##                    [1, 1, 0, 0, 0],
+##                    [1, 1, 1, 1, 1]
+##                  ]
+## Output: (2, 0, 3, 1, 4)
+##
+## The number of 1s in each row is:
+## - Row 0: 2
+## - Row 1: 4
+## - Row 2: 1
+## - Row 3: 2
+## - Row 4: 5
+#
+## Example 2
+##
+## Input: $matrix = [
+##                    [1, 0, 0, 0],
+##                    [1, 1, 1, 1],
+##                    [1, 0, 0, 0],
+##                    [1, 0, 0, 0]
+##                  ]
+## Output: (0, 2, 3, 1)
+##
+## The number of 1s in each row is:
+## - Row 0: 1
+## - Row 1: 4
+## - Row 2: 1
+## - Row 3: 1
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# First, we create a second array which contains the number of 1's
+# in each row and the row's index, then we sort that array by the
+# number of ones and the index - which means we only sort by the
+# number of ones since python's sort is stable, keeping the order
+# of elements intact if the values are the same, and the index is
+# therefore already sorted after sorting by number of ones.
+#
+
+from operator import itemgetter
+
+def number_of_ones( array: list) -> int:
+    output = 0
+    for value in array:
+        if value > 0:
+            output += 1
+    return output
+
+def weakest_row(matrix: list) -> None:
+    print("Input: [")
+    for array in matrix:
+        print("         [", ", ".join([str(x) for x in array]), "],")
+    print("       ]")
+    sort_by_me = []
+    i = 0
+    for array in matrix:
+        num = number_of_ones(array)
+        pos = i
+        i += 1
+        sort_by_me.append([num, pos])
+    sorted_array = sorted(sort_by_me, key=itemgetter(0))
+    print("Output: (", ", ".join([str(x[1]) for x in sorted_array]), ")", sep=None)
+
+
+weakest_row([ [1, 1, 0, 0, 0], [1, 1, 1, 1, 0], [1, 0, 0, 0, 0], [1, 1, 0, 0, 0], [1, 1, 1, 1, 1] ])
+weakest_row([ [1, 0, 0, 0], [1, 1, 1, 1], [1, 0, 0, 0], [1, 0, 0, 0] ])
+