Add solution 262.

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

2 files changed, 116 insertions, 0 deletions

diff --git a/challenge-262/jeanluc2020/python/ch-1.py b/challenge-262/jeanluc2020/python/ch-1.py
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+++ b/challenge-262/jeanluc2020/python/ch-1.py
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-262/#TASK1
+#
+# Task 1: Max Positive Negative
+# =============================
+#
+# You are given an array of integers, @ints.
+#
+# Write a script to return the maximum number of either positive or negative
+# integers in the given array.
+#
+## Example 1
+##
+## Input: @ints = (-3, 1, 2, -1, 3, -2, 4)
+## Output: 4
+##
+## Count of positive integers: 4
+## Count of negative integers: 3
+## Maximum of count of positive and negative integers: 4
+#
+## Example 2
+##
+## Input: @ints = (-1, -2, -3, 1)
+## Output: 3
+##
+## Count of positive integers: 1
+## Count of negative integers: 3
+## Maximum of count of positive and negative integers: 3
+#
+## Example 3
+##
+## Input: @ints = (1,2)
+## Output: 2
+##
+## Count of positive integers: 2
+## Count of negative integers: 0
+## Maximum of count of positive and negative integers: 2
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Simply count the negatives and the positives, then return
+# the bigger of the two numbers.
+
+def max_positive_negative(ints: list) -> None:
+    print("Input: (", ", ".join(str(x) for x in ints), ")", sep="")
+    pos = 0
+    neg = 0
+    for num in ints:
+        if num < 0:
+            neg += 1
+        else:
+            pos += 1
+    result = pos if pos > neg else neg
+    print(f"Output: {result}")
+
+max_positive_negative([-3, 1, 2, -1, 3, -2, 4]);
+max_positive_negative([-1, -2, -3, 1]);
+max_positive_negative([1,2]);
+
diff --git a/challenge-262/jeanluc2020/python/ch-2.py b/challenge-262/jeanluc2020/python/ch-2.py
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+++ b/challenge-262/jeanluc2020/python/ch-2.py
@@ -0,0 +1,53 @@
+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-262/#TASK2
+#
+# Task 2: Count Equal Divisible
+# =============================
+#
+# You are given an array of integers, @ints and an integer $k.
+#
+# Write a script to return the number of pairs (i, j) where
+#
+# a) 0 <= i < j < size of @ints
+# b) ints[i] == ints[j]
+# c) i x j is divisible by k
+#
+## Example 1
+##
+## Input: @ints = (3,1,2,2,2,1,3) and $k = 2
+## Output: 4
+##
+## (0, 6) => ints[0] == ints[6] and 0 x 6 is divisible by 2
+## (2, 3) => ints[2] == ints[3] and 2 x 3 is divisible by 2
+## (2, 4) => ints[2] == ints[4] and 2 x 4 is divisible by 2
+## (3, 4) => ints[3] == ints[4] and 3 x 4 is divisible by 2
+#
+## Example 2
+##
+## Input: @ints = (1,2,3) and $k = 1
+## Output: 0
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Have one index variable go from 0 to size of @ints - 1, then
+# another walk from the first index + 1 to size of @ints - 1. In
+# case ints[i] == ints[j], check if the product of i and j is
+# divisible by $k. Return the sum of all instances where this is true.
+
+def count_equal_divisible(ints: list, k: int) -> None:
+    print("Input: (", ", ".join(str(x) for x in ints), ")", sep="")
+    result = 0
+    for i in range(len(ints)):
+        for j in range(i+1, len(ints)):
+            if ints[i] == ints[j]:
+                if i * j % k == 0:
+                    result += 1
+    print(f"Output: {result}")
+
+count_equal_divisible( [3,1,2,2,2,1,3], 2);
+count_equal_divisible( [1,2,3], 1);
+