Merge branch 'master' of github.com:manwar/perlweeklychallenge-club into challenges

2 files changed, 131 insertions, 0 deletions

diff --git a/challenge-249/jeanluc2020/python/ch-1.py b/challenge-249/jeanluc2020/python/ch-1.py
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+++ b/challenge-249/jeanluc2020/python/ch-1.py
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+#!/usr/bin/env python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-249/#TASK1
+#
+# Task 1: Shortest Distance
+# =========================
+#
+# You are given an array of integers with even number of elements.
+#
+# Write a script to divide the given array into equal pairs such that:
+#
+# a) Each element belongs to exactly one pair.
+# b) The elements present in a pair are equal.
+#
+#
+## Example 1
+##
+## Input: @ints = (3, 2, 3, 2, 2, 2)
+## Output: (2, 2), (3, 3), (2, 2)
+##
+## There are 6 elements in @ints.
+## They should be divided into 6 / 2 = 3 pairs.
+## @ints is divided into the pairs (2, 2), (3, 3), and (2, 2) satisfying all the
+## conditions.
+#
+## Example 2
+##
+## Input: @ints = (1, 2, 3, 4)
+## Output: ()
+##
+## There is no way to divide @ints 2 pairs such that the pairs satisfy every
+## condition.
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# We use the elements of the array as keys for a hash in which we
+# count the amount of elements of this value in the array.
+# If in the end, all values in the hash are even, we can spit out the
+# correct number of arrays, otherwise we can only return an empty array.
+
+def shortest_distance(ints: list) -> list:
+    print("Input: (", ", ".join([str(x) for x in ints]), ")")
+    result = []
+    map = {}
+    for i in ints:
+        if i in map:
+            map[i] += 1
+        else:
+            map[i] = 1
+    for key in map.keys():
+        if map[key] % 2 != 0:
+            print("Output: ()")
+            return ()
+        i = int( map[key] / 2 )
+        for j in range(1,i+1):
+            result.append( [key, key] )
+    first = True
+    print("Output: ", end="")
+    for elem in result:
+        if first:
+            first = False
+        else:
+            print(", ", end="")
+        print(", ".join([str(x) for x in elem]), end="")
+    print("")
+
+shortest_distance([3, 2, 3, 2, 2, 2])
+shortest_distance([1, 2, 3, 4])
+
diff --git a/challenge-249/jeanluc2020/python/ch-2.py b/challenge-249/jeanluc2020/python/ch-2.py
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index 0000000000..504e70a26d
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+++ b/challenge-249/jeanluc2020/python/ch-2.py
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+#!/usr/bin/env python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-249/#TASK2
+#
+# Task 2: DI String Match
+# =======================
+#
+# You are given a string s, consisting of only the characters "D" and "I".
+#
+# Find a permutation of the integers [0 .. length(s)] such that for each
+# character s[i] in the string:
+#
+# s[i] == 'I' ⇒ perm[i] < perm[i + 1]
+# s[i] == 'D' ⇒ perm[i] > perm[i + 1]
+#
+## Example 1
+##
+## Input: $str = "IDID"
+## Output: (0, 4, 1, 3, 2)
+#
+## Example 2
+##
+## Input: $str = "III"
+## Output: (0, 1, 2, 3)
+#
+## Example 3
+##
+## Input: $str = "DDI"
+## Output: (3, 2, 0, 1)
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# We just count from 0 to length(s) for each "I" and from
+# length(s) to 0 for each "D", and in the end add the last
+# element we didn't use up yet.
+
+def DI_string_match(string: str) -> list:
+    print("Input: '", string, "'", sep="")
+    upper = len(string)
+    lower = 0
+    result = []
+    for char in list(string):
+        if char == "I":
+            result.append(lower)
+            lower += 1
+        else:
+            result.append(upper)
+            upper -= 1
+    result.append(lower)
+    print("Output: (", ", ".join([str(x) for x in result]), ")")
+    return result
+
+DI_string_match("IDID");
+DI_string_match("III");
+DI_string_match("DDI");
+