Add solution 261.

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

2 files changed, 135 insertions, 0 deletions

diff --git a/challenge-261/jeanluc2020/python/ch-1.py b/challenge-261/jeanluc2020/python/ch-1.py
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+++ b/challenge-261/jeanluc2020/python/ch-1.py
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-261/#TASK1
+#
+# Task 1: Element Digit Sum
+# =========================
+#
+# You are given an array of integers, @ints.
+#
+# Write a script to evaluate the absolute difference between element and digit
+# sum of the given array.
+#
+## Example 1
+##
+## Input: @ints = (1,2,3,45)
+## Output: 36
+##
+## Element Sum: 1 + 2 + 3 + 45 = 51
+## Digit Sum: 1 + 2 + 3 + 4 + 5 = 15
+## Absolute Difference: | 51 - 15 | = 36
+#
+## Example 2
+##
+## Input: @ints = (1,12,3)
+## Output: 9
+##
+## Element Sum: 1 + 12 + 3 = 16
+## Digit Sum: 1 + 1 + 2 + 3 = 7
+## Absolute Difference: | 16 - 7 | = 9
+#
+## Example 3
+##
+## Input: @ints = (1,2,3,4)
+## Output: 0
+##
+## Element Sum: 1 + 2 + 3 + 4 = 10
+## Digit Sum: 1 + 2 + 3 + 4 = 10
+## Absolute Difference: | 10 - 10 | = 0
+#
+## Example 4
+##
+## Input: @ints = (236, 416, 336, 350)
+## Output: 1296
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# This is pretty straight forward, calculate the sum for each
+# list, then calculate the digit sum by splitting each list
+# element into its digits and then calculating the sum from there.
+
+def element_digit_sum(ints: list) -> None:
+    print("Input: (", ", ".join(str(x) for x in ints), ")", sep="")
+    elem_sum = sum(ints)
+    digit_sum = 0
+    for elem in ints:
+        digits = [ int(x) for x in list(str(elem)) ]
+        digit_sum += sum(digits)
+    print("Output:", abs(elem_sum - digit_sum))
+
+element_digit_sum([1,2,3,45]);
+element_digit_sum([1,12,3]);
+element_digit_sum([1,2,3,4]);
+element_digit_sum([236, 416, 336, 350]);
+
diff --git a/challenge-261/jeanluc2020/python/ch-2.py b/challenge-261/jeanluc2020/python/ch-2.py
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+++ b/challenge-261/jeanluc2020/python/ch-2.py
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+#!/usr/bin/python3
+# https://theweeklychallenge.org/blog/perl-weekly-challenge-261/#TASK2
+#
+# Task 2: Multiply by Two
+# =======================
+#
+# You are given an array of integers, @ints and an integer $start.
+#
+# Write a script to do the followings:
+#
+# a) Look for $start in the array @ints, if found multiply the number by 2
+# b) If not found stop the process otherwise repeat
+#
+# In the end return the final value.
+#
+## Example 1
+##
+## Input: @ints = (5,3,6,1,12) and $start = 3
+## Output: 24
+##
+## Step 1: 3 is in the array so 3 x 2 = 6
+## Step 2: 6 is in the array so 6 x 2 = 12
+## Step 3: 12 is in the array so 12 x 2 = 24
+##
+## 24 is not found in the array so return 24.
+#
+## Example 2
+##
+## Input: @ints = (1,2,4,3) and $start = 1
+## Output: 8
+##
+## Step 1: 1 is in the array so 1 x 2 = 2
+## Step 2: 2 is in the array so 2 x 2 = 4
+## Step 3: 4 is in the array so 4 x 2 = 8
+##
+## 8 is not found in the array so return 8.
+#
+## Example 3
+##
+## Input: @ints = (5,6,7) and $start = 2
+## Output: 2
+##
+## 2 is not found in the array so return 2.
+#
+############################################################
+##
+## discussion
+##
+############################################################
+#
+# Instead of walking the list for every round, we create a
+# hash with the list elements as keys, which makes the lookup
+# easier. The we just check in a loop whether the current value
+# is still there.
+
+def multiply_by_two(ints: list, start: int) -> None:
+    print("Input: (", ", ".join(str(x) for x in ints), "), ", start, sep="")
+    hash: dict = {}
+    for x in ints:
+        hash[x] = 1
+    while start in hash:
+        start *= 2
+    print(f"Output: {start}")
+
+multiply_by_two( [5,3,6,1,12], 3);
+multiply_by_two( [1,2,4,3], 1);
+multiply_by_two( [5,6,7], 2);
+