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Diffstat (limited to 'challenge-262/jeanluc2020/python/ch-2.py')
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diff --git a/challenge-262/jeanluc2020/python/ch-2.py b/challenge-262/jeanluc2020/python/ch-2.py new file mode 100755 index 0000000000..c391bd37e6 --- /dev/null +++ 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); + |
