pwc243 solution in python

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diff --git a/challenge-243/pokgopun/python/ch-1.py b/challenge-243/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-243/
+"""
+
+Task 1: Reverse Pairs
+
+Submitted by: [50]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of integers.
+
+   Write a script to return the number of reverse pairs in the given
+   array.
+
+   A reverse pair is a pair (i, j) where: a) 0 <= i < j < nums.length and
+   b) nums[i] > 2 * nums[j].
+
+Example 1
+
+Input: @nums = (1, 3, 2, 3, 1)
+Output: 2
+
+(1, 4) => nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+
+Example 2
+
+Input: @nums = (2, 4, 3, 5, 1)
+Output: 3
+
+(1, 4) => nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+(2, 4) => nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+
+Task 2: Floor Sum
+"""
+### solution by pokgopun@gmail.com
+
+def rpCount(nums: tuple):
+    c, l = 0, len(nums)
+    for i in range(l-1):
+        for j in range(i,l):
+            if nums[i] > 2 * nums[j]: c += 1
+    return c
+
+for inpt,otpt in {
+        (1, 3, 2, 3, 1):2,
+        (2, 4, 3, 5, 1):3,
+        }.items():
+    print(rpCount(inpt)==otpt)
+
+
+
diff --git a/challenge-243/pokgopun/python/ch-2.py b/challenge-243/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-243/
+"""
+
+Task 2: Floor Sum
+
+Submitted by: [51]Mohammad S Anwar
+     __________________________________________________________________
+
+   You are given an array of positive integers (>=1).
+
+   Write a script to return the sum of floor(nums[i] / nums[j]) where 0 <=
+   i,j < nums.length. The floor() function returns the integer part of the
+   division.
+
+Example 1
+
+Input: @nums = (2, 5, 9)
+Output: 10
+
+floor(2 / 5) = 0
+floor(2 / 9) = 0
+floor(5 / 9) = 0
+floor(2 / 2) = 1
+floor(5 / 5) = 1
+floor(9 / 9) = 1
+floor(5 / 2) = 2
+floor(9 / 2) = 4
+floor(9 / 5) = 1
+
+Example 2
+
+Input: @nums = (7, 7, 7, 7, 7, 7, 7)
+Output: 49
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 19th November
+   2023.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+def floorSum(tup: tuple):
+    c, l = 0, len(tup)
+    for i in range(l):
+        for j in range(l):
+            c += tup[j]//tup[i]
+    return c
+
+for inpt,otpt in {
+        (2, 5, 9): 10,
+        (7, 7, 7, 7, 7, 7, 7): 49,
+        }.items():
+    print(floorSum(inpt)==otpt)
+
+