Add Python solution to challenge 101

2 files changed, 125 insertions, 0 deletions

diff --git a/challenge-101/paulo-custodio/python/ch-1.py b/challenge-101/paulo-custodio/python/ch-1.py
new file mode 100644
index 0000000000..ff71f11953
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+++ b/challenge-101/paulo-custodio/python/ch-1.py
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+#!/usr/bin/env python
+
+# You are given an array @A of items (integers say, but they can be anything).
+# 
+# Your task is to pack that array into an MxN matrix spirally counterclockwise, 
+# as tightly as possible.
+# 
+# 'Tightly' means the absolute value |M-N| of the difference has to be as small 
+# as possible.
+
+import sys
+import math
+
+def spiral(numbers):
+    
+    # find max width of elements, convert to int
+    def max_width(numbers):
+        num_width = 1
+        for i in range(0, len(numbers)):
+            if len(numbers[i]) < num_width:
+                num_width = len(numbers[i])
+            numbers[i] = int(numbers[i])
+        return num_width
+
+    # find out m,n
+    def smallest_rect(n):
+        l, h = 1, n
+        for i in range(1, int(math.sqrt(n)+1)):
+            if ((n % i) == 0):
+                l, h = i, int(n / i)
+        return l, h
+
+    # build empty rectangle
+    rect = []
+    def build_rect(numbers):
+        m, n = smallest_rect(len(numbers))
+        for r in range (0, m+1):
+            rect.append(["" for c in range(0, n+1)])
+        return m, n
+
+    # build spiral rectangle
+    num_width = max_width(numbers)
+    m, n = build_rect(numbers)
+    
+    r, c = m, 1
+    i = 0
+    while (i < len(numbers)):
+        # go East
+        while (c <= n):
+            if (rect[r][c] != ""):
+                break
+            rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
+            i += 1
+            c += 1
+        c -= 1 
+        r -= 1
+        # go North
+        while (r >= 1):
+            if (rect[r][c] != ""):
+                break
+            rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
+            i += 1
+            r -= 1
+        r += 1 
+        c -= 1
+        # go West
+        while (c >= 1):
+            if (rect[r][c] != ""):
+                break
+            rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
+            i += 1
+            c -= 1
+        c += 1 
+        r += 1
+        # go South
+        while (r <= m):
+            if (rect[r][c] != ""):
+                break
+            rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
+            i += 1
+            r += 1
+        r -= 1 
+        c += 1
+
+    for r in range (1, m+1):
+        output = ""
+        for c in range (1, n+1):
+            output += rect[r][c]
+        print(output)
+
+# main
+numbers = sys.argv[1:]
+spiral(numbers)
diff --git a/challenge-101/paulo-custodio/python/ch-2.py b/challenge-101/paulo-custodio/python/ch-2.py
new file mode 100644
index 0000000000..1ea5f422f7
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+++ b/challenge-101/paulo-custodio/python/ch-2.py
@@ -0,0 +1,32 @@
+#! /usr/bin/env python
+
+# TASK #2 > Origin-containing Triangle
+# Submitted by: Stuart Little
+# You are given three points in the plane, as a list of six co-ordinates: 
+# A=(x1,y1), B=(x2,y2) and C=(x3,y3).
+# 
+# Write a script to find out if the triangle formed by the given three 
+# co-ordinates contain origin (0,0).
+# 
+# Print 1 if found otherwise 0.
+
+import sys
+
+def point_in_triangle(xp,yp, x1,y1,x2,y2,x3,y3):
+    def sign(x1,y1,x2,y2,x3,y3):
+        return (x1 - x3) * (y2 - y3) - (x2 - x3) * (y1 - y3)
+
+    d1 = sign(xp,yp, x1,y1, x2,y2)
+    d2 = sign(xp,yp, x2,y2, x3,y3)
+    d3 = sign(xp,yp, x3,y3, x1,y1)
+
+    has_neg = (d1 < 0) or (d2 < 0) or (d3 < 0)
+    has_pos = (d1 > 0) or (d2 > 0) or (d3 > 0)
+
+    if (not (has_neg and has_pos)):
+        return 1
+    else:
+        return 0
+
+x1,y1,x2,y2,x3,y3 = (float(sys.argv[i]) for i in range(1, 7))
+print(point_in_triangle(0.0,0.0, x1,y1,x2,y2,x3,y3))