create solutions cleaner and solve 1,2,4,15

amazon_problems/problem_10.py

@@ -9,4 +9,4 @@
 0 0 0 0 0
 1 1 0 0 1
 1 1 0 0 1
-"""
+"""

amazon_problems/problem_11.py

@@ -10,4 +10,4 @@
      /  \
    22    35
 You can assume each node has a parent pointer.
-"""
+"""

amazon_problems/problem_12.py

@@ -4,20 +4,4 @@
 •	init(size): initialize the array with size
 •	set(i, val): updates index at i with val where val is either 1 or 0.
 •	get(i): gets the value at index i.
-"""
-
-class BitArray:
-
-	def __init__(self, size):
-		self.data=[]
-		self.size = size
-		for i in range(size):
-			self.data.append(0)
-
-	def set(i , val):
-		if 0 <= size < self.size
-			self.data[i] = val
-
-	def get(i):
-		if 0 <= size < self.size
-			return data[i]
+"""

amazon_problems/problem_15.py

@@ -5,19 +5,4 @@
 For example, carrace should return true, 
 since it can be rearranged to form racecar, which is a palindrome. 
 daily should return false, since there's no rearrangement that can form a palindrome.
-"""
-#idea : we can't have more than one letter that has an odd frequency
-def solve(s):
-    freq = [0] * 256
-    for c in s:
-        freq[ord(c)] += 1
-
-    cp = 0
-    for f in freq:
-        cp += f & 1
-
-    return cp <= 1
-
-
-print(solve('daily')) # returns False
-print(solve('carrace')) #returns True
+"""

amazon_problems/problem_2.py

@@ -4,36 +4,4 @@
 that contains at most k distinct characters.
 
 For example, given s = "abcba" and k = 2, the longest substring with k
-distinct characters is "bcb"."""
-
-
-def main(s, k):
-    start = 0
-    current = ""
-    max_lengths = 0
-    while start < len(s):
-        for i in range(start + k - 1, len(s)):
-            ss = s[start:i + 1]
-            counter = count_character(ss)
-            if counter > k:
-                break
-            else:
-                if len(ss) > max_lengths:
-                    max_lengths = len(ss)
-                    current = ss
-        start += 1
-    return current
-
-
-def count_character(string):
-    current = [string[0]]
-    counter = 1
-    for i in string:
-        if i not in current:
-            counter += 1
-            current.append(i)
-    return counter
-
-
-# print(count_character("abbbcaabca"))
-print(main("abbbcef", 3))
+distinct characters is "bcb"."""

amazon_problems/problem_3.py

@@ -6,23 +6,4 @@
 Implement run-length encoding and decoding.
 You can assume the string to be encoded have no digits
 and consists solely of alphabetic characters.
-You can assume the string to be decoded is valid """
-
-
-def main(string):
-    if len(string) == 0:
-        return ""
-    else:
-        current = string[0]
-        counter = 0
-        output = ""
-        for i in string:
-            if current == i:
-                counter += 1
-            else:
-                output += str(counter) + current
-                counter = 1
-                current = i
-        output += str(counter) + current
-        return output
-print(main("AAAABBBCCDAA"))
+You can assume the string to be decoded is valid """

amazon_problems/problem_30.py

@@ -4,11 +4,4 @@
 and returns the number of digits the input has.
 
 Constraint: don't use any loops.
-"""
-
-# idea: use the log of ten
-
-import numpy as np
-
-def number_of_digit(x):
-	return np.floor(np.log10(x,10)) + 1
+"""

amazon_problems/problem_4.py

@@ -6,30 +6,4 @@
 •	max(), which returns the maximum value in the stack currently. 
 If there are no elements in the stack, then it should throw an error or return null.
 Each method should run in constant time.
-"""
-
-
-class stack(object):
-    """stack object for the solution"""
-
-    def __init__(self):
-        self.max = None
-        self.count = 0
-        self.array = []
-
-    def push(self, val):
-        self.array.append(val)
-        if val >= self.max:
-            self.max = val
-
-    def pop(self):
-        if len(self.array) == 0:
-            return None
-        else:
-            value = self.array.pop()
-            if value == self.max:
-                self.max = max(self.array)
-            return value
-
-    def max(self):
-        return self.max
+"""

amazon_problems/problem_6.py

@@ -10,17 +10,4 @@
 not take any elements.
 
 Do this in O(N) time.
-"""
-
-
-def solve(arr):
-    mx = 0
-    sum = 0
-    for a in arr:
-        sum += a
-        if sum < 0: sum = 0
-        mx = max(mx, sum)
-    return mx
-
-print(solve([34, -50, 42, 14, -5, 86])) # returns 137
-print(solve([-5, -1, -8, -9])) # returns 0
+"""

amazon_problems/problem_9.py

@@ -29,71 +29,4 @@
 9
 14
 13
-12"""
-
-a = [[1, 2, 3, 4, 5],
-     [6, 7, 8, 9, 10],
-     [11, 12, 13, 14, 15],
-     [16, 17, 18, 19, 20]]
-
-b = [[1, 2, 3],
-     [4, 5, 6]]
-
-c = [[1, 2, 3]]
-
-d = [[1],
-     [2],
-     [3],
-     [4],
-     [5],
-     [6]]
-
-e = [[1, 2, 3, 4],
-     [1, 2, 3, 4],
-     [1, 2, 3, 4]]
-
-
-def main():
-    matrix = e
-    rs = 0
-    re = len(matrix) - 1
-    cs = 0
-    ce = len(matrix[0]) - 1
-    ci = 0
-    cj = 0
-    total = len(matrix) * len(matrix[0])
-    now = 0
-    ori = 0
-
-    while now < total:
-
-        if ori == 0:
-            for cj in range(cs, ce + 1):
-                print(matrix[ci][cj])
-                now += 1
-            rs += 1
-            ori = 1
-
-        elif ori == 1:
-            for ci in range(rs, re + 1):
-                print(matrix[ci][cj])
-                now += 1
-            ce -= 1
-            ori = 2
-
-        elif ori == 2:
-            for cj in range(ce, cs - 1, -1):
-                print(matrix[ci][cj])
-                now += 1
-            re -= 1
-            ori = 3
-
-        elif ori == 3:
-            for ci in range(re, rs - 1, -1):
-                print(matrix[ci][cj])
-                now += 1
-            cs += 1
-            ori = 0
-
-
-main()
+12"""

apple_problems/problem_1.py

@@ -1,18 +1,4 @@
 """This problem was asked by Apple.
 
 Implement a job scheduler which takes in a function
-f and an integer n, and calls f after n milliseconds."""
-
-import time as t
-
-def dummy():
-    print("I'm Called")
-
-def job_scheduller(f,n):
-    start = t.time()*1000
-    end = start + n
-    while start < end:
-        start = t.time()*1000
-    f()
-
-job_scheduller(dummy,1000)
+f and an integer n, and calls f after n milliseconds."""

apple_problems/problem_2.py

@@ -3,36 +3,4 @@
 Recall that a queue is a FIFO (first-in, first-out) 
 data structure with the following methods: enqueue, which 
 inserts an element into the queue, and dequeue, which removes it.
-."""
-
-
-class Queue:
-
-    def __init__(self):
-        self.inbounds = []
-        self.outbounds = []
-
-    def enqueue(self, data):
-        self.inbounds.append(data)
-
-    def dequeue(self):
-        if not self.outbounds:
-            while self.inbounds:
-                self.outbounds.append(self.inbounds.pop())
-        return self.outbounds.pop()
-
-
-# Testing the Queue.
-queue = Queue()
-queue.enqueue(5)
-queue.enqueue(6)
-queue.enqueue(7)
-print(queue.inbounds)
-
-queue.dequeue()
-print(queue.inbounds)
-print(queue.outbounds)
-
-queue.dequeue()
-print(queue.inbounds)
-print(queue.outbounds)
+."""

cisco_problems/problem_1.py

@@ -6,4 +6,4 @@
 For example, 10101010 should be 01010101. 11100010 should be 11010001.
 
 Bonus: Can you do this in one line?
-"""
+"""

clean_solutions.py

@@ -0,0 +1,35 @@
+import os 
+
+def get_files(skip=True):
+    files = os.listdir()
+    needed_folders = []
+    for file in files:
+        if skip:
+            if '.' in file or '0' in file or "LICENSE" == file or "data_structure" == file:
+                continue
+        needed_folders.append(file)
+    return needed_folders
+
+def clean_content(files):
+    for file in files:
+        if '.py' in file:
+            file_obj = open(file, 'r')
+            content = file_obj.readlines()
+            content = ''.join(content[:])
+            task = content.split('"""')[1]
+            file_obj = open(file, 'w')
+            file_obj.truncate()
+            task = '"""' + task + '"""'
+            file_obj.write(task)
+
+def remove_solutions():
+    needed_folders = get_files()
+    for folder in needed_folders:
+        os.chdir(folder)
+        files = get_files(skip=False)
+        clean_content(files)
+        os.chdir("..")
+
+if __name__ == '__main__':
+    remove_solutions()
+

context_logic_problems/problem_1.py

@@ -1,18 +1,4 @@
 """This question was asked by ContextLogic.
 
 Implement division of two positive integers without using the division, 
-multiplication, or modulus operators. Return the quotient as an integer, ignoring the remainder."""
-
-# main method for making division, quiet easy: just subtract second nomber for the first until 0. and count.
-def division(a,b):
-    i=0
-    while a >= b:
-        i += 1
-        a = a-b
-    return i
-
-print(division(6,2))
-print(division(10,5))
-print(division(20,5))
-print(division(120,20))
-
+multiplication, or modulus operators. Return the quotient as an integer, ignoring the remainder."""