Pylint refactor

algorithms/dp/__init__.py

@@ -20,4 +20,4 @@
 from .word_break import *
 from .int_divide import *
 from .k_factor import *
-from .planting_trees import *
+from .planting_trees import *

algorithms/dp/climbing_stairs.py

@@ -1,32 +1,36 @@
 """
 You are climbing a stair case.
-It takes n steps to reach to the top.
+It takes `steps` number of steps to reach to the top.
 
 Each time you can either climb 1 or 2 steps.
 In how many distinct ways can you climb to the top?
 
-Note: Given n will be a positive integer.
+Note: Given argument `steps` will be a positive integer.
 """
 
 
 # O(n) space
 
-def climb_stairs(n):
+def climb_stairs(steps):
     """
-    :type n: int
+    :type steps: int
     :rtype: int
     """
     arr = [1, 1]
-    for _ in range(1, n):
+    for _ in range(1, steps):
         arr.append(arr[-1] + arr[-2])
     return arr[-1]
 
 
 # the above function can be optimized as:
 # O(1) space
 
-def climb_stairs_optimized(n):
-    a = b = 1
-    for _ in range(n):
-        a, b = b, a + b
-    return a
+def climb_stairs_optimized(steps):
+    """
+    :type steps: int
+    :rtype: int
+    """
+    a_steps = b_steps = 1
+    for _ in range(steps):
+        a_steps, b_steps = b_steps, a_steps + b_steps
+    return a_steps

algorithms/dp/coin_change.py

@@ -1,31 +1,34 @@
 """
 Problem
-Given a value n, if we want to make change for N cents,
-and we have infinite supply of each of
-coins = {S1, S2, .. , Sm} valued coins, how many ways
-can we make the change?
-The order of coins doesn't matter.
-For example, for n = 4 and coins = [1, 2, 3], there are
-four solutions:
+Given a value `value`, if we want to make change for `value` cents, and we have infinite
+supply of each of coins = {S1, S2, .. , Sm} valued `coins`, how many ways can we make the change?
+The order of `coins` doesn't matter.
+For example, for `value` = 4 and `coins` = [1, 2, 3], there are four solutions:
 [1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3].
 So output should be 4.
 
-For n = 10 and coins = [2, 5, 3, 6], there are five solutions:
+For `value` = 10 and `coins` = [2, 5, 3, 6], there are five solutions:
+
 [2, 2, 2, 2, 2], [2, 2, 3, 3], [2, 2, 6], [2, 3, 5] and [5, 5].
 So the output should be 5.
 
-Time complexity: O(n * m) where n is the value and m is the number of coins
+Time complexity: O(n * m) where n is the `value` and m is the number of `coins`
 Space complexity: O(n)
 """
 
+def count(coins, value):
+    """ Find number of combination of `coins` that adds upp to `value`
 
-def count(coins, n):
+    Keyword arguments:
+    coins -- int[]
+    value -- int
+    """
     # initialize dp array and set base case as 1
-    dp = [1] + [0] * n
+    dp_array = [1] + [0] * value
 
     # fill dp in a bottom up manner
     for coin in coins:
-        for i in range(coin, n+1):
-            dp[i] += dp[i-coin]
+        for i in range(coin, value+1):
+            dp_array[i] += dp_array[i-coin]
 
-    return dp[n]
+    return dp_array[value]

algorithms/dp/combination_sum.py

@@ -27,34 +27,48 @@
 
 """
 
-dp = None
-
+DP = None
 
 def helper_topdown(nums, target):
-    global dp
-    if dp[target] != -1:
-        return dp[target]
+    """Generates DP and finds result.
+
+    Keyword arguments:
+    nums -- positive integer array without duplicates
+    target -- integer describing what a valid combination should add to
+    """
+    if DP[target] != -1:
+        return DP[target]
     res = 0
-    for i in range(0, len(nums)):
-        if target >= nums[i]:
-            res += helper_topdown(nums, target - nums[i])
-    dp[target] = res
+    for num in nums:
+        if target >= num:
+            res += helper_topdown(nums, target - num)
+    DP[target] = res
     return res
 
 
 def combination_sum_topdown(nums, target):
-    global dp
-    dp = [-1] * (target + 1)
-    dp[0] = 1
-    return helper_topdown(nums, target)
+    """Find number of possible combinations in nums that add up to target, in top-down manner.
 
+    Keyword arguments:
+    nums -- positive integer array without duplicates
+    target -- integer describing what a valid combination should add to
+    """
+    global DP
+    DP = [-1] * (target + 1)
+    DP[0] = 1
+    return helper_topdown(nums, target)
 
-# EDIT: The above solution is top-down. How about a bottom-up one?
 def combination_sum_bottom_up(nums, target):
-    comb = [0] * (target + 1)
-    comb[0] = 1
-    for i in range(0, len(comb)):
-        for j in range(len(nums)):
-            if i - nums[j] >= 0:
-                comb[i] += comb[i - nums[j]]
-    return comb[target]
+    """Find number of possible combinations in nums that add up to target, in bottom-up manner.
+
+    Keyword arguments:
+    nums -- positive integer array without duplicates
+    target -- integer describing what a valid combination should add to
+    """
+    combs = [0] * (target + 1)
+    combs[0] = 1
+    for i in range(0, len(combs)):
+        for num in nums:
+            if i - num >= 0:
+                combs[i] += combs[i - num]
+    return combs[target]

algorithms/dp/edit_distance.py

@@ -36,28 +36,34 @@
 indexes i and j, respectively.
 
 To find the edit distance between two words A and B,
-we need to find edit(m, n), where m is the length of A and n
-is the length of B.
+we need to find edit(length_a, length_b).
+
+Time: O(length_a*length_b)
+Space: O(length_a*length_b)
 """
 
 
-def edit_distance(A, B):
-    # Time: O(m*n)
-    # Space: O(m*n)
+def edit_distance(word_a, word_b):
+    """Finds edit distance between word_a and word_b
+
+    Kwyword arguments:
+    word_a -- string
+    word_b -- string
+    """
 
-    m, n = len(A) + 1, len(B) + 1
+    length_a, length_b = len(word_a) + 1, len(word_b) + 1
 
-    edit = [[0 for _ in range(n)] for _ in range(m)]
+    edit = [[0 for _ in range(length_b)] for _ in range(length_a)]
 
-    for i in range(1, m):
+    for i in range(1, length_a):
         edit[i][0] = i
 
-    for j in range(1, n):
+    for j in range(1, length_b):
         edit[0][j] = j
 
-    for i in range(1, m):
-        for j in range(1, n):
-            cost = 0 if A[i - 1] == B[j - 1] else 1
+    for i in range(1, length_a):
+        for j in range(1, length_b):
+            cost = 0 if word_a[i - 1] == word_b[j - 1] else 1
             edit[i][j] = min(edit[i - 1][j] + 1, edit[i][j - 1] + 1, edit[i - 1][j - 1] + cost)
 
-    return edit[-1][-1]  # this is the same as edit[m][n]
+    return edit[-1][-1]  # this is the same as edit[length_a][length_b]

algorithms/dp/egg_drop.py

@@ -26,9 +26,14 @@
 
 
 def egg_drop(n, k):
+    """
+    Keyword arguments:
+    n -- number of floors
+    k -- number of eggs
+    """
     # A 2D table where entery eggFloor[i][j] will represent minimum
     # number of trials needed for i eggs and j floors.
-    egg_floor = [[0 for x in range(k+1)] for x in range(n + 1)]
+    egg_floor = [[0 for _ in range(k + 1)] for _ in range(n + 1)]
 
     # We need one trial for one floor and 0 trials for 0 floors
     for i in range(1, n+1):

algorithms/dp/fib.py

@@ -35,8 +35,7 @@ def fib_recursive(n):
 
     if n <= 1:
         return n
-    else:
-        return fib_recursive(n-1) + fib_recursive(n-2)
+    return fib_recursive(n-1) + fib_recursive(n-2)
 
 # print(fib_recursive(35)) # => 9227465 (slow)
 
@@ -81,13 +80,13 @@ def fib_iter(n):
 
     fib_1 = 0
     fib_2 = 1
-    sum = 0
+    res = 0
     if n <= 1:
         return n
     for _ in range(n-1):
-        sum = fib_1 + fib_2
+        res = fib_1 + fib_2
         fib_1 = fib_2
-        fib_2 = sum
-    return sum
+        fib_2 = res
+    return res
 
 # print(fib_iter(100)) # => 354224848179261915075

algorithms/dp/hosoya_triangle.py

@@ -19,29 +19,38 @@
 
 """
 
+def hosoya(height, width):
+    """ Calculates the hosoya triangle
 
-def hosoya(n, m):
-    if ((n == 0 and m == 0) or (n == 1 and m == 0) or
-        (n == 1 and m == 1) or (n == 2 and m == 1)):
+    height -- height of the triangle
+    """
+    if (width == 0) and (height in (0,1)):
         return 1
-    if n > m:
-        return hosoya(n - 1, m) + hosoya(n - 2, m)
-    elif m == n:
-        return hosoya(n - 1, m - 1) + hosoya(n - 2, m - 2)
-    else:
-        return 0
-
-
-def print_hosoya(n):
-    for i in range(n):
+    if (width == 1) and (height in (1,2)):
+        return 1
+    if height > width:
+        return hosoya(height - 1, width) + hosoya(height - 2, width)
+    if width == height:
+        return hosoya(height - 1, width - 1) + hosoya(height - 2, width - 2)
+    return 0
+
+def print_hosoya(height):
+    """Prints the hosoya triangle
+
+    height -- height of the triangle
+    """
+    for i in range(height):
         for j in range(i + 1):
-            print(hosoya(i, j), end=" ")
-        print("\n", end="")
+            print(hosoya(i, j) , end = " ")
+        print ("\n", end = "")
 
+def hosoya_testing(height):
+    """Test hosoya function
 
-def hosoya_testing(n):
-    x = []
-    for i in range(n):
+    height -- height of the triangle
+    """
+    res = []
+    for i in range(height):
         for j in range(i + 1):
-            x.append(hosoya(i, j))
-    return x
+            res.append(hosoya(i, j))
+    return res

algorithms/dp/int_divide.py

@@ -1,5 +1,6 @@
 """
-Given positive integer n, find an algorithm to find the number of non-negative number division, or descomposition.
+Given positive integer decompose, find an algorithm to find the number of
+non-negative number division, or decomposition.
 
 The complexity is O(n^2).
 
@@ -36,15 +37,19 @@
 """
 
 
-def int_divide(n):
-    arr = [[0 for i in range(n + 1)] for j in range(n + 1)]
+def int_divide(decompose):
+    """Find number of decompositions from `decompose`
+
+    decompose -- integer
+    """
+    arr = [[0 for i in range(decompose + 1)] for j in range(decompose + 1)]
     arr[1][1] = 1
-    for i in range(1, n + 1):
-        for j in range(1, n + 1):
+    for i in range(1, decompose + 1):
+        for j in range(1, decompose + 1):
             if i < j:
                 arr[i][j] = arr[i][i]
             elif i == j:
                 arr[i][j] = 1 + arr[i][j - 1]
             else:
                 arr[i][j] = arr[i][j - 1] + arr[i - j][j]
-    return arr[n][n]
+    return arr[decompose][decompose]