remove sensitive question description

Author: kamyu104

Python/01-matrix.py

@@ -1,46 +1,13 @@
 # Time:  O(m * n)
 # Space: O(m * n)
 
-# Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
-# The distance between two adjacent cells is 1.
-#
-# Example 1:
-#
-# Input:
-# 0 0 0
-# 0 1 0
-# 0 0 0
-#
-# Output:
-# 0 0 0
-# 0 1 0
-# 0 0 0
-#
-# Example 2:
-#
-# Input:
-# 0 0 0
-# 0 1 0
-# 1 1 1
-#
-# Output:
-# 0 0 0
-# 0 1 0
-# 1 2 1
-#
-# Note:
-# The number of elements of the given matrix will not exceed 10,000.
-# There are at least one 0 in the given matrix.
-# The cells are adjacent in only four directions: up, down, left and right.
-
 import collections
 
 try:
     xrange          # Python 2
 except NameError:
     xrange = range  # Python 3
 
-
 class Solution(object):
     def updateMatrix(self, matrix):
         """
@@ -68,10 +35,6 @@ def updateMatrix(self, matrix):
 
         return matrix
 
-
-# Time:  O(m * n)
-# Space: O(m * n)
-# dp solution
 class Solution2(object):
     def updateMatrix(self, matrix):
         """
@@ -98,3 +61,4 @@ def updateMatrix(self, matrix):
                     if j < len(matrix[i])-1:
                         dp[i][j] = min(dp[i][j], dp[i][j+1]+1)
         return dp
+

Python/1-bit-and-2-bit-characters.py

@@ -1,40 +1,11 @@
 # Time:  O(n)
 # Space: O(1)
 
-# We have two special characters. The first character can be represented
-# by one bit 0.
-# The second character can be represented by two bits (10 or 11).
-#
-# Now given a string represented by several bits.
-#  Return whether the last character must
-# be a one-bit character or not. The given string will always end with a zero.
-#
-# Example 1:
-# Input:
-# bits = [1, 0, 0]
-# Output: True
-# Explanation:
-# The only way to decode it is two-bit character and one-bit character.
-# So the last character is one-bit character.
-#
-# Example 2:
-# Input:
-# bits = [1, 1, 1, 0]
-# Output: False
-# Explanation:
-# The only way to decode it is two-bit character and two-bit character.
-# So the last character is NOT one-bit character.
-# Note:
-#
-# 1 <= len(bits) <= 1000.
-# bits[i] is always 0 or 1.
-
 try:
     xrange          # Python 2
 except NameError:
     xrange = range  # Python 3
 
-
 class Solution(object):
     def isOneBitCharacter(self, bits):
         """
@@ -47,3 +18,4 @@ def isOneBitCharacter(self, bits):
                 break
             parity ^= bits[i]
         return parity == 0
+

Python/132-pattern.py

@@ -1,38 +1,11 @@
 # Time:  O(n)
 # Space: O(n)
 
-# Given a sequence of n integers a1, a2, ..., an,
-# a 132 pattern is a subsequence ai, aj, ak such that i < j < k and
-# ai < ak < aj. Design an algorithm that takes a list of n numbers as
-# input and checks whether there is a 132 pattern in the list.
-#
-# Note: n will be less than 15,000.
-#
-# Example 1:
-# Input: [1, 2, 3, 4]
-#
-# Output: False
-#
-# Explanation: There is no 132 pattern in the sequence.
-# Example 2:
-# Input: [3, 1, 4, 2]
-#
-# Output: True
-#
-# Explanation: There is a 132 pattern in the sequence: [1, 4, 2].
-# Example 3:
-# Input: [-1, 3, 2, 0]
-#
-# Output: True
-#
-# Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].
-
 try:
     xrange          # Python 2
 except NameError:
     xrange = range  # Python 3
 
-
 class Solution(object):
     def find132pattern(self, nums):
         """
@@ -49,3 +22,4 @@ def find132pattern(self, nums):
                     ak = st.pop()
             st.append(nums[i])
         return False
+

Python/2-keys-keyboard.py

@@ -1,29 +1,6 @@
 # Time:  O(sqrt(n))
 # Space: O(1)
 
-# Initially on a notepad only one character 'A' is present.
-# You can perform two operations on this notepad for each step:
-#
-# Copy All: You can copy all the characters present on the notepad
-# (partial copy is not allowed).
-# Paste: You can paste the characters which are copied last time.
-# Given a number n.
-# You have to get exactly n 'A' on the notepad by performing the minimum
-#  number of steps permitted.
-# Output the minimum number of steps to get n 'A'.
-#
-# Example 1:
-# Input: 3
-# Output: 3
-# Explanation:
-# Intitally, we have one character 'A'.
-# In step 1, we use Copy All operation.
-# In step 2, we use Paste operation to get 'AA'.
-# In step 3, we use Paste operation to get 'AAA'.
-# Note:
-# The n will be in the range [1, 1000].
-
-
 class Solution(object):
     def minSteps(self, n):
         """
@@ -41,3 +18,4 @@ def minSteps(self, n):
         if n > 1:
             result += n
         return result
+

Python/24-game.py

@@ -1,26 +1,6 @@
 # Time:  O(n^3 * 4^n) = O(1), n = 4
 # Space: O(n^2) = O(1)
 
-# You have 4 cards each containing a number from 1 to 9.
-# You need to judge whether they could operated through *, /, +, -, (, ) to get the value of 24.
-#
-# Example 1:
-# Input: [4, 1, 8, 7]
-# Output: True
-# Explanation: (8-4) * (7-1) = 24
-# Example 2:
-# Input: [1, 2, 1, 2]
-# Output: False
-# Note:
-# The division operator / represents real division, not integer division.
-# For example, 4 / (1 - 2/3) = 12.
-# Every operation done is between two numbers. In particular,
-# we cannot use - as a unary operator.
-# For example, with [1, 1, 1, 1] as input,
-# the expression -1 - 1 - 1 - 1 is not allowed.
-# You cannot concatenate numbers together. For example,
-# if the input is [1, 2, 1, 2], we cannot write this as 12 + 12.
-
 from operator import add, sub, mul, truediv
 from fractions import Fraction
 
@@ -29,7 +9,6 @@
 except NameError:
     xrange = range  # Python 3
 
-
 class Solution(object):
     def judgePoint24(self, nums):
         """
@@ -54,9 +33,6 @@ def judgePoint24(self, nums):
                     next_nums.pop()
         return False
 
-
-# Time:  O(n^3 * 4^n) = O(1), n = 4
-# Space: O(n^2) = O(1)
 class Solution2(object):
     def judgePoint24(self, nums):
         """
@@ -84,3 +60,4 @@ def dfs(nums):
             return False
 
         return dfs(map(Fraction, nums))
+

Python/3sum-closest.py

@@ -1,17 +1,6 @@
 # Time:  O(n^2)
 # Space: O(1)
 
-# Given an array S of n integers,
-# find three integers in S such that the sum is closest to a given number,
-# target.
-# Return the sum of the three integers.
-# You may assume that each input would have exactly one solution.
-#
-# For example, given array S = {-1 2 1 -4}, and target = 1.
-#
-# The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
-
-
 class Solution(object):
     def threeSumClosest(self, nums, target):
         """
@@ -36,3 +25,4 @@ def threeSumClosest(self, nums, target):
                         return target
             i += 1
         return result
+

Python/3sum-smaller.py

@@ -1,7 +1,6 @@
 # Time:  O(n^2)
 # Space: O(1)
 
-
 class Solution:
     # @param {integer[]} nums
     # @param {integer} target
@@ -22,3 +21,4 @@ def threeSumSmaller(self, nums, target):
             k += 1
 
         return count
+

Python/3sum.py

@@ -1,22 +1,8 @@
 # Time:  O(n^2)
 # Space: O(1)
 
-# Given an array S of n integers,
-# are there elements a, b, c in S such that a + b + c = 0?
-# Find all unique triplets in the array which gives the sum of zero.
-#
-# Note:
-# Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)
-# The solution set must not contain duplicate triplets.
-#    For example, given array S = {-1 0 1 2 -1 -4},
-#
-#    A solution set is:
-#    (-1, 0, 1)
-#    (-1, -1, 2)
-
 import collections
 
-
 class Solution(object):
     def threeSum(self, nums):
         """
@@ -62,3 +48,4 @@ def threeSum2(self, nums):
                         if sorted([j, y, 0 - j - y]) not in rtn:
                             rtn.append(sorted([j, y, 0 - j - y]))
         return rtn
+

Python/4-keys-keyboard.py

@@ -6,7 +6,6 @@
 except NameError:
     xrange = range  # Python 3
 
-
 class Solution(object):
     def maxA(self, N):
         """
@@ -26,9 +25,6 @@ def maxA(self, N):
         n4 = n - n3
         return 3**n3 * 4**n4
 
-
-# Time:  O(n)
-# Space: O(1)
 class Solution2(object):
     def maxA(self, N):
         """
@@ -41,3 +37,4 @@ def maxA(self, N):
         for i in xrange(7, N+1):
             dp[i % 6] = max(dp[(i-4) % 6]*3, dp[(i-5) % 6]*4)
         return dp[N % 6]
+

Python/4sum-ii.py

@@ -1,34 +1,8 @@
 # Time:  O(n^2)
 # Space: O(n^2)
 
-# Given four lists A, B, C, D of integer values,
-# compute how many tuples (i, j, k, l) there are
-# such that A[i] + B[j] + C[k] + D[l] is zero.
-#
-# To make problem a bit easier, all A, B, C, D have same length of N
-# where 0 <= N <= 500.
-# All integers are in the range of -228 to 228 - 1 and the result
-# is guaranteed to be at most 231 - 1.
-#
-# Example:
-#
-# Input:
-# A = [ 1, 2]
-# B = [-2,-1]
-# C = [-1, 2]
-# D = [ 0, 2]
-#
-# Output:
-# 2
-#
-# Explanation:
-# The two tuples are:
-# 1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
-# 2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
-
 import collections
 
-
 class Solution(object):
     def fourSumCount(self, A, B, C, D):
         """
@@ -40,3 +14,4 @@ def fourSumCount(self, A, B, C, D):
         """
         A_B_sum = collections.Counter(a+b for a in A for b in B)
         return sum(A_B_sum[-c-d] for c in C for d in D)
+