LeetCode – Page 8

Solution to Best Time to Buy and Sell Stock III by LeetCode

11 Jul

Question: https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/ Question Name: Best Time to Buy and Sell Stock III This problem is a simplified version of the challenge from Codility.

class Solution:

# @param prices, a list of integer

# @return an integer

def maxProfit(self, prices):

if len(prices) <= 1: return 0

# maxProfitEndingHere records the maximum profit, until here (inclusive)

# with zero or one transaction.

maxProfitEndingHere = [0] * len(prices)

lowestEndingHere = prices[0]

for index in xrange(1, len(prices)):

lowestEndingHere = min(lowestEndingHere, prices[index])

maxProfitEndingHere[index] = max(maxProfitEndingHere[index-1],

prices[index] - lowestEndingHere)

# maxProfitEndingHere records the maximum profit, after here (inclusive)

# with zero or one transaction.

maxProfitFromHere = [0] * len(prices)

highestFromHere = prices[-1]

for index in xrange(len(prices)-2, -1, -1):

highestFromHere = max(highestFromHere, prices[index])

maxProfitFromHere[index] = max(maxProfitFromHere[index+1],

highestFromHere - prices[index])

# Find the maximum profit, with zeor or one or two transactions.

maxProfit = maxProfitEndingHere[-1]

for index in xrange(len(prices)-1):

profit = maxProfitEndingHere[index] + maxProfitFromHere[index+1]

maxProfit = max(maxProfit, profit)

return maxProfit

Solution to Binary Tree Maximum Path Sum by LeetCode

10 Jul

Question: https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/ Question Name: Binary Tree Maximum Path Sum

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

class Solution:

# @param root, a tree node

# @return a tuple

def maxPathSumHelper(self, root):

maxPathSumInThisSubTree = root.val

maxPathSumToUpper = root.val # The maximum path sum, which could

# be used by its ancestors.

left, right = None, None

if root.left != None:

left = self.maxPathSumHelper(root.left)

# Any path, who has both the nodes of root's son(s) AND root's

# ancestors, must have the node "root" of this subtree.

maxPathSumToUpper = max(maxPathSumToUpper, root.val + left[1])

# The maximum path of current subtree may not cross its root.

maxPathSumInThisSubTree = max(maxPathSumInThisSubTree,

maxPathSumToUpper,

left[0]

)

if root.right != None:

# Same as the left son.

right = self.maxPathSumHelper(root.right)

maxPathSumToUpper = max(maxPathSumToUpper, root.val + right[1])

maxPathSumInThisSubTree = max(maxPathSumInThisSubTree,

maxPathSumToUpper,

right[0]

)

# If both the left son and right son exist, the maximum path of this

# subtree may across the left subtree, current root and the right

# substree.

if left != None and right != None:

maxPathSumInThisSubTree = max(maxPathSumInThisSubTree,

root.val + left[1] + right[1]

)

return maxPathSumInThisSubTree, maxPathSumToUpper

# @param root, a tree node

# @return an integer

def maxPathSum(self, root):

return self.maxPathSumHelper(root)[0]

Solution to Best Time to Buy and Sell Stock II by LeetCode

10 Jul

Question: https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/ Question Name: Best Time to Buy and Sell Stock II

class Solution:

# @param prices, a list of integer

# @return an integer

def maxProfit(self, prices):

maxProfit = 0

for index in xrange(len(prices)-1):

if prices[index+1] > prices[index]:

# As long as we can earn money, do the transaction.

maxProfit += prices[index+1] - prices[index]

return maxProfit

Solution to Best Time to Buy and Sell Stock by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock/ Question Name: Best Time to Buy and Sell Stock

lass Solution:

# @param prices, a list of integer

# @return an integer

def maxProfit(self, prices):

# Not enough time to earn profit

if len(prices) < 2: return 0

# The maximum price from here to the end

maxFromHere = prices[-1]

# The maximum profit, we could earn.

# Could initialize it with any valid profit.

maxProfit = prices[1] - prices[0]

for index in xrange(len(prices)-2, -1, -1):

price = prices[index]

# The maximum profit, if we buy it here.

profit = maxFromHere - price

maxProfit = max(maxProfit, profit)

maxFromHere = max(maxFromHere, price)

# If maxProfit is negative, we would not take

# any transaction.

# If we initialize the maxProfit with 0, this

# step is not necessary.

maxProfit = max(0, maxProfit)

return maxProfit

Solution to Triangle by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/triangle/ Question Name: Triangle

class Solution:

# @param triangle, a list of lists of integers

# @return an integer

def minimumTotal(self, triangle):

# Empty triangle

if len(triangle) == 0: return 0

# The length of triangle is equal to the length of its last row.

pathSum = [0] * len(triangle)

# Initialize the pathSum with the first row

pathSum[0] = triangle[0][0]

# Compute the remaining rows

for row in xrange(1, len(triangle)):

# The last element of current row can only be reached from the

# last element of previous row.

pathSum[row] = pathSum[row-1] + triangle[row][row]

# Each element can be reached from two adjacent nodes in the

# previous row.

for column in xrange(row-1, 0, -1):

pathSum[column] = min(pathSum[column], pathSum[column-1]) +

triangle[row][column]

# The first element of current row can only be reached from the

# first element of previous row.

pathSum[0] += triangle[row][0]

# Return the length of the shortest path.

return min(pathSum)

Solution to Populating Next Right Pointers in Each Node II by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/ Question Name: Populating Next Right Pointers in Each Node II The solution is completely the same as the one for Populating Next Right Pointers in Each Node.

Solution to Populating Next Right Pointers in Each Node by LeetCode

9 Jul

Question: https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/ Question Name: Populating Next Right Pointers in Each Node

# Definition for a binary tree node

# class TreeNode:

# def __init__(self, x):

# self.val = x

# self.left = None

# self.right = None

# self.next = None

class Solution:

# @param root, a tree node

# @return nothing

def connect(self, root):

fatherLayerCurrent = root

sonLayerHead = None

sonLayerCurrent = None

# Travel layer by layer as BFS travesal.

while fatherLayerCurrent != None:

# Travel inside a layer from left to right

while fatherLayerCurrent != None:

# Try to access the left son if there is

if fatherLayerCurrent.left != None:

if sonLayerHead == None:

sonLayerHead = fatherLayerCurrent.left

sonLayerCurrent = sonLayerHead

else:

sonLayerCurrent.next = fatherLayerCurrent.left

sonLayerCurrent = sonLayerCurrent.next

# Try to access the right son if there is

if fatherLayerCurrent.right != None:

if sonLayerHead == None:

sonLayerHead = fatherLayerCurrent.right

sonLayerCurrent = sonLayerHead

else:

sonLayerCurrent.next = fatherLayerCurrent.right

sonLayerCurrent = sonLayerCurrent.next

fatherLayerCurrent = fatherLayerCurrent.next

# Prepare to move down to the next layer.

fatherLayerCurrent = sonLayerHead

sonLayerHead =None

return

Solution to Pascal's Triangle II by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/pascals-triangle-ii/ Question Name: Pascal’s Triangle II

class Solution:

# @return a list of integers

def getRow(self, rowIndex):

''' Ref: http://en.m.wikipedia.org/wiki/Pascal%27s_triangle

Convert this question into:

how to compute the consecutive binomial coefficients.

'''

if rowIndex == 0: return [1]

if rowIndex == 1: return [1, 1]

result = [1]

# The final result is symmetrical. We could only compute the

# the left part, and then append the reversed array of left

# part to the end.

# COmpute the left part of the symmetrical result

nextDivisor = 1

nextMultiplier = rowIndex

for _ in xrange(rowIndex // 2):

nextVal = (result[-1] * nextMultiplier) / nextDivisor

result.append(nextVal)

nextDivisor += 1

nextMultiplier -= 1

# Append the reversed array of left part

if rowIndex % 2 == 1: result.extend(result[::-1])

else: result.extend(result[-2::-1])

return result

Solution to Pascal's Triangle by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/pascals-triangle/ Question Name: Pascal’s Triangle

class Solution:

# @return a list of lists of integers

def generate(self, numRows):

if numRows == 0: return []

result = [[1]]

while len(result) < numRows:

temp = [1] # Every row starts with 1

for index in xrange(len(result[-1])-1):

temp.append(result[-1][index] + result[-1][index+1])

temp.append(1) # Every row ends with 1

result.append(temp)

return result

Solution to Distinct Subsequences by LeetCode

7 Jul

Question: https://oj.leetcode.com/problems/distinct-subsequences/ Question Name: Distinct Subsequences This is a dynamic programming challenge. Firstly I solved it with recursion. But that solution did not satisfy the time requirement. These two posts give more details about the right solution: LeetCode: Distinct Subsequences（不同子序列的个数）(in … Read More »

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