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Added new problems and subjects
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24 files changed

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‎3sum/solution.py‎

Lines changed: 30 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,21 +1,40 @@
1-
class Solution:
2-
# @return a list of lists of length 3, [[val1,val2,val3]]
3-
def threeSum(self, num):
4-
num.sort()
1+
# -*- coding: utf-8 -*-
2+
"""
3+
Given an array S of n integers, are there elements a, b, c in S such that a +
4+
b + c = 0? Find all unique triplets in the array which gives the sum of zero.
5+
6+
Note:
7+
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
8+
The solution set must not contain duplicate triplets.
9+
For example, given array S = {-1 0 1 2 -1 -4},
10+
11+
A solution set is:
12+
(-1, 0, 1)
13+
(-1, -1, 2)
14+
"""
15+
16+
17+
class Solution(object):
18+
def threeSum(self, nums):
19+
"""
20+
:type nums: List[int]
21+
:rtype: List[List[int]]
22+
"""
23+
nums.sort()
524
res = []
6-
for i in range(len(num) - 2):
7-
if i == 0 or i > 0 and num[i - 1] != num[i]:
25+
for i in range(len(nums) - 2):
26+
if i == 0 or i > 0 and nums[i - 1] != nums[i]:
827
left = i + 1
9-
right = len(num) - 1
28+
right = len(nums) - 1
1029
while left < right:
11-
s = num[i] + num[left] + num[right]
30+
s = nums[i] + nums[left] + nums[right]
1231
if s == 0:
13-
res.append([num[i], num[left], num[right]])
32+
res.append([nums[i], nums[left], nums[right]])
1433
left += 1
1534
right -= 1
16-
while left < right and num[left] == num[left - 1]:
35+
while left < right and nums[left] == nums[left - 1]:
1736
left += 1
18-
while right > left and num[right] == num[right + 1]:
37+
while right > left and nums[right] == nums[right + 1]:
1938
right -= 1
2039
elif s < 0:
2140
left += 1

‎3sum_closest/solution.py‎

Lines changed: 22 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,29 @@
1-
class Solution:
2-
# @return an integer
3-
def threeSumClosest(self, num, target):
4-
num.sort()
5-
n = len(num)
6-
res = num[0] + num[1] + num[2]
1+
"""
2+
Given an array S of n integers, find three integers in S such that the sum is
3+
closest to a given number, target. Return the sum of the three integers. You
4+
may assume that each input would have exactly one solution.
5+
6+
For example, given array S = {-1 2 1 -4}, and target = 1.
7+
8+
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
9+
"""
10+
11+
12+
class Solution(object):
13+
def threeSumClosest(self, nums, target):
14+
"""
15+
:type nums: List[int]
16+
:type target: int
17+
:rtype: int
18+
"""
19+
nums.sort()
20+
n = len(nums)
21+
res = nums[0] + nums[1] + nums[2]
722
for i in range(n - 2):
823
l = i + 1
924
r = n - 1
1025
while l < r:
11-
s = num[i] + num[l] + num[r]
26+
s = nums[i] + nums[l] + nums[r]
1227
if abs(s - target) < abs(res - target):
1328
res = s
1429
if s == target:

‎4sum/solution.py‎

Lines changed: 34 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,27 +1,50 @@
1+
# -*- coding: utf-8 -*-
12
# Time Limit Exceeded
2-
class Solution:
3-
# @return a list of lists of length 4, [[val1,val2,val3,val4]]
4-
def fourSum(self, num, target):
5-
n = len(num)
6-
num.sort()
3+
"""
4+
Given an array S of n integers, are there elements a, b, c, and d in S such
5+
that a + b + c + d = target? Find all unique quadruplets in the array which
6+
gives the sum of target.
7+
8+
Note:
9+
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b
10+
≤ c ≤ d)
11+
The solution set must not contain duplicate quadruplets.
12+
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
13+
14+
A solution set is:
15+
(-1, 0, 0, 1)
16+
(-2, -1, 1, 2)
17+
(-2, 0, 0, 2)
18+
"""
19+
20+
21+
class Solution(object):
22+
def fourSum(self, nums, target):
23+
"""
24+
:type nums: List[int]
25+
:type target: int
26+
:rtype: List[List[int]]
27+
"""
28+
n = len(nums)
29+
nums.sort()
730
res = []
831
for a in range(n - 3):
9-
if a > 0 and num[a - 1] == num[a]:
32+
if a > 0 and nums[a - 1] == nums[a]:
1033
continue
1134
for b in range(a + 1, n - 2):
12-
if b > a + 1 and num[b - 1] == num[b]:
35+
if b > a + 1 and nums[b - 1] == nums[b]:
1336
continue
1437
c = b + 1
1538
d = n - 1
1639
while c < d:
17-
s = num[a] + num[b] + num[c] + num[d]
40+
s = nums[a] + nums[b] + nums[c] + nums[d]
1841
if s == target:
19-
res.append([num[a], num[b], num[c], num[d]])
42+
res.append([nums[a], nums[b], nums[c], nums[d]])
2043
c += 1
2144
d -= 1
22-
while c < d and num[c] == num[c - 1]:
45+
while c < d and nums[c] == nums[c - 1]:
2346
c += 1
24-
while c < d and num[d] == num[d + 1]:
47+
while c < d and nums[d] == nums[d + 1]:
2548
d -= 1
2649
elif s < target:
2750
c += 1

‎add_binary/solution.py‎

Lines changed: 15 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,19 @@
1-
class Solution:
2-
# @param a, a string
3-
# @param b, a string
4-
# @return a string
1+
"""
2+
Given two binary strings, return their sum (also a binary string).
3+
4+
For example,
5+
a = "11"
6+
b = "1"
7+
Return "100".
8+
"""
9+
10+
class Solution(object):
511
def addBinary(self, a, b):
12+
"""
13+
:type a: str
14+
:type b: str
15+
:rtype: str
16+
"""
617
res = []
718
# a is longer than b
819
if len(a) < len(b):

‎add_binary/solution2.py‎

Lines changed: 15 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,19 @@
1-
class Solution:
2-
# @param a, a string
3-
# @param b, a string
4-
# @return a string
1+
"""
2+
Given two binary strings, return their sum (also a binary string).
3+
4+
For example,
5+
a = "11"
6+
b = "1"
7+
Return "100".
8+
"""
9+
10+
class Solution(object):
511
def addBinary(self, a, b):
12+
"""
13+
:type a: str
14+
:type b: str
15+
:rtype: str
16+
"""
617
list_a = [int(i) for i in a[::-1]]
718
list_b = [int(i) for i in b[::-1]]
819
la = len(list_a)

‎add_binary/solution3.py‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -20,7 +20,7 @@ def addBinary(self, a, b):
2020
n = len(b)
2121
i = 0
2222
c = 0
23-
res = ['0' for k in range(max(m, n) + 1)]
23+
res = ['0' for _ in range(max(m, n) + 1)]
2424
while i < m or i < n or c > 0:
2525
tmp = c
2626
if i < m:

‎add_two_numbers/solution2.py‎

Lines changed: 16 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,25 @@
1+
"""
2+
You are given two linked lists representing two non-negative numbers. The
3+
digits are stored in reverse order and each of their nodes contain a single
4+
digit. Add the two numbers and return it as a linked list.
5+
6+
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
7+
Output: 7 -> 0 -> 8
8+
"""
9+
110
# Definition for singly-linked list.
2-
# class ListNode:
11+
# class ListNode(object):
312
# def __init__(self, x):
413
# self.val = x
514
# self.next = None
615

7-
class Solution:
8-
# @return a ListNode
16+
class Solution(object):
917
def addTwoNumbers(self, l1, l2):
18+
"""
19+
:type l1: ListNode
20+
:type l2: ListNode
21+
:rtype: ListNode
22+
"""
1023
carry = 0
1124
res = None
1225
res_end = None

‎balanced_binary_tree/solution.py‎

Lines changed: 7 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -5,17 +5,19 @@
55
which the depth of the two subtrees of every node never differ by more than 1.
66
"""
77

8-
# Definition for a binary tree node
9-
# class TreeNode:
8+
# Definition for a binary tree node.
9+
# class TreeNode(object):
1010
# def __init__(self, x):
1111
# self.val = x
1212
# self.left = None
1313
# self.right = None
1414

15-
class Solution:
16-
# @param root, a tree node
17-
# @return a boolean
15+
class Solution(object):
1816
def isBalanced(self, root):
17+
"""
18+
:type root: TreeNode
19+
:rtype: bool
20+
"""
1921
if root is None:
2022
return True
2123
else:

‎balanced_binary_tree/solution2.py‎

Lines changed: 7 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -5,17 +5,19 @@
55
which the depth of the two subtrees of every node never differ by more than 1.
66
"""
77

8-
# Definition for a binary tree node
9-
# class TreeNode:
8+
# Definition for a binary tree node.
9+
# class TreeNode(object):
1010
# def __init__(self, x):
1111
# self.val = x
1212
# self.left = None
1313
# self.right = None
1414

15-
class Solution:
16-
# @param root, a tree node
17-
# @return a boolean
15+
class Solution(object):
1816
def isBalanced(self, root):
17+
"""
18+
:type root: TreeNode
19+
:rtype: bool
20+
"""
1921
if root is None:
2022
return True
2123
else:

‎best_time_to_buy_and_sell_stock/solution.py‎

Lines changed: 16 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,24 @@
1-
class Solution:
2-
# @param prices, a list of integer
3-
# @return an integer
1+
"""
2+
Say you have an array for which the ith element is the price of a given stock
3+
on day i.
4+
5+
If you were only permitted to complete at most one transaction (ie, buy one
6+
and sell one share of the stock), design an algorithm to find the maximum
7+
profit.
8+
"""
9+
10+
11+
class Solution(object):
412
def maxProfit(self, prices):
13+
"""
14+
:type prices: List[int]
15+
:rtype: int
16+
"""
517
if not prices:
618
return 0
719
max_profit = 0
820
min_price = prices[0]
9-
for i, p in enumerate(prices, start=0):
21+
for i, p in enumerate(prices):
1022
max_profit = max(max_profit, (p - min_price))
1123
min_price = min(min_price, p)
1224
return max_profit

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