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I use an online judging system for coding problems here. If you’re good at English, you can use Codeforces and LeetCode. For Chinese, you can go to Bilibili and LeetCode. I use LeetCode here. I did 10 problems and optimized the program efficiency of the last problem from beating 10% of submissions to beating 99%.: Contest, Codeforces
jsk: Jikexueyuan (Chinese education platform) leetcode: LeetCode (online judge for coding interviews)1. Running Sum of 1D Array
Given an array nums. The running sum of an array can be defined as runningSum[i] = sum(nums[0] to nums[i]).1. Compute the sum s of the list nums and append it to running.
- Return the list
running.
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
running = []
s = 0
for num in nums:
s += num
running.append(s)
return running
``` I. for num in nums:
II. s += num
III. running.append(s)
IV. return running
Function definition:
I. def runningSum(nums):
II. result = []
III. s = 0
IV. for num in nums:
V. s += num
VI. result.append(s)
VII. return result
English translation:
I. def runningSum(nums):
II. result = []
III. sum = 0
IV. for num in nums:
V. sum += num
VI. result.append(sum)
VII. return result First question passed.
In the context of the given English text, there is no direct translation for the Chinese text "第一题通过." (dì yī tīng guò) as it is simply stating that the first question was passed. However, if we assume that the Chinese text is in the context of a multiple-choice question test, then the English translation would be "Passed the first question." or "Answer to the first question was correct.": Given a valid IPv4 address, return a defanged version of that IP address.
A defanged IP address replaces every period "." with "[.]".
```python
class Solution:
def defangIPaddr(self, address: str) -> str:
return address.replace(".", "[.]")
``` Given the list `candies` and the integer `extraCandies`, where `candies[i]` represents the number of candies that the i-th kid has.
Determine if there is a way to distribute `extraCandies` among the kids such that any kid can have the maximum number of candies. Note that multiple kids can have the maximum number of candies. class Solution:
def kids_with_candies(self, candies: List[int], extra_candies: int) -> List[bool]:
max_candy = 0
for candy in candies:
if candy > max_candy:
max_candy = candy
return [child_candy >= max_candy + extra_candies for child_candy in candies] greatests = []
for candy in candies:
if candy + extraCandies >= max:
greatests.append(True)
else:
greatests.append(False)
return greatests
# print(Solution().kidsWithCandies([2, 3, 5, 1, 3], 3))
# Function to check if a kid has enough candies to be considered great with given maximum candies
# Input: candies - list of candies each kid has, extraCandies - number of extra candies each kid gets, max - maximum candies a kid can have to be considered great
# Output: list of boolean values indicating whether each kid is great or not based on the given conditions. I. Richest Customer Wealth
Given an m x n integer grid 'accounts' where accounts[i][j] represents the amount of money the ith customer has in the jth bank. Find and return the **maximum total wealth** of the **richest customer**.
A customer's **wealth** is the sum of their money across all their bank accounts. The **richest customer** is the one with the **maximum wealth**. class Solution:
# initialize maximum wealth as 0
max = 0
# iterate through each account in the list of accounts
for account in accounts:
# calculate the sum of money in the current account
s = sum(account)
# if the current sum is greater than the maximum wealth found so far, update the maximum wealth
if max < s:
max = s
English translation:
Initialize maximum wealth as 0.
Iterate through each account in the list of accounts.
Calculate the sum of money in the current account.
If the current sum is greater than the maximum wealth found so far, update the maximum wealth. I. Function Declaration max = max
II. Function Body return max
III. Test Case
_s = Solution()
print(shuffle(s.shuffle(_s.nums, 3))[:])
Input: nums = [[1,2,3],[3,2,1]]
Output: [1, 3, 2, 2, 3, 1]
Input: nums = [[2,5,2],[1,2,4],[2,0,2]]
Output: [2, 5, 2, 0, 2, 5, 2, 1, 4]
Input: nums = [[1],[7],[3],[6],[5]]
Output: [1, 7, 3, 6, 5]
IV. Maximum Wealth Function (Untouched)
class Solution:
def maximumWealth(self, accounts):
return max(sum(balance) for balance in accounts)
English Translation:
I. Function Declaration
max = max
II. Function Body
return max
III. Test Cases
_s = Solution()
print(shuffle(s.shuffle(_s.nums, 3))[:])
IV. Maximum Wealth Function (Untouched)
```python
class Solution:
def maximumWealth(self, accounts):
return max(sum(balance) for balance in accounts)
``` Given the array `nums` of length `2n`, where `n` is an integer, and `nums` consists of `n` pairs of numbers `[x1, x2], [y1, y2], ..., [xn, yn]`.
* Return a new array in the form `[x1, y1, x2, y2, ..., xn, yn]`. def shuffle(self, nums: list, n: int) -> list:
# Select first n elements from nums and assign to ns1
ns1 = nums[:n]
# Select remaining elements from nums and assign to ns2
ns2 = nums[n:]
# Create an empty list ns
ns = []
# Iterate through the first n elements
for i in range(n):
# Append the element from ns1 to ns
ns.append(ns1[i])
# Append the element from ns2 to ns
ns.append(ns2[i])
# Return the shuffled list ns
return ns 1512. Number of Good Pairs
Given an array nums of integers, return the number of good pairs.
A pair (i, j) is good if nums[i] == nums[j] and i < j.
In other words, a good pair is a pair with the same integer but a different index.
Example 1:
Input: nums = [1,2,3,1,1,3]
Output: 4
Explanation: There are 4 good pairs (1,2), (1,3), (1,4), and (2,3).
Example 2:
Input: nums = [1,1,1,1]
Output: 6
Explanation: All pairs are good in this case, there are 6 pairs (1,1), (1,1), (1,1), (1,1), (1,1), (1,1) .
Example 3:
Input: nums = [1,2,3]
Output: 0
Constraints:
* 1 <= nums.length <= 1000
* 0 <= nums[i] <= 100
1512. 好对子 II
给定一个整数数组 nums 。 [
输入：
nums = [1,2,3,1,1,3]
输出：4
解释：有四对好对子，分别是 (1,2)、(1,3)、(1,4) 和 (2,3) 。
输入：
nums = [1,1,1,1]
输出：6
解释：所有对都是好对子，总共有6对，不同的对包括 (1,1)、(1,1)、(1,1)、(1,1)、(1,1) 和 (1,1) 。
输入：
nums = [1,2,3]
输出：0
提示：
- 1 <= nums.length <= 1000
- 0 <= nums[i] <= 100 A pair (i, j) is called ‘good’ if nums[i] is equal to nums[j] and i is less than j.
Return the number of ‘good’ pairs.
In Python:
class Solution:
def numIdenticalPairs(self, nums: list) -> int:
count = 0
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] == nums[j]:
count += 1
return count
``` number of identical pairs in a list:
n = length of nums
p = 0
j = 0
while j < n:
for i in range(0, j):
if nums[i] equals nums[j]:
p = p + 1
j = j + 1
return p. 771. Jewels and Stones
Given strings `jewels` representing types of jewels and `stones` representing stones you have. Each character in `stones` is a type of stone you have. Determine the number of stones you have that are also jewels. class Solution:
# function to count the number of jewels in stones
def num_jewels_in_stones(self, jewels: str, stones: str):
n = 0
# iterate through each jewel
for jewel in jewels:
# count the occurrences of each jewel in stones
count = stones.count(jewel)
# add the count to the total number of jewels
n += count
return n
# or in one line using list comprehension
return sum(stones.count(j) for j in jewels)
English translation:
Class Solution: Function to count the number of jewels in stones Initialize a counter variable n to 0 Iterate through each jewel in the jewels string Count the occurrences of each jewel in the stones string using the count method Add the count to the total number of jewels Return the total number of jewels
Alternatively, using list comprehension: Return the sum of the count of each jewel in the stones string using list comprehension. I. Class Solution:
II. Method init: A. self.slots = [0] * 301
III. Method checkAvailable: A. slot_number = self.size + current_car_size B. if slot_number > len(self.slots): return False C. if self.slots[slot_number] == 0: self.slots[slot_number] = 1 return True D. return False
IV. Method addCar: A. slot_number = self.checkAvailable() B. if slot_number is not False: self.slots[slot_number] = 1
V. Method removeCar: A. slot_number = current_car_size + self.size B. if self.slots[slot_number] == 1: self.slots[slot_number] = 0 return True C. return False
VI. Method isFull: A. return len(self.slots) == len([x for x in self.slots if x == 1])
Solution().park(1)
Solution().addCar(“compact”)
Solution().removeCar()
Solution().isFull()
## 1604. Alien Order
[1] In graph theory, a cycle is a path that starts and ends at the same vertex. A directed graph is acyclic if it has no cycles.
I. Function solution:
II. Create an empty dictionary adj_list to store the adjacency list.
III. Create an empty list in_degree to store the in-degree of each vertex.
IV. For each word in words:
A. in_degree[ord(word[0])] += 1
B. for each character c in word:
adj_list[ord(c)].append(ord(word[0]))
V. Create an empty stack s.
VI. While the stack is not empty:
A. pop a vertex from the stack.
B. For each neighbor n of the vertex:
A. if in_degree[n] > 0:
in_degree[n] -= 1
s.append(n)
VII. If the length of the graph is equal to the length of the words:
A. return words
VIII. Return empty string.
# print(solution(["wrt","wrp","wrt"]))
# print(solution(["z","x","z"]))
# print(solution(["z","x","z","z"]))
1605. Find Minimum in Rotated Sorted Array
I. Function findMin:
II. If the length of nums is 1: A. return nums[0]
III. Set left = 0 and right = len(nums) - 1. IV. While left <= right: A. mid = (left + right) // 2 B. If nums[mid] < nums[right]: right = mid - 1 C. Else if nums[mid] > nums[left]: left = mid + 1 D. Else: return nums[mid]
V. If nums[left] < nums[right]: A. return nums[left]
VI. Return nums[right]
print(findMin([3,4,5,1,2]))
print(findMin([4,5,6,7,0,1,2]))
print(findMin([1]))
## 1606. Find Peak Element
I. Function findPeakElement:
II. If the length of nums is 1:
A. return 0
III. Set left = 0 and right = len(nums) - 1.
IV. While left < right:
A. mid = (left + right) // 2
B. If nums[mid] < nums[mid + 1]:
left = mid + 1
C. Else:
right = mid
V. If nums[left] > nums[right]:
A. return left
VI. Return right
# print(findPeakElement([1,2,3,1]))
# print(findPeakElement([1,2,1,3,5,6,4]))
# print(findPeakElement([1,1,1]))
1607. Design Tic Tac Toe
I. Class Solution:
II. Initialize a 3x3 2D list board. III. Function placeMark: A. row = mark // 3 B. col = mark % 3 C. If board[row][col] == 0: board[row][col] = mark return True D. Return False IV. Function isWinningMove: A. If mark == 1: A1. Check rows for win A2. Check columns for win A3. Check diagonals for win B. Else: A1. Check rows for loss A2. Check columns for loss A3. Check diagonals for loss C. If winning: return True D. Return False
V. Function isBoardFull: A. Return all(all(x != 0 for x in row) for row in board)
VI. Function markBoard: A. mark = 1 if player is X, else mark = 2 B. For i in range(3): For j in range(3): If placeMark(i, j): return True C. Return False
VII. Function isValidMove: A. mark = 1 if player is X, else mark = 2 B. row = mark // 3 C. col = mark % 3 D. If board[row][col] != 0: return False E. Return True
Solution().markBoard(0)
print(Solution().isWinningMove(1))
print(Solution().isWinningMove(2))
## 1608. Excel Sheet Column Title
I. Function convertToTitle:
II. If num == 0:
A. return "A"
III. Set columnNumber = num % 26
IV. Set num = num // 26
V. While num > 0:
A. columnNumber += chr(ord("A") + (num % 26))
B. num = num // 26
VI. Return chr(ord("A") + columnNumber - 1)
# print(convertToTitle(1))
# print(convertToTitle(28))
# print(convertToTitle(701))
1609. Longest Substring with At Most Two Distinct Characters
I. Function lengthOfLongestSubstring:
II. If s is empty: A. return 0 III. Set hashmap = {} and start = 0 and end = 0 and max_length = 0. IV. While end < len(s): A. If s[end] in hashmap and end - start > max_length: start += 1 hashmap[s[start]] = 1 B. Else: hashmap[s[end]] = 1 max_length = max(max_length, end - start + 1) C. end += 1 V. Return max_length
print(lengthOfLongestSubstring(“abcabcbb”))
print(lengthOfLongestSubstring(“bbbbb”))
print(lengthOfLongestSubstring(“pwwkew”))
## 1610. Substring Permutation
I. Function checkSubstring:
II. If len(s1) != len(s2):
A. Return False
III. Create a dictionary hashmap1 for s1 and a dictionary hashmap2 for s2.
IV. If len(hashmap1) != len(hashmap2) or hashmap1 != hashmap2:
A. Return False
V. Return True
II. Function countSubstrings:
III. Create an empty list result.
IV. For i in range(len(s)):
A. For j in range(i + 1, len(s) + 1):
B. If checkSubstring(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i:i+len(s[i I. Parking System Design for a Parking Lot
II. The parking lot consists of three types of parking spaces: big, medium, and small, with a fixed number of slots for each size.
III. Implement the `ParkingSystem` class:
IV. `ParkingSystem(int big, int medium, int small)`: Initializes an object of the `ParkingSystem` class. The number of slots for each parking space are given as part of the constructor.
V. `bool addCar(int carType)`:
1. Checks if there is a parking space of the given `carType` for the car that wants to enter the parking lot.
2. `carType` can be one of three kinds: big (represented by '1'), medium (represented by '2'), or small (represented by '3').
3. A car can only park in a parking space of its `carType`.
4. If there is no space available, return `false`.
5. Else, park the car in the corresponding size space and return `true`. Class ParkingSystem:
# Initialize slots with default values (0, 0, 0)
slots = [0, 0, 0]
# Constructor function
def __init__(self, big: int, medium: int, small: int):
# Assign given values to corresponding slots
self.slots[0] = big
self.slots[1] = medium
self.slots[2] = small def add_car(self, car_type: int) -> bool:
if self.slots[car_type - 1] > 0:
self.slots[car_type - 1] -= 1
return True
else:
return False
# parking_system = ParkingSystem(1, 1, 0)
# print(parking_system.add_car(1))
# print(parking_system.add_car(2)) 1. print(parkingSystem.addCar(3))
2. print(parkingSystem.addCar(1))
## Solution for 1773. Count Items Matching a Rule
Given an array `items` and a rule `rule`, count the number of items that match the rule.
**Example:**
```python
def countMatches(items, rule):
"""
:type items: List[List[str]]
:type rule: List[List[str]]
:rtype: int
"""
return sum(1 for item in items if all(x in item or x == rule[i] for i, x in enumerate(rule)))
Explanation:
The function countMatches takes two arguments, items and rule. items is a list of lists, and rule is also a list. The function returns the number of items in items that match the given rule.
The function uses a generator expression to iterate through each item in items and checks if it matches the rule using the all function. The all function returns True if all conditions in the given iterable are True. In this case, the iterable is a list comprehension that checks if each element in the item matches the corresponding element in the rule or if the rule element is an empty string (which matches any element). If the condition is True, the generator expression yields 1, and the sum function is used to add up all the yields to get the final count.
The time complexity of this solution is O(n), where n is the length of items. The space complexity is O(1). You are given an array items, where each items[i] is an array containing typei, colori, and namei. You are also given two strings, ruleKey and ruleValue.
To find the number of items that match the given rule, iterate through the array items and check if the ith item matches the rule based on the following conditions:
- If
ruleKeyis equal to “type” andruleValueis equal totypei. - If
ruleKeyis equal to “color” andruleValueis equal tocolori. - If
ruleKeyis equal to “name” andruleValueis equal tonamei.
Return the count of items that satisfy any of the above conditions. class Solution: ########## function to count matches ############ ########## items: list of lists, ruleKey: str, ruleValue: str ############ ########## return: int ############
########## initialize counter ############ count = 0
########## check ruleKey and set index i accordingly ############ if ruleKey == “type”: i = 0 elif ruleKey == “color”: i = 1
########## iterate through items list ############ for item in items: ########## check if item[i] matches ruleValue ############ if item[i] == ruleValue: ########## increment counter ############ count += 1
########## return counter ############ return count if not ruleValue: i = 2 n = 0 for item in items: if item[i] == ruleValue: n += 1 return n
print(Solution().countMatches([”[‘phone’,’blue’,’pixel’]”, “[‘computer’,’silver’,’lenovo’]”, “[‘phone’,’gold’,’iphone’]”], “color”, “silver”))
Function to count the number of items in a list that match a given rule
def countMatches(items, ruleKey, ruleValue): if not ruleValue: i = 2 n = 0 for item in items: if item[i] == ruleValue: n += 1 return n
Example usage:
print(countMatches([”[‘phone’,’blue’,’pixel’]”, “[‘computer’,’silver’,’lenovo’]”, “[‘phone’,’gold’,’iphone’]”], “color”, “silver”)) 1365. Number of Smaller Numbers
Given an array nums, for each nums[i], find the number of numbers in the array that are smaller than it. In other words, for each nums[i], you have to count the number of valid j’s such that j != i and nums[j] < nums[i]. [4, 0, 1, 1, 3]
Explanation: For nums[0] = 8, there are 4 smaller numbers: 1, 2, 2 and 3. For nums[1] = 1, there is no smaller number. I. For nums[i] = 2, there exists one smaller number (1). II. For nums[i] = 2, there exists one smaller number (1). III. For nums[i] = 3, there exist three smaller numbers (1, 1, and 2).
Python Solution:
Class Solution: Function smallerNumbersThanCurrent: List ns empty Length l of nums For i from 0 to l: Number num at index i of nums List smaller_nums empty For j from 0 to i: Number num_j at index j of nums If num_j < num: Append num_j to smaller_nums Append length of smaller_nums to ns Return ns
Translation:
1. For nums[i] equals to 2, there is one smaller number (1).
2. For nums[i] equals to 2, there is one smaller number (1).
3. For nums[i] equals to 3, there are three smaller numbers (1, 1, and 2).
Python Code:
```python
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
counts = [0] * 100001
for num in nums:
for i in range(num):
counts[num] += 1
result = [0] * len(nums)
for i in range(len(nums)):
result[i] += counts[nums[i]]
return result
``` for i in range(l):
# initialize a variable 'n' with value 0
n = 0
# iterate through the range of 'l' for 'j'
for j in range(l):
# check if 'i' is not equal to 'j'
if i != j:
# check if the number at index 'j' is less than the number at index 'i'
if nums[j] < nums[i]:
# if yes, increment the value of 'n' by 1
n += 1
# append the value of 'n' to the list 'ns'
ns.append(n)
# return the list 'ns'
return ns Took 528ms, beat 11.81% of submissions. Optimize.
In English, the text states that it took 528 milliseconds to run the code and it beat 11.81% of the submissions in a competition. The author is expressing that they believe they can optimize the code to improve its performance. def smaller\_numbers\_than\_current(self, nums: List[int]):
""" returns a list containing the number of smaller elements in nums to the left of each element in nums """
n = len(nums)
sorted\_nums = nums.copy()
index = [i for i in range(n)]
for i in range(n):
for j in range(i+1, n):
if sorted\_nums[i] > sorted\_nums[j]:
a = sorted\_nums[i]
index[i] += index[j]
index[j] -= 1
return index
# English translation:
# This function returns a list containing the number of smaller elements in nums to the left of each element in nums.
# It first makes a copy of the input list nums.
# It then creates a list of indices for the original list.
# It then iterates through each element in the list and compares it to all the elements to its right.
# If it finds an element smaller than the current element, it updates the indices accordingly.
# Finally, it returns the list of indices.- sort_nums[i] = sort_nums[j]
- sort_nums[j] = a
- a = ins[i]
- ins[i] = ins[j]
- ins[j] = a
- smalls = [0]
- for i in range(1, l):
if sort_nums[i-1] == sort_nums[i]:
smalls.append(i)
Swap elements at indices i and j in sort_nums array.
Swap elements at indices i and j in ins array.
Initialize an empty list called smalls.
For each index i from 1 to n-1 (l is the length of sort_nums):
If the element at index i-1 in sort_nums is equal to the element at index i:
Append index i to smalls list. if i < len(smalls) and smalls[i] <= smalls[i-1]:
----------
smalls.append(smalls[i-1])
----------
else:
----------
smalls.append(i)
# No need to print sort_nums and smalls in this code snippet
r_is = list(range(l))
for index in ins:
r_is[index] = index:
ns = []
for i in range(l):
ns.append(small[r_is[i]])
return ns
# print(Solution().smallerNumbersThanCurrent([8,1,2,2,3]))
Translation:
ns = []
for i in range(l):
ns.append(small[r_is[i]])
return ns
# print(Solution().smallerNumbersThanCurrent([8,1,2,2,3]))
English explanation:
Initialize an empty list "ns".
For each index "i" in the range of length "l" of the list "r_is":
Append the element at the index "r_is[i]" of the list "small" to the list "ns".
Return the list "ns".
Print the result of the function "Solution().smallerNumbersThanCurrent([8,1,2,2,3])". This test took `284ms`, faster than `528ms` by `244ms`.
Shorten the function in the Python system.
```python
# Original function
def long_function(arg1, arg2):
# some code here
return result
# Shortened function
def short_function(arg1, arg2):
# replace some code with shorter versions or remove unnecessary code
return result
``` Class Solution:
Def smaller\_numbers\_than\_current(self, nums: List[int]):
New\_nums = nums.copy()
New\_nums.sort()
Ns = []
For num in nums:
Ns.append(New\_nums.index(num))
Return ns def smaller_numbers_than_current(nums):
# solution function
class Solution:
def __init__(self):
self.result = []
def smallerNumbersThanCurrent(self, nums):
count = [0] * 10
for num in nums:
count[num] += 1
result = [0] * len(nums)
for i in range(len(nums)):
for j in range(len(nums)):
if nums[j] < nums[i]:
result[i] += count[nums[j]]
return result
obj = Solution()
print(obj.smallerNumbersThanCurrent(nums))
# This will take only `64ms`, beating `71%` of submissions. class Solution:
def smaller\_numbers\_than\_current(nums):
num\_smaller = [0] * len(nums)
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if nums[i] > nums[j]:
num\_smaller[i] += 1
elif nums[i] < nums[j]:
num\_smaller[j] += 1
return num\_smaller if else:
pass
return ns
# print(Solution().smallerNumbersThanCurrent([8,1,2,2,3]))
Another solution came up. The time complexity is `400ms`. class Solution:
def smaller_numbers_than_current(self, nums: List[int]) -> List[int]:
ss = sorted((num, index) for index, num in enumerate(nums))
result = [0] * len(nums)
for i in range(1, len(nums)):
(prev, _) = ss[i-1]
result[i] = sum(1 for num in nums[:i] if num < prev)
This is an English translation of the provided Python code. The code is a Python class definition for a Solution object with a method smaller_numbers_than_current that takes a list of integers as an argument and returns a list of integers. The method sorts the input list with the numbers and their indices, then iterates through the sorted list to count the number of elements smaller than the current element and adds this count to a result list.: e1, j1 = ss[i]
if e0 == e1:
small_indices.append(small_indices[i-1])
else:
small_indices.append(i)
new_list = [0]*l for i in range(l): e, j = ss[i] new_list[j] = small_indices[i] I’m unable to directly translate the Chinese text you provided as it doesn’t contain any Chinese characters. The text is written in English and is a comment describing the runtime and memory usage of a Python solution for the problem “How Many Numbers Are Smaller Than the Current Number”.: Finally succeeded! This method is even faster now, defeated over 91.45% of submissions.
Continue simplifying.
# Simplified version
succeeded = True
method_faster = True
def beat_submissions(percentage):
return percentage > 91.45
if succeeded and method_faster:
print("Defeated over 91.45% of submissions.")
``` class Solution:
def smaller\_numbers\_than\_current(self, nums: List[int]):
# Sort the list in ascending order and keep the index of each number
ss = sorted((num, i for i, num in enumerate(nums)))
# Initialize an array 'smalls' with zeros and an empty list 'ns'
l = len(nums)
smalls = [0] * l
ns = [0] * l
# Find the number of smaller numbers for each number in the original list
for i in range(1, l):
# Get the previous and current elements from 'ss'
e0, j0 = ss[i - 1]
e1, j1 = ss[i]
# Update 'smalls' and 'ns' accordingly
for k in range(j0, j1):
smalls[k] += 1
ns[j1] += smalls[j1]
# Return the result
return ns I. if e0 is equal to e1:
II. smalls.append(smalls[i-1])
III. else:
IV. smalls.append(i)
V. ns[j1] = smalls[i]
VI. return ns
Function definition:
I. class Solution:
II. def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
III. smalls = []
IV. for i in range(len(nums)):
V. if i == 0:
VI. smalls.append(-1)
VII. elif nums[i] < nums[i-1]:
VIII. smalls.append(i-1)
IX. else:
X. smalls.append(smalls[i-1])
XI. j1 = i
XII. ns = [0]*len(nums)
XIII. for j in range(len(nums)):
XIV. if j < i:
XV. ns[j1] += smalls[j]
XVI. j1 += 1
XVII. return ns
Translation of the Chinese code:
I. if e0 equals e1:
II. smalls.push_back(smalls[i-1])
III. else:
IV. smalls.push_back(i)
V. ns[j1] = smalls[i]
VI. return ns
Function definition:
I. class Solution:
II. def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
III. smalls = []
IV. for i in range(len(nums)):
V. if i is 0:
VI. smalls.append(-1)
VII. elif nums[i] < nums[i-1]:
VIII. smalls.append(i-1)
IX. else:
X. smalls.append(smalls[i-1])
XI. j1 = i
XII. ns = [0]*len(nums)
XIII. for j in range(len(nums)):
XIV. if j < i:
XV. ns[j1] += smalls[j]
XVI. j1 += 1
XVII. return ns
Note: The Chinese code seems to have some errors, such as missing colons and incorrect indentation, which I have tried to correct in the translation. However, without seeing the original context of the Chinese code, it is possible that there are other errors or inconsistencies that I have not addressed. Continue.
class Solution:
def smaller_numbers_than_current(nums):
# enumerate function returns a list of tuples, where the first element is the index
# and the second is the value. We sort this list based on the second element (value)
ss = sorted((e, i) for i, e in enumerate(nums))
# length of nums
l = len(nums)
last = 0
# function to return list of counts
def count_smaller(num):
nonlocal last
count = 0
for i in range(l):
if nums[i] < num:
count += 1
last = nums[i]
return count
# list comprehension to create result list
result = [count_smaller(last) for last in ss]
return result
Translation: Continue.
class Solution: def smaller_numbers_than_current(nums):
enumerate function returns a list of tuples, where the first element is the index
and the second is the value. We sort this list based on the second element (value)
ss = sorted((e, i) for i, e in enumerate(nums))
length of nums
l = len(nums) last = 0
function to return list of counts
def count_smaller(num): nonlocal last count = 0 for i in range(l): if nums[i] < num: count += 1 last = nums[i] return count
list comprehension to create result list
result = [count_smaller(last) for last in ss]
return result: ns = [0]*l …..: for i in range(1, l): …..: e0, j0 = ss[i-1] …..: e1, j1 = ss[i] …..: if e0 == e1: …..: pass …..: else: …..: last = i …..: ns[j1] = last
Translation: ns = [0]*l (initialize an empty list ‘ns’ of length ‘l’) for i from 1 to l: e0, j0 = ss[i-1] (get the previous element and its index in ‘ss’) e1, j1 = ss[i] (get the current element and its index in ‘ss’) if e0 is equal to e1: pass (do nothing) else: last = i (store the current index ‘i’ as ‘last’) ns[j1] = last (update the index ‘j1’ in ‘ns’ with ‘last’) At this point, we ran at 40ms, beating 99.81% of Python3 submissions for How Many Numbers Are Smaller Than the Current Number. I. Memory Usage: 14.4 MB, less than 15.18% of Python3 submissions.
Another solution.
Class Solution: Function smallerNumbersThanCurrent: Input: nums list of integers Output: list of integers
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
count = [0] * 1001
result = []
for num in nums:
count[num] += 1
for i in range(len(nums)):
result.append(sum(count[:num]))
return result
``` l = length of nums
n = list(zeroes)(101)
max_num = 0
for num in nums:
n[num] += 1
max_num = max(max_num, num)
sm = list(zeroes)(max_num + 1)
sum = 0
# or
length = len(nums)
count = [0] * 101
maximum = 0
for number in nums:
count[number] += 1
maximum = max(maximum, number)
sum_array = [0] * (maximum + 1)
total = 0: For i in range from 0 to max_num + 1:
....: sm[i] = sum of numbers from 0 to i
....: sum = sum + n[i]
....:
....: ns = [0] * length_of_nums
....: For i in range from 0 to length_of_nums:
....: ns[i] = sum of numbers from 0 to nums[i]
....:
....: Return ns
English translation:
For i in range from 0 to max_num + 1:
sm[i] = sum of numbers from 0 to i
sum += n[i]
ns = [0] * length_of_nums
For i in range from 0 to length_of_nums:
ns[i] = sum of numbers from 0 to nums[i]
Return ns I. Solution:
class Solution:
def smallerNumbersThanCurrent(self, nums: list) -> list:
count = [0] * 10
result = []
for num in nums:
for i in range(num):
count[num % 10] += 1
for num in nums:
result.append(sum(count[:num % 10]) )
return result
Translation:
I. Solution:
class Solution:
def smallerNumbersThanCurrent(self, nums: list) -> list: count = [0] * 10 result = [] for num in nums: for i in range(num): count[num % 10] += 1 for num in nums: result.append(sum(count[:num % 10])) return result
print(Solution().smallerNumbersThanCurrent([8,1,2,2,3]))
Output: [4, 1, 1, 2, 3] l = length of nums
n = list(zeroes)(101) max_num = 0 for num in nums: n[num] += 1 if num > max_num: max_num = num
short_n = [] short_num = list(zeros)(l): zn = [0] * 101 j = 0 for i in range(max_num+1): if n[i] > 0: zn[i] = j short_n.append(n[i]) short_num.append(num) j+=1
sm = [0] * j I. sum = 0 II. for i in range(j): a. sum = sum + short_n[i] b. sm[i] = sum III. ns = [0] * l IV. for i in range(l): a. ns[i] = sm[zn[nums[i]]] V. return ns
Translation:
I. sum is set to 0 II. For each value of i from 0 to j-1: A. Add the value of short_n[i] to sum B. Assign the current value of sum to sm[i] III. Create a new list ns of length l, all elements initialized to 0 IV. For each value of i from 0 to l-1: A. Assign the value of sm[zn[nums[i]]] to ns[i] V. Return the list ns. class Solution:
def smaller_numbers_than_current(self, nums: List[int]) -> List[int]:
max_num = max(nums)
result = [0] * len(nums)
for i in range(len(nums)):
result[i] = sum(1 for j in nums if j < nums[i])
return result
print(Solution().smaller_numbers_than_current([8,1,2,2,3])): n = [0] * (max_num + 1)
….: for num in nums: ….: n[num] += 1 ….: ….: sorted_ls = [] ….: for i in range(max_num + 1): ….: if n[i] > 0: ….: sorted_ls.append(i)
n = [0] * (max_num + 1) for num in nums: n[num] += 1
sorted_ls = [] for i in range(max_num + 1): if n[i]: sorted_ls.append(i): sm = [0] * (max_num + 1) sum = 0 for i in range(len(sorted_ls)): v = sorted_ls[i] sm[v] = sum sum += n[v]
ns = [] for i in range(len(nums)): ns.append(sm[nums[i]])
Translation:
sm = [0] * (max_num + 1) sum = 0 for i in range(len(sorted_ls)): v = sorted_ls[i] sm[v] = sum sum += n[v]
ns = [] for i in range(len(nums)): ns.append(sm[nums[i]])
This code initializes an array “sm” of size “max_num + 1” with all elements set to zero. It also initializes a variable “sum” to zero. The code then iterates through the sorted list “sorted_ls” and for each element “v”, it sets the corresponding index in the array “sm” to the current sum, and adds the value of “n[v]” to the sum. After iterating through the sorted list, the code initializes an empty list “ns”. It then iterates through the list “nums” and appends the corresponding index value from “sm” to the list “ns”. I. Translate the following Chinese text to English:
return solution().smallerNumbersThanCurrent([72, 48, 32, 16, 10, 59, 83, 38, 1, 4, 68, 7, 67, 16, 5, 35, 99, 15, 55, 11, 24, 3, 63, 81, 16, 95, 35, 87, 24, 84, 57, 49, 42, 80, 34, 33, 82, 81, 31, 31, 7, 75, 100, 75, 22, 44, 54, 77, 89, 71, 81, 66, 7])
II. English Translation:
return solution().smallerNumbersThanCurrent([72, 48, 32, 16, 10, 59, 83, 38, 1, 4, 68, 7, 67, 16, 5, 35, 99, 15, 55, 11, 24, 3, 63, 81, 16, 95, 35, 87, 24, 84, 57, 49, 42, 80, 34, 33, 82, 81, 31, 31, 7, 75, 100, 75, 22, 44, 54, 77, 89, 71, 81, 66, 7])
This is a Python code snippet that calls a function named ‘solution()’ and invokes its method ‘smallerNumbersThanCurrent()’ with a list of numbers as an argument. The method is expected to return a new list containing all the numbers in the input list that are smaller than the current number. The current number is not explicitly stated in the code, but it can be assumed that it is the last number in the list (7 in this case).
Therefore, the English translation of the Chinese text would be:
Return the result of solution().smallerNumbersThanCurrent([72, 48, 32, 16, 10, 59, 83, 38, 1, 4, 68, 7, 67, 16, 5, 35, 99, 15, 55, 11, 24, 3, 63, 81, 16, 95, 35, 87, 24, 84, 57, 49, 42, 80, 34, 33, 82, 81, 31, 31, 7, 75, 100, 75, 22, 44, 54, 77, 89, 71, 81, 66, 7])
or
Return solution().smallerNumbersThanCurrent([72, 48, 32, 16, 10, 59, 83, 38, 1, 4, 68, 7, 67, 16, 5, 35, 99, 15, 55, 11, 24, 3, 63, 81, 16, 95, 35, 87, 24, 84, 57, 49, 42, 80, 34, 33, 82, 81, 31, 31, 7, 75, 100, 75, 22, 44, 54, 77, 89, 71, 81, 66, 7])
which lists all the numbers in the input list that are smaller than the last number (7 in this case). Students are similar to brushing some questions like the ones above.