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class Solution:
def separateDigits(self, nums: List[int]) -> List[int]:
return list(map(int,list(''.join(map(str,nums)))))
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The LeetCode problem "2553. Separate the Digits in an Array" requires transforming an array of positive integers into a new array containing each digit of those integers in the order they appear.
Problem Description:
Given an array of positive integers
nums, return an array answer that consists of the digits of each integer in nums after separating them in the same order they appear in nums.
*Example 1:*
Input: nums = [13, 25, 83, 77] Output: [1, 3, 2, 5, 8, 3, 7, 7] Explanation: - The separation of 13 is [1, 3]. - The separation of 25 is [2, 5]. - The separation of 83 is [8, 3]. - The separation of 77 is [7, 7]. answer = [1, 3, 2, 5, 8, 3, 7, 7].*Example 2:*
Input: nums = [7, 1, 3, 9] Output: [7, 1, 3, 9] Explanation: The separation of each integer in nums is itself. answer = [7, 1, 3, 9].Approach: To solve this problem, iterate through each number in the input array, extract its digits, and append them to the result array in the correct order. This can be achieved by converting each number to a string, splitting it into individual characters, and then converting those characters back to integers. Python Solution:
class Solution:
def separateDigits(self, nums: List[int]) -> List[int]:
result = []
for num in nums:
digits = list(str(num)) # Convert number to string and split into characters
result.extend(int(digit) for digit in digits) # Convert characters back to integers and add to result
return result
Explanation:
1. Initialize an empty list result to store the separated digits.
2. Iterate over each number num in the input list nums.
3. Convert the number to a string and split it into individual characters using list(str(num)).
4. Convert each character back to an integer and extend the result list with these integers.
5. After processing all numbers, return the result list containing all the separated digits.
This approach ensures that the digits are added to the result list in the same order they appear in each number and maintains the order of numbers as they appear in the input list.
For more information and alternative solutions, you can refer to the [LeetCode problem discussion](https://leetcode.com/problems/separate-the-digits-in-an-array/description/).1 262
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Repost from Codeforces Official
Educational Codeforces Round 174
(rated for Div. 2) starts on the 18th of February at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2069
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How to Set Up and Use Competitive Programming Tools for Codeforces
Many of you have asked how to use Codeforces efficiently for running and submitting code. It can be difficult to debug submissions when errors occur on random test cases without showing the input and output. Below is the method I use to make the process easier.
### Step-by-Step Guide
1. Install Required Extensions
- Open Google Chrome and install the Competitive Companion extension.
- Install CPH Submit from the Chrome Web Store.
2. Set Up VS Code
- Open VS Code and install the Competitive Programming Helper extension.
3. Organizing Your Workspace
- Open VS Code and create a new folder, e.g.,
Competitive Programming.
- You only need to create this folder once; after that, you can use it for all your problems without needing to create new folders repeatedly.
4. Using Competitive Companion
- Go to a Codeforces problem.
- Click the Competitive Companion extension in your browser.
- This will automatically send the problem statement and test cases to your VS Code environment.
5. Running and Submitting Code
- Write your solution in VS Code.
- Use the Competitive Programming Helper to run test cases locally.
- Submit the solution using the CPH Submit extension.
This setup makes running and debugging Codeforces problems much easier, especially when dealing with hidden test cases. I hope this helps!1 262
t = int(input())
while t>0:
t-=1
lenght = int(input())
arr1 = list(map(int,input().split()))
arr2 = list(map(int,input().split()))
hash1 = set()
hash2 = set()
for i in range(len(arr1)):
hash1.add(arr1[i])
hash2.add(arr2[i])
if (len(hash2)<=1 and len(hash1) <=2 )or (len(hash1)<=1 and len(hash2) <=2 ):
print("NO")
else:
print("YES")1 262
▎Problem A: Milya and Two Arrays
Contest: Codeforces Round #2059
Time Limit: 1 second per test
Memory Limit: 256 megabytes
▎Problem Description
An array is called good if every element x that appears in the array appears at least twice. For example, the arrays:
• [1, 2, 1, 1, 2]
• [3, 3]
• [1, 2, 4, 1, 2, 4]
are good, while the arrays:
• [1]
• [1, 2, 1]
• [2, 3, 4, 4]
are not good.
Milya has two good arrays a and b of length n . She can rearrange the elements in array a in any way she wants. After that, she obtains an array c of length n , where:
cᵢ = aᵢ + bᵢ (1 ≤ i ≤ n)Your task is to determine whether Milya can rearrange the elements in array a such that there are at least 3 distinct elements in array c . ▎Input Format • The first line contains an integer t ( 1 ≤ t ≤ 1000 ) — the number of test cases. • For each test case: • The first line contains an integer n ( 3 ≤ n ≤ 50 ) — the length of arrays a and b . • The second line contains n integers a₁, a₂, ..., aₙ ( 1 ≤ aᵢ ≤ 10⁹ ) — the elements of array a . • The third line contains n integers b₁, b₂, ..., bₙ ( 1 ≤ bᵢ ≤ 10⁹ ) — the elements of array b . ▎Output Format For each test case, output "YES" if it is possible to obtain at least 3 distinct elements in array c by rearranging array a ; otherwise, output "NO". You can output each letter in any case (for example, "YES", "Yes", "yes", etc.). ▎Example Input
5 4 1 2 1 2 1 2 1 2 6 1 2 3 3 2 1 1 1 1 1 1 1 3 1 1 1 1 1 1 6 1 52 52 3 1 3 59 4 3 59 3 4 4 100 1 100 1 2 2 2 2▎Example Output
YES YES NO YES NO
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i dont know how it ranked div-2 it is pretty easy question https://codeforces.com/contest/2059/problem/A
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Solution 4: Digit-by-Digit Using a Dictionary
This method converts the number to a string and processes each digit individually. It uses a dictionary to map base values and computes the appropriate power of ten for each digit. The solution handles subtractive cases (i.e., when a digit is 4 or 9) separately.
class Solution:
def intToRoman(self, num: int) -> str:
roman = {1: "I", 5: "V", 10: "X", 50: "L", 100: "C", 500: "D", 1000: "M"}
s = ""
st = str(num)
length = len(st)
for ch in st:
i = int(ch)
# Calculate the current power of ten (e.g., 1000, 100, 10, 1)
place = 10 ** (length - 1)
if i == 4 or i == 9:
# For subtractive cases, e.g., 4 -> "IV" and 9 -> "IX"
s += roman[place] + roman[place * (i + 1)]
else:
if i >= 5:
s += roman[place * 5]
s += roman[place] * (i - 5)
else:
s += roman[place] * i
length -= 1
return s
# Example usage:
if __name__ == '__main__':
sol = Solution()
print(sol.intToRoman(3749)) # Output: "MMMDCCXLIX"1 262
▎Approach 1: Greedy Algorithm
class Solution:
def intToRoman(self, num: int) -> str:
# Define the mapping of integers to Roman numerals
val = [
1000, 900, 500, 400,
100, 90, 50, 40,
10, 9, 5, 4,
1
]
syms = [
"M", "CM", "D", "CD",
"C", "XC", "L", "XL",
"X", "IX", "V", "IV",
"I"
]
roman_num = ''
i = 0
while num > 0:
for _ in range(num // val[i]):
roman_num += syms[i]
num -= val[i]
i += 1
return roman_num
# Example usage:
sol = Solution()
print(sol.intToRoman(1994)) # Output: "MCMXCIV"
▎Approach 2: Using a Dictionary
class Solution:
def intToRoman(self, num: int) -> str:
# Mapping of integer values to Roman numerals
roman_map = {
1000: 'M',
900: 'CM',
500: 'D',
400: 'CD',
100: 'C',
90: 'XC',
50: 'L',
40: 'XL',
10: 'X',
9: 'IX',
5: 'V',
4: 'IV',
1: 'I'
}
roman_numeral = ''
for value in roman_map.keys():
while num >= value:
roman_numeral += roman_map[value]
num -= value
return roman_numeral
# Example usage:
sol = Solution()
print(sol.intToRoman(58)) # Output: "LVIII"
▎Approach 3: Recursive Method
class Solution:
def intToRoman(self, num: int) -> str:
# Define the mapping of integers to Roman numerals
roman_map = [
(1000, 'M'),
(900, 'CM'),
(500, 'D'),
(400, 'CD'),
(100, 'C'),
(90, 'XC'),
(50, 'L'),
(40, 'XL'),
(10, 'X'),
(9, 'IX'),
(5, 'V'),
(4, 'IV'),
(1, 'I')
]
def convert(n):
if n == 0:
return ""
for value, symbol in roman_map:
if n >= value:
return symbol + convert(n - value)
return convert(num)
# Example usage:
sol = Solution()
print(sol.intToRoman(3)) # Output: "III"
▎Summary of Approaches
1. Greedy Algorithm: This approach involves repeatedly subtracting the largest possible Roman numeral value from the integer until the number is reduced to zero.
2. Using a Dictionary: Similar to the greedy approach but utilizes a dictionary to map values to their Roman numeral representations for clarity.
3. Recursive Method: This approach uses recursion to build the Roman numeral string by checking against the mapping and reducing the number accordingly.1 262
Problem Statement
Write a function
int_to_roman(num: int) -> str that converts an integer (1 to 3999) to its Roman numeral representation. The Roman numeral system uses the following symbols:
• I = 1
• V = 5
• X = 10
• L = 50
• C = 100
• D = 500
• M = 1000
▎Example Test Cases
1. Input: int_to_roman(1994)
Output: "MCMXCIV"
Explanation: 1994 = 1000 (M) + 900 (CM) + 90 (XC) + 4 (IV).
2. Input: int_to_roman(58)
Output: "LVIII"
Explanation: 58 = 50 (L) + 5 (V) + 3 (III).
3. Input: int_to_roman(4)
Output: "IV"
Explanation: 4 is represented as IV.
4. Input: int_to_roman(9)
Output: "IX"
Explanation: 9 is represented as IX.
5. Input: int_to_roman(3999)
Output: "MMMCMXCIX"
Explanation: 3999 = 3000 (MMM) + 900 (CM) + 90 (XC) + 9 (IX).
▎Constraints
• Input integer num will be in the range [1, 3999].1 262
Codeforces (Div. 3)
▎A. Candies
Time limit per test: 1 second
Memory limit per test: 256 megabytes
▎Problem Statement
Recently, Vova found n candy wrappers. He remembers that he bought x candies during the first day, 2x candies during the second day, 4x candies during the third day, and so on.
On the k -th day, he bought 2⁽ᵏ⁻¹⁾ ⋅ x candies. However, Vova doesn't remember x or k , but he is sure that:
• x and k are positive integers
• k > 1
Vova will be satisfied if you find any positive integer x such that there exists some integer k > 1 where:
x + 2x + 4x + … + 2ᵏ⁻¹x = nIt is guaranteed that at least one solution exists. ▎Input The first line contains one integer t (1 ≤ t ≤ 10⁴) — the number of test cases. Each of the next t lines contains one integer n (3 ≤ n ≤ 10⁹) — the number of candy wrappers Vova found. ▎Output For each test case, print one integer — any valid x such that there exists an integer k > 1 satisfying the given equation. ▎Examples ▎Input
7 3 6 7 21 28 999999999 999999984▎Output
1 2 1 7 4 333333333 333333328▎Explanation • For n = 3 , one possible solution is x = 1, k = 2 since 1 ⋅ 1 + 2 ⋅ 1 = 3 . • For n = 6 , one possible solution is x = 2, k = 2 since 1 ⋅ 2 + 2 ⋅ 2 = 6 . • For n = 7 , one possible solution is x = 1, k = 3 since 1 ⋅ 1 + 2 ⋅ 1 + 4 ⋅ 1 = 7 . --- ▎Solution Code
t = int(input())
while t > 0:
num = int(input())
k = 3
n = 4
while num % k:
k += n
n *= 2
print(num // k)
🔗 Problem link: Codeforces 1343A - Candies