🚀 Challenge for Aspiring Weightlifters! 🏋️♂️
Yaman, an aspiring weightlifter, is gearing up for his next tournament and needs your help! He has n barbell plates, each with a specific weight aᵢ and width bᵢ . Yaman's goal is to distribute these plates into two subsets: one for the left side of the barbell and one for the right.
His coach has defined a balancing function f(l, r) as follows:
f(l, r) = (| ∑(i ∈ l) bᵢ - ∑(i ∈ r) bᵢ )|
Yaman has to ensure that the absolute difference in widths between the two sides falls within a specified range during his training.
▎Your Task:
For each test case, determine the maximum total weight Yaman can lift while keeping the balancing function's value within the range [L, R]. If it's impossible to achieve this balance, you should return 0 .
▎Input Format:
• The first line contains the number of test cases t ( 1 ≤ t ≤ 10⁵ ).
• Each test case starts with two integers n (number of plates) and q (number of tests).
• The next line lists n integers representing the weights of the plates.
• The following line lists n integers representing the widths of the plates.
• The next q lines each contain two integers Lᵢ and Rᵢ , specifying the desired range for the balancing function.
▎Output Format:
For each test, output the maximum total weight Yaman can lift while maintaining the balancing function's value within the specified range. If it's not possible, output 0 .
▎Example:
Input:
Output:
Can you help Yaman achieve his training goals? Share your solutions and strategies! 💪
🔗 GotoCodeForce
Yaman, an aspiring weightlifter, is gearing up for his next tournament and needs your help! He has n barbell plates, each with a specific weight aᵢ and width bᵢ . Yaman's goal is to distribute these plates into two subsets: one for the left side of the barbell and one for the right.
His coach has defined a balancing function f(l, r) as follows:
f(l, r) = (| ∑(i ∈ l) bᵢ - ∑(i ∈ r) bᵢ )|
Yaman has to ensure that the absolute difference in widths between the two sides falls within a specified range during his training.
▎Your Task:
For each test case, determine the maximum total weight Yaman can lift while keeping the balancing function's value within the range [L, R]. If it's impossible to achieve this balance, you should return 0 .
▎Input Format:
• The first line contains the number of test cases t ( 1 ≤ t ≤ 10⁵ ).
• Each test case starts with two integers n (number of plates) and q (number of tests).
• The next line lists n integers representing the weights of the plates.
• The following line lists n integers representing the widths of the plates.
• The next q lines each contain two integers Lᵢ and Rᵢ , specifying the desired range for the balancing function.
▎Output Format:
For each test, output the maximum total weight Yaman can lift while maintaining the balancing function's value within the specified range. If it's not possible, output 0 .
▎Example:
Input:
1
5 4
5 2 3 2 5
1 4 9 12 20
1 2
7 7
99 100
3 3
Output:
17
12
0
14
Can you help Yaman achieve his training goals? Share your solutions and strategies! 💪
🔗 GotoCodeForce
Codeforces
Problem - A - Codeforces
Codeforces. Programming competitions and contests, programming community
👍2
Leetcode with dani
🚀 Challenge for Aspiring Weightlifters! 🏋️♂️ Yaman, an aspiring weightlifter, is gearing up for his next tournament and needs your help! He has n barbell plates, each with a specific weight aᵢ and width bᵢ . Yaman's goal is to distribute these plates…
in Short Yaman is training with a barbell that has plates of varying weights and widths. He needs to distribute these plates between the left and right sides of the barbell. The goal is to maximize the total weight lifted while ensuring that the difference in total widths between the two sides falls within a specified range [L, R]. For each test case, determine the maximum weight Yaman can lift under these conditions. If it's not possible to achieve the desired width balance, the result should be 0.
🌟 Day 11 Challenge: Find Pivot Index 🌟
🔗 Problem Link: 724. Find Pivot Index
📝 Problem Statement:
Given an array of integers
The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.
If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.
Return the leftmost pivot index. If no such index exists, return -1.
---
▎🔍 Approach: Prefix Sum
We can use a prefix sum technique to efficiently check for the pivot index.
▎Steps to Solve:
1. Compute prefix sums: We maintain a prefix sum array
2. Iterate through the array: For each index
3. Return the first valid index: If we find an index where both sums match, return it.
4. Return
---
▎📝 Code Implementation:
---
▎📝 Explanation of the Code:
1. Building Prefix Sum Array:
• We initialize
• Iterate through
2. Finding Pivot Index:
• For each
• Left sum:
• Right sum:
• If both sums are equal, return
3. Edge Case Handling:
• If no such index exists, return
---
▎✅ Complexity Analysis:
• Time Complexity: O(N) (single pass for prefix sum + single pass for checking pivot index).
• Space Complexity: O(N) (extra space for prefix sum array).
🔗 Problem Link: 724. Find Pivot Index
📝 Problem Statement:
Given an array of integers
nums, calculate the pivot index of this array. The pivot index is the index where the sum of all the numbers strictly to the left of the index is equal to the sum of all the numbers strictly to the index's right.
If the index is on the left edge of the array, then the left sum is 0 because there are no elements to the left. This also applies to the right edge of the array.
Return the leftmost pivot index. If no such index exists, return -1.
---
▎🔍 Approach: Prefix Sum
We can use a prefix sum technique to efficiently check for the pivot index.
▎Steps to Solve:
1. Compute prefix sums: We maintain a prefix sum array
p, where p[i] stores the sum of elements from index 0 to i-1.2. Iterate through the array: For each index
i, check if the left sum (p[i] - nums[i-1]) equals the right sum (p[-1] - p[i]).3. Return the first valid index: If we find an index where both sums match, return it.
4. Return
-1 if no pivot index is found. ---
▎📝 Code Implementation:
class Solution:
def pivotIndex(self, nums: List[int]) -> int:
# Step 1: Compute prefix sum
p = [0] # Prefix sum array initialized with 0
for i in range(len(nums)):
p.append(p[-1] + nums[i]) # Store cumulative sum
# Step 2: Iterate through the array and check pivot condition
for i in range(1, len(p)):
left_sum = p[i] - nums[i - 1] # Sum of elements to the left of index i-1
right_sum = p[-1] - p[i] # Sum of elements to the right of index i-1
if left_sum == right_sum:
return i - 1 # Return the pivot index
return -1 # If no pivot index is found
---
▎📝 Explanation of the Code:
1. Building Prefix Sum Array:
• We initialize
p with [0] to store cumulative sums.• Iterate through
nums, appending the cumulative sum to p.2. Finding Pivot Index:
• For each
i, compute:• Left sum:
p[i] - nums[i-1] (sum of elements before index i-1).• Right sum:
p[-1] - p[i] (sum of elements after index i-1).• If both sums are equal, return
i-1.3. Edge Case Handling:
• If no such index exists, return
-1.---
▎✅ Complexity Analysis:
• Time Complexity: O(N) (single pass for prefix sum + single pass for checking pivot index).
• Space Complexity: O(N) (extra space for prefix sum array).
👍2
I really love the story! Please check it out when you have some free time and let me know what you think.
watch the video
watch the video
YouTube
The Egg - A Short Story
The Egg
Story by Andy Weir
Animated by Kurzgesagt
A Big Thanks to Andy Weir for allowing us to use his story.
The original was released here: http://www.galactanet.com/oneoff/theegg_mod.html
Visit his website here: http://www.andyweirauthor.com/
If you…
Story by Andy Weir
Animated by Kurzgesagt
A Big Thanks to Andy Weir for allowing us to use his story.
The original was released here: http://www.galactanet.com/oneoff/theegg_mod.html
Visit his website here: http://www.andyweirauthor.com/
If you…
❤🔥2✍1
🌟 Day 11 Challenge: Walking Robot Simulation 🤖
🔗 Problem Link: 874. Walking Robot Simulation
Hey everyone! 👋 Today, we’re solving an interesting robot movement problem where the robot moves on an infinite grid while avoiding obstacles! 🚀
---
▎📝 Problem Statement:
A robot starts at (0,0) facing north and follows a series of commands:
•
•
•
• If the robot hits an obstacle, it stops moving in that direction but continues executing the next command.
🛑 Goal: Find the maximum squared Euclidean distance the robot reaches from the origin (0,0).
💡 Key Notes:
• North → +Y
• East → +X
• South → -Y
• West → -X
---
▎🔍 Understanding with Examples
📌 Example 1:
Input:
Output:
✅ The robot moves to (3,4), and the max squared distance is 3² + 4² = 25.
📌 Example 2 (With Obstacles):
Input:
Output:
✅ The robot gets blocked at (2,4), then continues moving north to (1,8), reaching a max squared distance of 1² + 8² = 65.
---
▎🏗 Approach to Solve the Problem
1️⃣ Track directions efficiently:
• Use
2️⃣ Use a set for obstacles for quick lookup (O(1) checks).
3️⃣ Loop through commands and update position accordingly:
• If moving forward → Check if the next step is an obstacle.
• If turning → Update direction accordingly.
---
▎💻 Code (Python)
---
▎📝 Explanation of the Code:
1. Tracking Directions:
• We use an array
• Turning left (
2. Efficient Obstacle Lookup:
• We store obstacles in a set (O(1) lookup) instead of a list to quickly check if a position is blocked.
3. Processing Commands:
• For k steps forward, check if the next step is an obstacle:
• ✅ If not blocked, update x, y.
• ❌ If blocked, stop moving in that direction.
• Keep track of the maximum squared Euclidean distance reached.
---
▎✅ Complexity Analysis:
• Time Complexity: O(N + M), where N = len(commands), M = len(obstacles)
• Space Complexity: O(M) (for storing obstacles in a set)
🔗 Problem Link: 874. Walking Robot Simulation
Hey everyone! 👋 Today, we’re solving an interesting robot movement problem where the robot moves on an infinite grid while avoiding obstacles! 🚀
---
▎📝 Problem Statement:
A robot starts at (0,0) facing north and follows a series of commands:
•
-2: Turn left 90° •
-1: Turn right 90° •
1 ≤ k ≤ 9: Move forward k units • If the robot hits an obstacle, it stops moving in that direction but continues executing the next command.
🛑 Goal: Find the maximum squared Euclidean distance the robot reaches from the origin (0,0).
💡 Key Notes:
• North → +Y
• East → +X
• South → -Y
• West → -X
---
▎🔍 Understanding with Examples
📌 Example 1:
Input:
commands = [4,-1,3], obstacles = [] Output:
25 ✅ The robot moves to (3,4), and the max squared distance is 3² + 4² = 25.
📌 Example 2 (With Obstacles):
Input:
commands = [4,-1,4,-2,4], obstacles = [[2,4]] Output:
65 ✅ The robot gets blocked at (2,4), then continues moving north to (1,8), reaching a max squared distance of 1² + 8² = 65.
---
▎🏗 Approach to Solve the Problem
1️⃣ Track directions efficiently:
• Use
dx, dy arrays to manage the direction changes when turning left or right. 2️⃣ Use a set for obstacles for quick lookup (O(1) checks).
3️⃣ Loop through commands and update position accordingly:
• If moving forward → Check if the next step is an obstacle.
• If turning → Update direction accordingly.
---
▎💻 Code (Python)
class Solution:
def robotSim(self, commands: List[int], obstacles: List[List[int]]) -> int:
# Step 1: Define movement directions (North, East, South, West)
directions = [(0,1), (1,0), (0,-1), (-1,0)]
x, y, max_dist, d = 0, 0, 0, 0 # Robot starts at (0,0), facing North
# Step 2: Convert obstacles list into a set for quick lookups
obstacle_set = set(map(tuple, obstacles))
# Step 3: Process each command
for command in commands:
if command == -2: # Turn left
d = (d - 1) % 4
elif command == -1: # Turn right
d = (d + 1) % 4
else: # Move forward k steps
for _ in range(command):
next_x, next_y = x + directions[d][0], y + directions[d][1]
# Stop moving if an obstacle is hit
if (next_x, next_y) in obstacle_set:
break
x, y = next_x, next_y # Update position
max_dist = max(max_dist, x*x + y*y) # Update max distance
return max_dist # Return max squared Euclidean distance
---
▎📝 Explanation of the Code:
1. Tracking Directions:
• We use an array
directions = [(0,1), (1,0), (0,-1), (-1,0)] to handle North, East, South, and West movements. • Turning left (
-2) moves us backwards in the array, while turning right (-1) moves us forward. 2. Efficient Obstacle Lookup:
• We store obstacles in a set (O(1) lookup) instead of a list to quickly check if a position is blocked.
3. Processing Commands:
• For k steps forward, check if the next step is an obstacle:
• ✅ If not blocked, update x, y.
• ❌ If blocked, stop moving in that direction.
• Keep track of the maximum squared Euclidean distance reached.
---
▎✅ Complexity Analysis:
• Time Complexity: O(N + M), where N = len(commands), M = len(obstacles)
• Space Complexity: O(M) (for storing obstacles in a set)
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:
It 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
▎Output
▎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
🔗 Problem link: Codeforces 1343A - Candies
▎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 = n
It 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:
t-=1
num = int(input())
k = 3
n = 4
while num % k:
k += n
n *= 2
print(num // k)
🔗 Problem link: Codeforces 1343A - Candies
Problem Statement
Write a function
• I = 1
• V = 5
• X = 10
• L = 50
• C = 100
• D = 500
• M = 1000
▎Example Test Cases
1. Input:
Output:
Explanation: 1994 = 1000 (M) + 900 (CM) + 90 (XC) + 4 (IV).
2. Input:
Output:
Explanation: 58 = 50 (L) + 5 (V) + 3 (III).
3. Input:
Output:
Explanation: 4 is represented as IV.
4. Input:
Output:
Explanation: 9 is represented as IX.
5. Input:
Output:
Explanation: 3999 = 3000 (MMM) + 900 (CM) + 90 (XC) + 9 (IX).
▎Constraints
• Input integer
Try it
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].Try it
LeetCode
Integer to Roman - LeetCode
Can you solve this real interview question? Integer to Roman - Seven different symbols represent Roman numerals with the following values:
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
Roman numerals are formed by appending the conversions of decimal…
Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000
Roman numerals are formed by appending the conversions of decimal…
Leetcode with dani
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 •…
▎Approach 1: Greedy Algorithm
▎Approach 2: Using a Dictionary
▎Approach 3: Recursive Method
▎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.
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.
Leetcode with dani
▎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, …
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.
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"
▎Problem A: Milya and Two Arrays
Contest: Codeforces Round #2059
Time Limit: 1 second per test
Memory Limit: 256 megabytes
▎Problem Denoscription
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:
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
▎Example Output
Contest: Codeforces Round #2059
Time Limit: 1 second per test
Memory Limit: 256 megabytes
▎Problem Denoscription
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
Codeforces
Problem - A - Codeforces
Codeforces. Programming competitions and contests, programming community
Leetcode with dani
▎Problem A: Milya and Two Arrays Contest: Codeforces Round #2059 Time Limit: 1 second per test Memory Limit: 256 megabytes ▎Problem Denoscription An array is called good if every element x that appears in the array appears at least twice. For example…
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")
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.,
- 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!
https://news.1rj.ru/str/zethioprograming
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!
https://news.1rj.ru/str/zethioprograming
🙏6
Forwarded 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
(rated for Div. 2) starts on the 18th of February at 14:35 UTC.
Please, join by the link https://codeforces.com/contests/2069
Codeforces
Educational Codeforces Round 174 (Rated for Div. 2) - Codeforces
Codeforces. Programming competitions and contests, programming community
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🌟 I see a lot of people talking about ALX and recommending it, so it might be worth checking out! Before feburary 24
for African student
🔥 ProDev Front-End Developer
📅 Duration: 8 Months | ⏳ 30 hrs/week | 🌍 Online
💡 Skills: TypeScript, TailwindCSS, Next.js, React Native, Figma
🔥 Front-End Web Development
📅 Duration: 8 Months | ⏳ 20 hrs/week | 🌍 Online
💡 Skills: HTML, CSS, JavaScript, GitHub, Figma
🔥 ProDev Back-End Developer
📅 Duration: 8 Months | ⏳ 30 hrs/week | 🌍 Online
💡 Skills: Django, Python, SQL, Git, GitHub
🔥 Back-End Web Development
📅 Duration: 8 Months | ⏳ 20 hrs/week | 🌍 Online
💡 Skills: Django, Python, SQL, Git, GitHub
🔥 AWS Cloud Computing
📅 Duration: 10 Months | ⏳ 30-40 hrs/week | 🌍 Online
💡 Skills: AWS Console, Linux Terminal
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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 Denoscription:
Given an array of positive integers
*Example 1:*
*Example 2:*
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:
Explanation:
1. Initialize an empty list
2. Iterate over each number
3. Convert the number to a string and split it into individual characters using
4. Convert each character back to an integer and extend the
5. After processing all numbers, return the
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]
Problem Denoscription:
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]
LeetCode
Separate the Digits in an Array - LeetCode
Can you solve this real interview question? Separate the Digits in an Array - 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.…