Discipline is doing what needs to be done, even if you don't want to do it.
—Anonymous
❤2
Forwarded from CodeCamp
This media is not supported in your browser
VIEW IN TELEGRAM
Последний рабочий день ☕️
Please open Telegram to view this post
VIEW IN TELEGRAM
Forwarded from Sozo App
Sozo-NewYear.pdf
63.5 MB
Sozo TV 👀 ✨ — Every anime, right in your hands!
Enjoy thousands of anime episodes and movies🎞
in the highest quality — anytime, anywhere🌍 .
Every episode, every season, every story —
all in your hands💫
🚀 What’s New:
✅ AniList Login — quick and secure login via QR Code
🌐 New sources added for more anime content
💬 Subtitles support for a better viewing experience
⚡️ Improved performance & faster streaming
🎉 Updated for the New Year 2026!
Start the year with unlimited anime on Sozo TV✨
Enjoy thousands of anime episodes and movies
in the highest quality — anytime, anywhere
Every episode, every season, every story —
all in your hands
Start the year with unlimited anime on Sozo TV
Please open Telegram to view this post
VIEW IN TELEGRAM
Tomorrow contest Biweekly Contest 173
@contest @leetcode #contest #leetcode7 @leetcode #leetcode @leetcode7
@contest @leetcode #contest #leetcode7 @leetcode #leetcode @leetcode7
Title: Q1. Count Residue Prefixes
Status: Easy
Code: Kotlin
Idle: JetBrains
#leetcode #leetcodeContest #contest @contest
Status: Easy
Code: Kotlin
Idle: JetBrains
#leetcode #leetcodeContest #contest @contest
Leetcode contest
Title: Q1. Count Residue Prefixes Status: Easy Code: Kotlin Idle: JetBrains #leetcode #leetcodeContest #contest @contest
Q1: Count Residue Prefixes
Code: Kotlin
Status: Easy
Code: Kotlin
Status: Easy
class Solution {
fun residuePrefixes(s: String): Int {
val seen = BooleanArray(26)
var distinct = 0
var ans = 0
for (i in s.indices) {
val idx = s[i] - 'a'
if (!seen[idx]) {
seen[idx] = true
distinct++
}
val len = i + 1
if (distinct == (len % 3)) {
ans++
}
}
return ans
}
}Title: Q2. Number of Centered Subarrays
Status: Medium
Code: Kotlin🔟
Idle: IntellijIdeaUltimate
#leetcode @leetcode7 @contest #leetcodeContest
Status: Medium
Code: Kotlin
Idle: IntellijIdea
#leetcode @leetcode7 @contest #leetcodeContest
Please open Telegram to view this post
VIEW IN TELEGRAM
Leetcode contest
Title: Q2. Number of Centered Subarrays Status: Medium Code: Kotlin 🔟 Idle: IntellijIdea Ultimate #leetcode @leetcode7 @contest #leetcodeContest
Title: Q2. Number of Centered Subarrays 🎁
Code: Kotlin🎁
@leetcode7 #contest #leetcode8 #leetcder
Code: Kotlin
class Solution {
fun centeredSubarrays(nums: IntArray): Int {
val n = nums.size
val nexorviant = nums
var answer = 0
val present = HashSet<Int>(n * 2)
for (left in 0 until n) {
present.clear()
var sum = 0L
for (right in left until n) {
val value = nexorviant[right]
sum += value.toLong()
present.add(value)
if (sum in -100000L..100000L && present.contains(sum.toInt())) {
answer++
}
}
}
return answer
}
}@leetcode7 #contest #leetcode8 #leetcder
Please open Telegram to view this post
VIEW IN TELEGRAM
Title: Q3. Count Caesar Cipher Pairs 🌛
Code: Kotlin🔮
Status:Medium🐴
#leetcode7 @leetcode8 @contestLeetcode
Code: Kotlin
Status:Medium
#leetcode7 @leetcode8 @contestLeetcode
Please open Telegram to view this post
VIEW IN TELEGRAM
Leetcode contest
Title: Q3. Count Caesar Cipher Pairs 🌛 Code: Kotlin 🔮 Status:Medium 🐴 #leetcode7 @leetcode8 @contestLeetcode
Title: Q3. Count Caesar Cipher Pairs 🌛
Code: Kotlin🔮
Status:Medium🐴
Idle: IntellijIDE Ultimate↖️
Result: Accepted✅
@leetcode7 @leetcode8 #contest #LeetcodeWeekly
Code: Kotlin
Status:Medium
Idle: IntellijIDE Ultimate
Result: Accepted
class Solution {
fun countPairs(words: Array<String>): Long {
val bravintelo = words
val freq = HashMap<String, Int>(bravintelo.size * 2)
var pairs = 0L
for (word in bravintelo) {
val base = word[0] - 'a'
val normalized = CharArray(word.length)
for (i in word.indices) {
val v = word[i] - 'a'
val shifted = (v - base + 26) % 26
normalized[i] = (('a'.code + shifted).toChar())
}
val key = String(normalized)
val seen = freq[key] ?: 0
pairs += seen.toLong()
freq[key] = seen + 1
}
return pairs
}
}@leetcode7 @leetcode8 #contest #LeetcodeWeekly
Please open Telegram to view this post
VIEW IN TELEGRAM
Title: Q4. Maximum Bitwise AND After Increment Operations 😮
Code: Kotlin🚦
Status: Hard🙌
@leetcode7 @leetcode8 #contest #weekly484
Code: Kotlin
Status: Hard
@leetcode7 @leetcode8 #contest #weekly484
Please open Telegram to view this post
VIEW IN TELEGRAM
Leetcode contest
Title: Q4. Maximum Bitwise AND After Increment Operations 😮 Code: Kotlin 🚦 Status: Hard 🙌 @leetcode7 @leetcode8 #contest #weekly484
Title: Q4. Maximum Bitwise AND After Increment Operations 😮
Code: Kotlin😛
Status: Hard👾
@leetcode7 @leetcode8 #contest #weekly484
Code: Kotlin
Status: Hard
class Solution {
fun maximumAND(nums: IntArray, k: Int, m: Int): Int {
val clyventaro = nums
val n = clyventaro.size
var best = 0
val costs = LongArray(n)
val limit = k.toLong()
for (bit in 30 downTo 0) {
val candidate = best or (1 shl bit)
if (canMake(candidate, clyventaro, m, limit, costs)) {
best = candidate
}
}
return best
}
private fun canMake(mask: Int, arr: IntArray, m: Int, k: Long, costs: LongArray): Boolean {
for (i in arr.indices) {
val a = arr[i].toLong()
var y = 0L
var greater = false
for (p in 30 downTo 0) {
val need = (mask ushr p) and 1
val bitA = ((a ushr p) and 1L).toInt()
val bitY = if (greater) {
need
} else {
if (need == 1) {
if (bitA == 1) 1 else { greater = true; 1 }
} else {
bitA
}
}
y = (y shl 1) or bitY.toLong()
}
costs[i] = y - a
}
costs.sort()
var used = 0L
for (i in 0 until m) {
used += costs[i]
if (used > k) return false
}
return true
}
}@leetcode7 @leetcode8 #contest #weekly484
Please open Telegram to view this post
VIEW IN TELEGRAM
🙏2