ACCENTURE EXAM SOLUTIONS
Kanalga Telegram’da o‘tish
🔥Guys plz Stop fearing for daily exams 📝 👨💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk
Ko'proq ko'rsatish📈 Telegram kanali ACCENTURE EXAM SOLUTIONS analitikasi
ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 13 954 obunachidan iborat bo'lib, Taʼlim toifasida 14 642-o'rinni va Hindiston mintaqasida 28 153-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 13 954 obunachiga ega bo‘ldi.
03 Noyabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -151 ga, so‘nggi 24 soatda esa -6 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.70% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.38% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 516 marta ko‘riladi; birinchi sutkada odatda 193 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.
📝 Tavsif va kontent siyosati
Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥Guys plz Stop fearing for daily exams 📝
👨💻 @srksvk is here to help you all at lowest cost possible.💪
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company
🔥Effort from our side = 💯
📱Main Channel: @coding_are
📱Tel I'd : @srks...”
Yuqori yangilanish chastotasi (oxirgi ma’lumot 04 Noyabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
13 954
Obunachilar
-624 soatlar
-377 kun
-15130 kun
Postlar arxiv
13 954
def variantsCount(n, s0, k, b, m, a):
if n <=0:
return 0
sum_array = [s0]
for i in range(1,n):
s_i = ((k * sum_array[i - 1] + b) % m) + 1 + sum_array[i - 1]
sum_array.append(s_i)
# two pointer to calculate the count
left = 0
right = len(sum_array) -1
count = 0
while left <= right:
area = sum_array[left] * sum_array[right]
if area <= a:
count += 2*(right - left) + 1
left +=1
elif area > a:
right -=1
return count
Paint the ceiling code
Python3
13 954
#include <bits/stdc++.h>
using namespace std;
vector<int> batonPass(int n, long t){
vector<int> ans;
int oddEven = 1,count=0;
while(true){
if(oddEven & 1){
if(count==0){
for(int i=0 ; i<n ; i++){
count++;
if(t == count){
if(i!=n-1){
// cout << i+1 << " " << i+2;
ans.push_back(i+1);
ans.push_back(i+2);
return ans;
}else{
ans.push_back(i+1);
ans.push_back(i);
return ans;
}
}
}
}else{
for(int i=1 ; i<n ; i++){
count++;
if(t == count){
if(i!=n-1){
ans.push_back(i+1);
ans.push_back(i+2);
return ans;
}else{
ans.push_back(i+1);
ans.push_back(i);
return ans;
}
}
}
}
oddEven++;
}else{
for(int i = n-2 ; i>=0 ; i--){
count++;
if(t == count){
if(i!=0){
ans.push_back(i+1);
ans.push_back(i);
return ans;
}else{
ans.push_back(i);
ans.push_back(i+1);
return ans;
}
}
}
oddEven++;
}
}
}
int main()
{
int friends,timee;
cin >> friends >> timee;
vector<int> ans = batonPass(friends,timee);
for(auto it : ans) cout << it << " ";
return 0;
}
Pass the baton code 👆👆
C++14
13 954
vector<int> batonPass(int friends, long time) {
vector<int> f;
int k = 1;
bool r = false;
long i = 1;
while (i <= time) {
if (!r) {
if (k <= friends)
k++;
else {
r = true;
}
} else {
if (k >= 2) {
k--;
} else {
r = false;
}
}
i++;
}
if (r) {
f.push_back(k);
f.push_back(k - 1);
} else {
f.push_back(k - 1);
f.push_back(k);
}
return f;
}
Pass the baton code- c++14
13 954
#include <bits/stdc++.h>
using namespace std;
vector<int> batonPass(int n, long t){
vector<int> ans;
int oddEven = 1,count=0;
while(true){
if(oddEven & 1){
if(count==0){
for(int i=0 ; i<n ; i++){
count++;
if(t == count){
if(i!=n-1){
// cout << i+1 << " " << i+2;
ans.push_back(i+1);
ans.push_back(i+2);
return ans;
}else{
ans.push_back(i+1);
ans.push_back(i);
return ans;
}
}
}
}else{
for(int i=1 ; i<n ; i++){
count++;
if(t == count){
if(i!=n-1){
ans.push_back(i+1);
ans.push_back(i+2);
return ans;
}else{
ans.push_back(i+1);
ans.push_back(i);
return ans;
}
}
}
}
oddEven++;
}else{
for(int i = n-2 ; i>=0 ; i--){
count++;
if(t == count){
if(i!=0){
ans.push_back(i+1);
ans.push_back(i);
return ans;
}else{
ans.push_back(i);
ans.push_back(i+1);
return ans;
}
}
}
oddEven++;
}
}
}
int main()
{
int friends,timee;
cin >> friends >> timee;
vector<int> ans = batonPass(friends,timee);
for(auto it : ans) cout << it << " ";
return 0;
}
1st code -c++
13 954
Atlassian free code help 🔥🔥
Join here guys
Join this group guys for free code
Atlassian code free 🤘 free
h:https://t.me/rituyr
https://t.me/rituyr
https://t.me/rituyr
share ✅share✅ share ✅
13 954
Atlassian free code help 🔥🔥
Join here guys
Join this group guys for free code
Atlassian code free 🤘 free
h:https://t.me/rituyr
https://t.me/rituyr
https://t.me/rituyr
share ✅share✅ share ✅
13 954
def FindRemainingBalls(direction, strength):
asteroids = []
for i in range(0, len(direction)):
if direction[i] < 0:
asteroids.append(strength[i]*-1)
else:
asteroids.append(strength[i])
stack = []
collided = False
for asteroid in asteroids:
while stack and asteroid < 0 < stack[-1]:
defending_asteroid = stack.pop()
if defending_asteroid != abs(asteroid):
asteroid = asteroid if abs(asteroid) > defending_asteroid else defending_asteroid
else:
asteroid = 0
if asteroid != 0:
stack.append(asteroid)
return stack
Ball collision-python
13 954
Atlassian free code help 🔥🔥
Join here guys
Join this group guys for free code
Atlassian code free 🤘 free
h:https://t.me/rituyr
https://t.me/rituyr
https://t.me/rituyr
share ✅share✅ share ✅
13 954
int mex(vectorarr,int K)
{
int n=arr.size();
unordered_map mp;
for (int i = 0; i < n; i++) {
mp[arr[i] % K]++;
}
for (int i = 0; i < n; i++) {
if (mp.find(i % K) == mp.end()) {
return i;
}
mp[i % K]--;
if (mp[i % K] == 0)
mp.erase(i % K);
}
return n;
}
Mex problem guys ✅✅
Note -type as it is it will work
C++
13 954
def FindRemainingBalls(direction, strength):
asteroids = []
for i in range(0, len(direction)):
if direction[i] < 0:
asteroids.append(strength[i]*-1)
else:
asteroids.append(strength[i])
stack = []
collided = False
for asteroid in asteroids:
while stack and asteroid < 0 < stack[-1]:
defending_asteroid = stack.pop()
if defending_asteroid != abs(asteroid):
asteroid = asteroid if abs(asteroid) > defending_asteroid else defending_asteroid
else:
asteroid = 0
if asteroid != 0:
stack.append(asteroid)
return stack
Ball collision -Python
