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allcoding1_official (@allcoding1_official) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 85 467 obunachidan iborat bo'lib, Texnologiyalar & Aralashmalar toifasida 1 502-o'rinni va Hindiston mintaqasida 3 471-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 85 467 obunachiga ega bo‘ldi.
25 Iyun, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -1 422 ga, so‘nggi 24 soatda esa -71 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 2.42% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 0.88% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 2 071 marta ko‘riladi; birinchi sutkada odatda 749 ta ko‘rish yig‘iladi.
- Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 4 ta reaksiya keladi.
- Tematik yo‘nalishlar: Kontent dsa, stack, namaste, javascript, dev kabi asosiy mavzularga jamlangan.
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Yuqori yangilanish chastotasi (oxirgi ma’lumot 26 Iyun, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Texnologiyalar & Aralashmalar toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
85 467
Obunachilar
-7124 soatlar
-3077 kunlar
-1 42230 kunlar
Postlar arxiv
85 467
Cognizant Off Campus Hiring | Data Analyst – Fresher | Upto 7.5 LPA
Job Role Data Analyst
Qualification B.E/B.Tech
Batch Any Batch
CTC/Salary Upto 7.5 LPA
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Telegram:- @allcoding1_official
85 467
def count(a, X):
a.sort()
ans = 0
n = len(a)
for i in range(n):
if (i < n // 2):
ans += max(0, a[i] - X)
elif (i == n // 2):
ans += abs(X - a[i])
else:
ans += max(0, X - a[i]);
return ans
IBM
Telegram:-
85 467
int solve(vector& nums) {
vector s(nums);
sort(s.begin(), s.end());
int i = 0, j = nums.size() - 1;
while (i < nums.size() && nums[i] == s[i]) {
i++;
}
while (j > i && nums[j] == s[j]) {
j--;
}
return j - i + 1;
}
Arrange the heights
Apple ✅
Telegram:-
85 467
void solve(string x){
ll n = x.size();
vector> dp(n,vector(n,0));
for(ll i=0;i
85 467
#include<bits/stdc++.h>
using namespace std;
void dfs(int node, vector<int>& vis, vector<vector<int>>& adj) {
vis[node] = 1;
for(auto it : adj[node]) {
if(!vis[it]) {
dfs(it, vis, adj);
}
}
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> adj(n);
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
int x;
cin >> x;
if(x == 1) {
adj[i].push_back(j);
adj[j].push_back(i);
}
}
}
vector<int> vis(n, 0);
int cc = 0;
for(int i = 0; i < n; i++) {
if(!vis[i]) {
dfs(i, vis, adj);
cc++;
}
}
cout << cc << endl;
return 0;
}
85 467
int getMinOperations(vector arr) {
int n=arr.size();
int operations = 0;
for (int i = 1; i < n - 1; i++)
{
if (arr[i] < 0 && arr[i - 1] < 0)
{
arr[i] = 1e9;
operations++;
}
else if (arr[i] < 0 && arr[i - 1] > 0 && arr[i + 1] > 0)
{
if ((arr[i] + arr[i - 1] < 0) || (arr[i] + arr[i + 1] < 0))
{
arr[i] = 1e9;
operations++;
}
}
}
return operations;
}
Make The Array Postive ✅
85 467
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<vector<int>> freq(n, vector<int> (26, 0));
string s;
for(int i = 0; i < n; i++) {
string s;
cin >> s;
for(auto &ch : s)
freq[i][ch - 'a']++;
}
long long ans = 0;
for(int i = 0; i < 26; i++) {
int minValue = INT_MAX;
for(int j = 0; j < n; j++)
minValue = min(minValue, freq[j][i]);
ans += minValue;
}
cout << ans << endl;
}
Service Now ✅
85 467
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(ll k, ll s) {
auto comb = [](ll n) {
return n * (n - 1) / 2;
};
if (k > 3 * s) return 0;
ll ans = comb(k + 2);
if (k > s) ans -= 3 * comb(k - s + 1);
if (k - 2 >= 2 * s) ans += 3 * comb(k - 2 * s);
return ans;
}
int main() {
ll n, s;
cin >> n >> s;
cout << solve(s,n) << endl;
return 0;
}
Distribute Car Toy
Service Now ✅
85 467
🎯Amazon off Campus Drive 2024 Hiring Freshers As Associate Quality Services | 2.52-4 LPA
Job Role : Associate Quality Services
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Experience : Freshers
Salary CTC : Rs 2.52-4 LPA
Apply Now:- www.allcoding1.com
85 467
def decodeSequence(binarySequence):
nucleobases = {'001': 'C', '011': 'A', '101': 'T','010': 'G', '110': 'U'}
if len(binarySequence) % 3 != 0:
print("Error: The length of the input string should be a multiple of 3.")
return
decodedSequence = ""
for i in range(0, len(binarySequence), 3):
chunk = binarySequence[i:i+3]
if i == 0 and chunk == '000':
identifier = 'DNA'
elif i == 0 and chunk == '111':
identifier = 'RNA'
else:
nucleobase = nucleobases.get(chunk, 'X')
if nucleobase == 'U' and identifier == 'DNA':
nucleobase = 'T'
elif nucleobase == 'T' and identifier == 'RNA':
nucleobase = 'U'
decodedSequence += nucleobase
print(decodedSequence)
binaryInput = input("Enter the binary sequence: ")
decodeSequence(binaryInput)
DNA
IBM
Telegram:- @allcoding1_official
85 467
string solve(string bs) {
map nb = {
{"001", "C"},
{"010", "G"},
{"011", "A"},
{"101", "T"},
{"110", "U"},
{"000", "DNA"},
{"111", "RNA"}
};
string ds = "";
string t = nb[bs.substr(0, 3)];
for(int i = 3; i < bs.length(); i += 3) {
string b = bs.substr(i, 3);
if(nb.find(b) != nb.end()) {
string x = nb[b];
if(t == "DNA" && x == "U") {
x = "T";
}
ds += x;
} else {
ds += "Error";
}
}
return ds;
}
DNA✅
IBM
85 467
bool isPal(int n) {
int r, s = 0, t;
t = n;
while (n > 0) {
r = n % 10;
s = (s * 10) + r;
n = n / 10;
}
return (t == s);
}
int firstPal(int n) {
int i = 1;
while (true) {
if (isPal(i)) {
int d = 1 + log10(i);
if (d == n)
return i;
}
i++;
}
}
void login(int d, string u, string p) {
map users = {
{"user1", "pass1"},
{"user2", "pass2"},
{"user3", "pass3"},
{"user4", "pass4"},
{"user5", "pass5"}
};
if (users.find(u) != users.end() && users[u] == p) {
int t = firstPal(d);
cout << "Welcome " << u << " and the generated token is: token-" << t << endl;
} else {
cout << "UserId or password is not valid, please try again." << endl;
}
}
IBM✅
85 467
Here's a Python program to simulate the given problem:
`python
def print_terrain(terrain):
for row in terrain:
print(''.join(row))
def flow_water(terrain, n):
water_level = int(terrain[n // 2][n // 2])
terrain[n // 2][n // 2] = 'W'
def can_flow(x, y, direction):
if direction == 'N':
return x > 0 and terrain[x-1][y] != 'W' and int(terrain[x-1][y]) <= water_level
elif direction == 'S':
return x < n - 1 and terrain[x+1][y] != 'W' and int(terrain[x+1][y]) <= water_level
elif direction == 'E':
return y < n - 1 and terrain[x][y+1] != 'W' and int(terrain[x][y+1]) <= water_level
elif direction == 'W':
return y > 0 and terrain[x][y-1] != 'W' and int(terrain[x][y-1]) <= water_level
def flow(x, y):
if can_flow(x, y, 'N'):
terrain[x-1][y] = 'W'
return True
if can_flow(x, y, 'S'):
terrain[x+1][y] = 'W'
return True
if can_flow(x, y, 'E'):
terrain[x][y+1] = 'W'
return True
if can_flow(x, y, 'W'):
terrain[x][y-1] = 'W'
return True
return False
while True:
print_terrain(terrain)
has_flown = False
for i in range(n):
for j in range(n):
if terrain[i][j] == 'W':
if flow(i, j):
has_flown = True
if not has_flown:
water_level += 1
print(f"Cannot flow, increasing water level to {water_level}")
break
if any(cell == 'W' and (i == 0 or j == 0 or i == n - 1 or j == n - 1) for i, row in enumerate(terrain) for j, cell in enumerate(row)):
print("Reached edge, exiting.")
break
n = 7
terrain = [
[494, 88, 89, 778, 984, 726, 587],
[340, 959, 220, 301, 639, 280, 290],
[666, 906, 632, 824, 127, 505, 787],
[673, 499, 843, 172, 193, 613, 154],
[544, 211, 124, 60, 575, 572, 389],
[635, 170, 174, 946, 593, 314, 300],
[620, 167, 931, 780, 416, 954, 275]
]
flow_water(terrain, n)