allcoding1
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📈 Analytical overview of Telegram channel allcoding1
Channel allcoding1 (@allcoding1) in the English language segment is an active participant. Currently, the community unites 21 533 subscribers, ranking 9 159 in the Education category and 19 101 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 21 533 subscribers.
According to the latest data from 01 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -367 over the last 30 days and by -14 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 7.16%. Within the first 24 hours after publication, content typically collects 1.25% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 543 views. Within the first day, a publication typically gains 270 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 0.
- Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, learning.
📝 Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 02 September, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
21 533
Subscribers
-1424 hours
-797 days
-36730 days
Posts Archive
21 529
string solve(string bs) {
map<string, string> 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
Telegram:- @allcoding1
21 529
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<string, string> 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✅
Telegram:- @allcoding1
21 529
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)
Python
Telegram:- @allcoding1_official
21 529
from itertools import permutations
def unique_permutations(nums):
unique_perms = set(permutations(nums))
return [list(perm) for perm in unique_perms]
# Take input for nums
nums_input = input("Enter a list of numbers separated by spaces: ")
nums = [int(num) for num in nums_input.split()]
# Get and print unique permutations
output = unique_permutations(nums)
print(output)
21 529
Saks subarray product✅
long long solve(vector& nums, int k) {
if (k <= 1) return 0;
int n = nums.size();
long long p = 1;
int i = 0, j = 0;
long long ans = 0;
while (j < n) {
p *= nums[j];
while (i <= j && p > k) {
p /= nums[i];
i++;
}
ans += j - i + 1;
j++;
}
return ans;
}
21 529
#include <iostream>
#include <vector>
#include <unordered_map>
class Main {
public:
static long getZeroBitSubarrays(const std::vector<int>& arr) {
int n = arr.size();
long totalSubarrayCount = static_cast<long>(n) * (n + 1) / 2;
long nonzeroSubarrayCount = 0;
std::unordered_map<int, int> windowBitCounts;
int leftIdx = 0;
for (int rightIdx = 0; rightIdx < n; rightIdx++) {
int rightElement = arr[rightIdx];
if (rightElement == 0) {
windowBitCounts.clear();
leftIdx = rightIdx + 1;
continue;
}
std::vector<int> setBitIndices = getSetBitIndices(rightElement);
for (int index : setBitIndices) {
windowBitCounts[index]++;
}
while (leftIdx < rightIdx && isBitwiseAndZero(rightIdx - leftIdx + 1, windowBitCounts)) {
for (int index : getSetBitIndices(arr[leftIdx])) {
windowBitCounts[index]--;
if (windowBitCounts[index] == 0) {
windowBitCounts.erase(index);
}
}
leftIdx++;
}
nonzeroSubarrayCount += (rightIdx - leftIdx + 1);
}
return totalSubarrayCount - nonzeroSubarrayCount;
}
private:
static std::vector<int> getSetBitIndices(int x) {
std::vector<int> setBits;
int pow2 = 1;
int exponent = 0;
while (pow2 <= x) {
if ((pow2 & x) != 0) {
setBits.push_back(exponent);
}
exponent++;
pow2 *= 2;
}
return setBits;
}
static bool isBitwiseAndZero(int windowLength, const std::unordered_map<int, int>& bitCounts) {
for (const auto& entry : bitCounts) {
if (entry.second >= windowLength) {
return false;
}
}
return true;
}
};
DE Shaw ✅
C++
Telegram:- @allcoding1
21 529
🎯TCS National Qualifier Test (TCS NQT) 2024
Location: Across India
Qualification: B.E / B.Tech / M.E / M.Tech / M.Sc / MCA / Any Graduate / Under Graduate / Diploma
Batch: 2018/2019/2020/2021/2022/2023/2024
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
21 529
package Graph;
import java.util.*;
public class Largest_Sum_Cycle
{
public static int solution(int arr[])
{
ArrayList<Integer>sum=new ArrayList<>();
for(int i=0;i<arr.length;i++)
{
ArrayList<Integer>path=new ArrayList<>();
int j=i;
int t=0;
while(arr[j]<arr.length&&arr[j]!=i&&arr[j]!=-1&&!path.contains(j))
{
path.add(j);
t+=j;
j=arr[j];
if(arr[j]==i)
{
t+=j;
break;
}
}
if(j<arr.length&&i==arr[j])
sum.add(t);
}
if(sum.isEmpty())
return -1;
return Collections.max(sum);
}
public static void main(String[] args)
{
// TODO Auto-generated method stub
Scanner sc=new Scanner(System.in);
int testcases=sc.nextInt();
for(int loop=0;loop<testcases;loop++)
{
int numofBlocks=sc.nextInt();
int arr[]=new int[numofBlocks];
int src,dest;
for(int i=0;i<numofBlocks;i++)
{
arr[i]=sc.nextInt();
}
System.out.println(solution(arr));
}
}
}
Juspay ✅
Telegram:- @allcoding1
21 529
import java.util.Scanner;
import java.util.*;
public class metting
{
public static void helperFunction()
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[] edges = new int[n];
for (int i = 0; i < n; i++)
{
edges[i] = sc.nextInt();
}
int C1 = sc.nextInt();
int C2 = sc.nextInt();
// int ans=minimumWeight(n,edges,C1,C2);
// System.out.println(ans);
// public static int minimumWeight(int n, int[] edges, int C1, int C2) {
List<List<Integer>> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
list.add(new ArrayList<Integer>());
}
for (int i = 0; i < n; i++) {
if (edges[i] != -1) {
list.get(i).add(edges[i]);
}
}
long[] array1 = new long[n];
long[] array2 = new long[n];
Arrays.fill(array1, Long.MAX_VALUE);
Arrays.fill(array2, Long.MAX_VALUE);
juspay(C1, list, array1);
juspay(C2, list, array2);
int node = 0;
long dist = Long.MAX_VALUE;
for (int i = 0; i < n; i++) {
if (array1[i] == Long.MAX_VALUE || array2[i] == Long.MAX_VALUE)
continue;
if (dist > array1[i] + array2[i]) {
dist = array1[i] + array2[i];
node = i;
}
}
if (dist == Long.MAX_VALUE)
System.out.print(-1);
//return -1;
// return node;
System.out.print(node);
}
private static void juspay(int start, List<List<Integer>> graph, long[] distances)
{
PriorityQueue<Integer> pq = new PriorityQueue<>();
pq.offer(start);
distances[start] = 0;
while (!pq.isEmpty())
{
int curr = pq.poll();
for (int neighbor : graph.get(curr))
{
long distance = distances[curr] + 1;
if (distance < distances[neighbor])
{
distances[neighbor] = distance;
pq.offer(neighbor);
}
}
}
}
public static void main(String[] args) {
metting m = new metting();
metting.helperFunction();
}
}
Nearest meeting Cell
Juspay ✅
Telegram:- @allcoding1
21 529
🎯TCS National Qualifier Test (TCS NQT) 2024
Location: Across India
Qualification: B.E / B.Tech / M.E / M.Tech / M.Sc / MCA / Any Graduate / Under Graduate / Diploma
Batch: 2018/2019/2020/2021/2022/2023/2024
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
21 529
string make_string_S_to_T(string S) {
string T=“programming”;
bool possible = false;
int M = T.length();
int N = S.length();
for (int i = 0; i <= M; i++) {
int prefix_length = i;
int suffix_length = M - i;
string prefix = S.substr(0, prefix_length);
string suffix = S.substr(N - suffix_length, suffix_length);
if (prefix + suffix == T) {
possible = true;
break;
}
}
if (possible)
return "YES";
else
return "NO";
}
Deleting substring ✅
Zeta
Telegram:- @allcoding1
21 529
#include <bits/stdc++.h>
using namespace std;
vector<int> solution(vector<int> a, int n, int k) {
vector<int> v;
deque<int> dq;
for (int i = 0; i < n; i++) {
while (!dq.empty() && dq.front() <= i - k)
dq.pop_front();
while (!dq.empty() && a[dq.back()] <= a[i])
dq.pop_back();
dq.push_back(i);
if (i >= k - 1)
v.push_back(a[dq.front()]);
}
return v;
}
int main() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++)
cin >> a[i];
vector<int> result = solution(a, n, k);
for (int i = 0; i < result.size(); i++)
cout << result[i] << " ";
return 0;
}.
//cricket match ✅
Zeta
Telegram:- @allcoding1
