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π Analytical overview of Telegram channel allcoding1_official
Channel allcoding1_official (@allcoding1_official) in the English language segment is an active participant. Currently, the community unites 84 831 subscribers, ranking 1 497 in the Technologies & Applications category and 3 505 in the India region.
π Audience metrics and dynamics
Since its creation on Π½Π΅Π²ΡΠ΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 84 831 subscribers.
According to the latest data from 06 July, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -1 554 over the last 30 days and by -61 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 2.83%. Within the first 24 hours after publication, content typically collects 0.90% reactions from the total number of subscribers.
- Post reach: On average, each post receives 2 405 views. Within the first day, a publication typically gains 762 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 1.
- Thematic interests: Content is focused on key topics such as dsa, stack, namaste, javascript, dev.
π Description and content policy
Channel description not provided.
Thanks to the high frequency of updates (latest data received on 07 July, 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 Technologies & Applications category.
84 831
Subscribers
-6124 hours
-3767 days
-1 55430 days
Posts Archive
84 831
Guys β€οΈ
TCS NQT
Answers are available
Instagram
Once check it
https://instagram.com/allcoding_1?igshid=YmMyMTA2M2Y=
https://instagram.com/allcoding_1?igshid=YmMyMTA2M2Y=
Telegram:-@allcoding1
84 831
Guys β€οΈ
Answers are available
Instagram
https://instagram.com/allcoding_1?igshid=YmMyMTA2M2Y=
https://instagram.com/allcoding_1?igshid=YmMyMTA2M2Y=
Telegram:-@allcoding1
84 831
TCS NQT Verbal
Q)I advised him
Ans) d
Q) 2020 was
Ans) no error
Q) neither of the four sons
Ans) neither or nor
Q)The adverse impacts
Ans) C and d
Q) With the spread
Ans) QR
Q) pranay
Ans) has been working
Telegram - @allcoding1
84 831
TCS NQT 9 am Answers
1- X-5
2-14
3-8
4-50
5-91.2
6-735
7-166.66%
9-12000
10-4200
11-1900,2100
14-3
17-22 min
Telegram:-@allcoding1
84 831
π― Accenture Associate Software Engineer Hiring | 4.5 LPA
Job Title : Associate Software Engineer
Qualification : B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
Batch : 2022 / 2021
Salary : Rs 4.5 LPA
Apply Now:- https://www.allcoding1.com/2022/07/accenture-associate-software-engineer.html
Telegram:-@allcoding1
84 831
π― Accenture Associate Software Engineer Hiring | 4.5 LPA
Job Title : Associate Software Engineer
Qualification : B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
Batch : 2022 / 2021
Salary : Rs 4.5 LPA
Apply Now:- https://www.allcoding1.com/2022/07/accenture-associate-software-engineer.html
Telegram:-@allcoding1
84 831
#include <bits/stdc++.h>
#define n 3
using namespace std;
int findLongestFromACell(int i, int j, int mat[n][n], int dp[n][n])
{
if (i < 0 i >= n j < 0 || j >= n)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int x = INT_MIN, y = INT_MIN, z = INT_MIN, w = INT_MIN;
if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
x = 1 + findLongestFromACell(i, j + 1, mat, dp);
if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
y = 1 + findLongestFromACell(i, j - 1, mat, dp);
if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
z = 1 + findLongestFromACell(i - 1, j, mat, dp);
if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
w = 1 + findLongestFromACell(i + 1, j, mat, dp);
return dp[i][j] = max({x, y, z, w, 1});
}
int LongestPath(int mat[n][n])
{
int result = 1;
int dp[n][n];
memset(dp, -1, sizeof dp);
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
result = max(result, dp[i][j]);
}
}
return result;
}
int main()
{
int mat[n][n] = {{1, 2, 9}, {5, 3, 8}, {4, 6, 7}};
cout << "Length of the longest path is "
<< LongestPath(mat);
return 0;
}
Telegram:-@allcoding1
84 831
#include <bits/stdc++.h>
#define n 3
using namespace std;
int findLongestFromACell(int i, int j, int mat[n][n], int dp[n][n])
{
if (i < 0 i >= n j < 0 || j >= n)
return 0;
if (dp[i][j] != -1)
return dp[i][j];
int x = INT_MIN, y = INT_MIN, z = INT_MIN, w = INT_MIN;
if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
x = 1 + findLongestFromACell(i, j + 1, mat, dp);
if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
y = 1 + findLongestFromACell(i, j - 1, mat, dp);
if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
z = 1 + findLongestFromACell(i - 1, j, mat, dp);
if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
w = 1 + findLongestFromACell(i + 1, j, mat, dp);
return dp[i][j] = max({x, y, z, w, 1});
}
int LongestPath(int mat[n][n])
{
int result = 1;
int dp[n][n];
memset(dp, -1, sizeof dp);
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
result = max(result, dp[i][j]);
}
}
return result;
}
int main()
{
int mat[n][n] = {{1, 2, 9}, {5, 3, 8}, {4, 6, 7}};
cout << "Length of the longest path is "
<< LongestPath(mat);
return 0;
}
Telegram:-@allcoding1
84 831
// Java program to find next greater
// number with same set of digits.
import java.util.Arrays;
public class nextGreater
{
// Utility function to swap two digit
static void swap(char ar[], int i, int j)
{
char temp = ar[i];
ar[i] = ar[j];
ar[j] = temp;
}
// Given a number as a char array number[],
// this function finds the next greater number.
// It modifies the same array to store the result
static void findNext(char ar[], int n)
{
int i;
// I) Start from the right most digit
// and find the first digit that is smaller
// than the digit next to it.
for (i = n - 1; i > 0; i--)
{
if (ar[i] > ar[i - 1]) {
break;
}
}
// If no such digit is found, then all
// digits are in descending order means
// there cannot be a greater number with
// same set of digits
if (i == 0)
{
System.out.println("Not possible");
}
else
{
int x = ar[i - 1], min = i;
// II) Find the smallest digit on right
// side of (i-1)'th digit that is greater
// than number[i-1]
for (int j = i + 1; j < n; j++)
{
if (ar[j] > x && ar[j] < ar[min])
{
min = j;
}
}
// III) Swap the above found smallest
// digit with number[i-1]
swap(ar, i - 1, min);
// IV) Sort the digits after (i-1)
// in ascending order
Arrays.sort(ar, i, n);
System.out.print("Next number with same" +
" set of digits is ");
for (i = 0; i < n; i++)
System.out.print(ar[i]);
}
}
public static void main(String[] args)
{
char digits[] = { '5','3','4','9','7','6' };
int n = digits.length;
findNext(digits, n);
}
}
Java
Find next greater
Telegram - @allcoding1
