allcoding1_official
前往频道在 Telegram
📈 Telegram 频道 allcoding1_official 的分析概览
频道 allcoding1_official (@allcoding1_official) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 84 838 名订阅者,在 技术与应用 类别中位列第 1 497,并在 印度 地区排名第 3 505 位。
📊 受众指标与增长动态
自 невідомо 创建以来,项目保持高速增长,吸引了 84 838 名订阅者。
根据 06 七月, 2026 的最新数据,频道保持稳定运转。过去 30 天订阅人数变化为 -1 554,过去 24 小时变化为 -61,整体触达仍然可观。
- 认证状态: 未认证
- 互动率 (ER): 平均受众互动率为 2.83%。内容发布后 24 小时内通常能获得 0.90% 的反应,占订阅者总量。
- 帖子覆盖: 每篇帖子平均可获得 2 405 次浏览,首日通常累积 762 次浏览。
- 互动与反馈: 受众积极参与,单帖平均反应数为 1。
- 主题关注点: 内容集中在 dsa, stack, namaste, javascript, dev 等核心主题上。
📝 描述与内容策略
尚未提供频道描述。
凭借高频更新(最新数据采集于 07 七月, 2026),频道始终保持新鲜度与高覆盖。分析显示受众积极互动,使其成为 技术与应用 类别中的关键影响点。
84 838
订阅者
-6124 小时
-3767 天
-1 55430 天
帖子存档
84 831
Guys ❤️
TCS NQT
Answers are available
Instagram
Once check it
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84 831
Guys ❤️
Answers are available
Instagram
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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
🎯 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
🎯 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;
}
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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;
}
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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
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