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allcoding1_official

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📈 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

📊 受众指标与增长动态

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
-1 55430
帖子存档
Guys ❤️ 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

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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

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#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

#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

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// 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