ACCENTURE EXAM SOLUTIONS
前往频道在 Telegram
🔥Guys plz Stop fearing for daily exams 📝 👨💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk
显示更多📈 Telegram 频道 ACCENTURE EXAM SOLUTIONS 的分析概览
频道 ACCENTURE EXAM SOLUTIONS (@coding_are) 英语 语言赛道中的 是活跃参与者。目前社区聚集了 14 041 名订阅者，在 教育 类别中位列第 14 513，并在 印度 地区排名第 28 436 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 14 041 名订阅者。
根据 13 十月, 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 -124，过去 24 小时变化为 -6，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 3.63%。内容发布后 24 小时内通常能获得 1.83% 的反应，占订阅者总量。
- 帖子覆盖： 每篇帖子平均可获得 510 次浏览，首日通常累积 257 次浏览。
- 互动与反馈： 受众积极参与，单帖平均反应数为 2。
- 主题关注点： 内容集中在 placement, gaurntee, suree, capgemini, infosy 等核心主题上。
📝 描述与内容策略
作者将该频道定位为表达主观观点的平台：
“🔥Guys plz Stop fearing for daily exams 📝
👨💻 @srksvk is here to help you all at lowest cost possible.💪
🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company
🔥Effort from our side = 💯
📱Main Channel: @coding_are
📱Tel I'd : @srks...”
凭借高频更新（最新数据采集于 14 十月, 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 教育 类别中的关键影响点。
14 042
订阅者
-624 小时
-377 天
-12430 天
帖子存档
14 041
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elitmus , Deloitte (26, 27, 28, All slots), HCL, IBM, any other exams
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Only codeing help also available ✅
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int f(int i, int j, int flag, vector& arr, int sum, vector>>& dp) {
if (i >= j) return 0;
if (dp[i][j][flag] != -1) return dp[i][j][flag];
int firsttwo = 0;
int lasttwo = 0;
int firstlast = 0;
if (i + 1 < arr.size() && arr[i] + arr[i + 1] == sum) {
firsttwo = 1 + f(i + 2, j, flag, arr, sum, dp);
}
if (arr[i] + arr[j] == sum) {
firstlast = 1 + f(i + 1, j - 1, flag, arr, sum, dp);
}
if (j >= 0 && arr[j] + arr[j - 1] == sum) {
lasttwo = 1 + f(i, j - 2, flag, arr, sum, dp);
}
return dp[i][j][flag] = max({firsttwo, lasttwo, firstlast});
}
int solution(vector& arr) {
int n = arr.size();
vector>> dp(n, vector>(n, vector(3, -1)));
int sum1 = arr[0] + arr[n - 1];
int sum2 = arr[0] + arr[1];
int sum3 = arr[n - 1] + arr[n - 2];
int ans1 = f(0, n - 1, 0, arr, sum1, dp);
int ans2 = f(0, n - 1, 1, arr, sum2, dp);
int ans3 = f(0, n - 1, 2, arr, sum3, dp);
int res = max(ans1, max(ans2, ans3));
return res;
}
14 041
#include <bits/stdc++.h>
using namespace std;
int solution(vector<int>& A) {
sort(A.begin(), A.end());
if (A.size() <= 2) {
return A.size();
}
int maxCount = 0;
unordered_map<int, int> differences[A.size()];
for (int i = 1; i < A.size(); i++) {
for (int j = 0; j < i; j++) {
int currentDiff = A[i] - A[j];
int currentCnt = 1;
if (differences[j].count(currentDiff) != 0) {
currentCnt += differences[j][currentDiff];
differences[i][currentDiff] = currentCnt;
} else {
differences[i][currentDiff] = 1 + currentCnt;
}
maxCount = max(maxCount, differences[i][currentDiff]);
}
}
return maxCount;
}
task 2
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