KPIT 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

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印度28 046 教育14 030...

📈 Telegram 频道 KPIT EXAM SOLUTIONS 的分析概览

频道 KPIT 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 14 152 名订阅者，在教育类别中位列第 14 030，并在印度地区排名第 28 046 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 14 152 名订阅者。

根据 17 九月, 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 -126，过去 24 小时变化为 -13，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 4.44%。内容发布后 24 小时内通常能获得 1.62% 的反应，占订阅者总量。
帖子覆盖： 每篇帖子平均可获得 629 次浏览，首日通常累积 229 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 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 18 九月, 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。

14 152

订阅者

-1324 小时

-117 天

-12630 天

629

帖子浏览量

~ 22924 小时

~ 30948 小时

4.44%

参与率

~ 3

每日帖子数

Ads index

beta

帖子存档

14 148

13.3k after i will post ✅ All tests passed code will Upload again

14 148

https://t.me/coding_are Share this group with your friends and college group.. I will upload next answer in 5min

14 148

import sys input = sys.stdin.readline def solve(n: int, l: int, a: list) -> int: if len(a) < n: a += list(map(int, sys.stdin.read().split())) cur = 1 ans = 1 for i in range(1, n): if abs(a[i] - a[i - 1]) <= l: cur += 1 else: cur = 1 ans = max(ans, cur) return ans Write with function

14 148

Next answer in 5min Give ♥️

14 148

https://t.me/coding_are Share this group with your friends and college group.. I will upload next answer in 5min

14 148

import sys input = sys.stdin.readline def solve(S: str) -> int: def cost(ch): pos = [i for i, c in enumerate(S) if c == ch] k = len(pos) if k <= 1: return 0 q = [pos[i] - i for i in range(k)] m = q[k // 2] return sum(abs(x - m) for x in q) return min(cost('0'), cost('1')) Full passed ✅

14 148

https://t.me/coding_are Share this group with your friends and college group.. I will upload next answer in 5min

14 148

#include <bits/stdc++.h> using namespace std; int solve(int N, int K, vector<int>& A) { sort(A.begin(), A.end()); int i = 0; int c = 0; while (i < N - 1) { if (abs(A[i] - A[i + 1]) <= K) { c++; i += 2; } else { i += 1; } } return c; } int main() { int N, K; cin >> N >> K; vector<int> A(N); for (int i = 0; i < N; i++) { cin >> A[i]; } cout << solve(N, K, A) << endl; return 0; } Full passed ✅ Check function name the. Write 👍

14 148

Ready for next code ♥️

14 148

import sys input = sys.stdin.readline def solve(n: int, a: list, b: list) -> int: a.sort() b.sort() i = j = 0 wins = 0 while i < n and j < n: if a[i] > b[j]: wins += 1 i += 1 j += 1 else: i += 1 return wins n = int(input()) a = [int(input()) for _ in range(n)] b = [int(input()) for _ in range(n)] print(solve(n, a, b)) https://t.me/coding_are Share with you frd

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Wait 5min everyone...

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