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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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🔥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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📈 Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 13 949 subscribers, ranking 14 646 in the Education category and 28 164 in the India region.

📊 Audience metrics and dynamics

Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 13 949 subscribers.

According to the latest data from 03 November, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -151 over the last 30 days and by -6 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.70%. Within the first 24 hours after publication, content typically collects 1.38% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 516 views. Within the first day, a publication typically gains 193 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

📝 Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
“🔥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...”

Thanks to the high frequency of updates (latest data received on 04 November, 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 Education category.

13 949
Subscribers
-624 hours
-377 days
-15130 days
Posts Archive
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In c language

#include #define MAX_SIZE 100 int main() {     char str[MAX_SIZE];     char toSearch;  &nb
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#include <stdio.h> #define MAX_SIZE 100 int main() {     char str[MAX_SIZE];     char toSearch;     int i, count;       gets(str);        toSearch = getchar();     count = 0;     i=0;     while(str[i] != '\0')     {         if(str[i] == toSearch)         {             count++;         }         i++;     }     printf("%d",count);     return 0; } Tech Mahindra

#include using namespace std; #include using namespace std; int main() { &nbsp; int n; &nbsp; cin &gt;&gt; n; &nbsp; vector i
#include <bits/stdc++.h> using namespace std; #include <bits/stdc++.h> using namespace std; int main() {   int n;   cin >> n;   vector<int> inputArr, answer;   for(int i=0; i<n; ++i) {     int temp;     cin >> temp;     inputArr.push_back(temp);   }   for(int i=0; i<n; ++i)     if(inputArr[i]%2==0)       answer.push_back(inputArr[i]);   for(int i=0; i<n; ++i)     if(inputArr[i]%2!=0)       answer.push_back(inputArr[i]);   for(int i=0; i<n; ++i)     cout << answer[i] << " ";   return 0; } 2nd code -C++ langugage

#include int main() { &nbsp;&nbsp;&nbsp; int num,tNum,digit,cnt; &nbsp;&nbsp;&nbsp; int rem; &nbsp;&nbsp;&nbsp; scanf("%d",&a
#include <stdio.h> int main() {     int num,tNum,digit,cnt;     int rem;     scanf("%d",&num);     scanf("%d",&digit);     cnt=0;     tNum=num;     while(tNum>0)     {         rem=tNum%10;         if(rem==digit)             cnt++;         tNum/=10;     }     printf("%d",cnt);          return 0; } Tech Mahindra first code

Tech Mahindra help done💯💯💯💥💥
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Tech Mahindra help done💯💯💯💥💥

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tcs nqt free help to all if i have 500 members

Please share the group everyone. In this group you will get help for free coding questions

def solution(a,b): res=a.count(b) if res>0: return res else: return 0 a=input() b=input() print(solution(a,b))
def solution(a,b):     res=a.count(b)     if res>0:         return res     else:         return 0     a=input()  b=input() print(solution(a,b))

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