COGNIZANT EXAM HELP GROUP
3.55K subscribers
5.3K photos
21 videos
10 files
3.22K links
🚀 Placement Preparation Hub

🎓 From Preparation to Placement

OA Support • Coding • Aptitude • Technical • HR

🌟 Trusted by Hundreds of Students

🏆 372+ Placement Successes

📩 DM: @Mrtrueliving_ix

💙 Turning Aspirations into Offer Letters.
Download Telegram
https://careers.morningstar.com/us/en/job/REQ-036608/MDP-Associate


4-5Lpa

Batch 20/21/22/23/24

Test mail confirmed 👍
All placement help available

I will guide today test

DM
@mrtrueliving_ix

@mrtrueliving_ix


💯 Test Clearance
▶️
Please open Telegram to view this post
VIEW IN TELEGRAM
👍2
Successfully cleared Accenture on campus placement

Prev:

https://t.me/code_alphix/2970?single

https://t.me/code_alphix/2975

All placement help available

@mrtrueliving_ix

@mrtrueliving_ix


#Accenture #shortlisted #communication

#Remote access #100% Help
@mrtrueliving_ix☄️

@mrtrueliving_ix▶️


😎All placement help available 😔


👨‍💻🔘 MCQ + Coding Help 🔘 👨‍💻

All Coding+MCQs Exams help Available ❤️

Aptitude + coding Help 👑


Remote Access Available ✔️

Proofs will be provided ✔️✔️

💯😔Genuine Help

🔥All Companies help Available (Genuine + Clearance guarantee) ▶️


Accenture (MCQ+ Coding)
Capgemini
Cognizant
Wipro
Deloitte
Infosys
Tech Mahindra
Paytm
Samsung
Linkdin

All Mnc help available

@mrtrueliving_ix☄️

@mrtrueliving_ix▶️


💯 Test Clearance ✔️


😎All placement help available 😔


👨‍💻🔘 MCQ + Coding Help 🔘 👨‍💻
Juspay coding available

Authentic coding

Verfied codes

DM
@mrtrueliving_ix

💯 Test Clearance 🤗
End of an Era.. RIP Sir 💔💐
💔31🕊1🫡1
Help available

🔈 Accenture

🔈 Cognizant (Communication) -5pm

🔈 Juspay coding

🔈 Accenture (communication)

Any Kind Of OA { on // off}

DM
@mrtrueliving_ix

Test Clearance 🥇
Please open Telegram to view this post
VIEW IN TELEGRAM
IBM// Data science

CODING & SQL are done
▶️

All placement help available

@mrtrueliving_ix

Test accomplished ❤️
Please open Telegram to view this post
VIEW IN TELEGRAM
Please open Telegram to view this post
VIEW IN TELEGRAM
IBM// Data science

CODING & SQL are done ▶️

All placement help available

@mrtrueliving_ix
Please open Telegram to view this post
VIEW IN TELEGRAM
Juspay coding help done


Server down 👎▶️▶️

DM for juspay coding

@mrtrueliving_ix📣
Please open Telegram to view this post
VIEW IN TELEGRAM
Locking the tree of space code available

DM @mrtrueliving_ix 💯👍

Only paid help
Please open Telegram to view this post
VIEW IN TELEGRAM
👍1🔥1
Juspay coding help available

End time :8 pm

Asap complete your coding round

CTS communication help

DM @mrtrueliving_ix ▶️
Please open Telegram to view this post
VIEW IN TELEGRAM
Codes available

DM @mrtrueliving_ix 👍💯
Please open Telegram to view this post
VIEW IN TELEGRAM
string largestMagical(string binString) {
    if (binString.empty()) return binString;
    vector<string> ans;
    int cnt = 0, j = 0;
    for (int i = 0; i < binString.size(); ++i) {
        cnt += binString[i] == '1' ? 1 : -1;
        if (cnt == 0) {
            ans.push_back("1" + largestMagical(binString.substr(j + 1, i - j - 1)) + "0");
            j = i + 1;
        }
    }
    sort(ans.begin(), ans.end(), greater<string>());
    return accumulate(ans.begin(), ans.end(), string{});
}

Nvidia
def max_distance(n, intervals):
intervals.sort(key=lambda x: x[1])
def possible(distance):
prev = intervals[0][0]
for start, end in intervals[1:]:
prev = max(prev + distance, start)
if prev > end:
return False
return True

low, high = 0, intervals[-1][1] - intervals[0][0]
while low <= high:
mid = (low + high) // 2
if possible(mid):
low = mid + 1
else:
high = mid - 1

return low - 1

n = int(input())
intervals = [list(map(int, input().split())) for _ in range(n)]
print(max_distance(n, intervals))

Juspay//The social distancing
#include <stdio.h>
#include <regex.h>
int isPowerOfTwo(const char *binaryString) {
    regex_t regex;
    int result;
    result = regcomp(&regex, "^0*10*$", REG_EXTENDED);
    if (result) {
        printf("Could not compile regex\n");
        return 0;
    }
    result = regexec(&regex, binaryString, 0, NULL, 0);
    regfree(&regex);
    if (!result) {
        return 1;
    } else {
        return 0;
    }
}

int main() {
    int t;
    char binaryString[1000];
    scanf("%d", &t);
    while (t--) {
        scanf("%s", binaryString);
        if (isPowerOfTwo(binaryString)) {
            printf("True\n");
        } else {
            printf("False\n");
        }
    }

    return 0;
}

Nvidia (Regex)