COGNIZANT EXAM HELP GROUP
3.55K subscribers
5.3K photos
21 videos
10 files
3.22K links
๐Ÿš€ Placement Preparation Hub

๐ŸŽ“ From Preparation to Placement

โšก OA Support โ€ข Coding โ€ข Aptitude โ€ข Technical โ€ข HR

๐ŸŒŸ Trusted by Hundreds of Students

๐Ÿ† 372+ Placement Successes โœ…

๐Ÿ“ฉ DM: @Mrtrueliving_ix

๐Ÿ’™ Turning Aspirations into Offer Letters.
Download Telegram
Codes available

DM @mrtrueliving_ix ๐Ÿ‘๐Ÿ’ฏ
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 โœ…
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) โœ…
def battery_conversion(s):
e = 0
result = []
i = 0

while i < len(s):
if i < len(s) - 1 and s[i] == 'D' and s[i + 1] == 'E':
result.append('E')
result.append('C')
e += 1
i += 2
elif i < len(s) - 1 and s[i] == 'E' and s[i + 1] == 'D':
result.append('C')
result.append('E')
e += 1
i += 2
else:
result.append(s[i])
i += 1

for _ in range(1):
pass

return e

s = input()
print(battery_conversion(s))

Juspay// Robot battery
Accenture help available with remote access

โค๏ธ 10 am

โค๏ธ 2 pm

โค๏ธ 7 pm
Previous:

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


https://t.me/code_alphix/2959?single


https://t.me/code_alphix/2964

๐Ÿ’ฏ Test Clearance
โ–ถ๏ธ

๐Ÿ˜ญDon't miss the opportunity ๐Ÿ˜ญ

DM @mrtrueliving_ix โœ…

@mrtrueliving_ix โœ…
Please open Telegram to view this post
VIEW IN TELEGRAM
๐Ÿพ1
Apply asap

Batch 23/24/25/26

Multiple roles

Limited application are accepted


DM for test clearance

https://whatsapp.com/channel/0029VahiS3p2v1IyoS891Y1g/713
All placement help available

LinkedIn

Rupeek sde

Accenture -7pm

Meesho sde

DM for clearance


@Mrtrueliving_ix

@Mrtrueliving_ix

๐Ÿ’ฏ Remote access { Success}

Don't Miss the opportunity
๐Ÿ“ฃ
Check prev proofs โ–ถ๏ธ
Please open Telegram to view this post
VIEW IN TELEGRAM
long minOperations(long n) {
if (n == 0)
return 0;
long x = 1;
while (x * 2 <= n)
x <<= 1;
return minOperations(n ^ (x | (x >> 1))) + 1 + x - 1;
}

Oracle // minOperationsโœ”๏ธ
Please open Telegram to view this post
VIEW IN TELEGRAM
โœ”๏ธ๐Ÿ’ฏ
Please open Telegram to view this post
VIEW IN TELEGRAM
๐Ÿ˜ญ1
All placement help available

0๏ธโƒฃCognizant { aptitude+ Coding}

0๏ธโƒฃLinkedIn

0๏ธโƒฃDe shaw

0๏ธโƒฃRupeek

0๏ธโƒฃAccenture -On campus

0๏ธโƒฃOracle


Any other OA

{ on // Off campus}

With remote access { high success rate}
โ–ถ๏ธ

๐Ÿ’ฏ Test Clearance
โ–ถ๏ธโ˜„๏ธ
-
@mrtrueliving_ix๐Ÿ“ฃ
Please open Telegram to view this post
VIEW IN TELEGRAM
Get ready for

LinkedIn

De Shaw

Math works


@mrtrueliving_ix
COGNIZANT EXAM HELP GROUP pinned ยซAll placement help available 0๏ธโƒฃCognizant { aptitude+ Coding} 0๏ธโƒฃLinkedIn 0๏ธโƒฃDe shaw 0๏ธโƒฃRupeek 0๏ธโƒฃAccenture -On campus 0๏ธโƒฃOracle Any other OA { on // Off campus} With remote access { high success rate}โ–ถ๏ธ ๐Ÿ’ฏ Test Clearance โ–ถ๏ธโ˜„๏ธ -@mrtrueliving_ix๐Ÿ“ฃยป
Done ๐Ÿ‘


def simpleCipher(encrypted, k):
k = k % 26
decrypted = []
for char in encrypted:
old_pos = ord(char) - ord('A')
new_pos = (old_pos - k) % 26
new_char = chr(new_pos + ord('A'))
decrypted.append(new_char)
return ''.join(decrypted)
Please open Telegram to view this post
VIEW IN TELEGRAM
def countPairs(projectCosts, target):
projectCosts.sort()
count = 0
n = len(projectCosts)
i = 0
j = 0
while i < n:
while j < n and projectCosts[j] - projectCosts[i] <= target:
j += 1
count += j - i - 1
i += 1
return count
def minimalOperations(words):
    def count_substitutions(word):
        count = 0
        n = len(word)
        i = 0
        while i < n - 1:
            if word[i] == word[i + 1]:
                count += 1
                i += 2
            else:
                i += 1
        return count
    return [count_substitutions(word) for word in words]
This media is not supported in your browser
VIEW IN TELEGRAM
def findMinincrease(threadSize):
n = len(threadSize)
m = n - 2
if m <= 0:
return 0

groupA_cost = 0
groupB_cost = 0

for f in range(2, n):
index = f - 1
left = threadSize[index - 1]
right = threadSize[index + 1]
target = max(left, right) + 1
cost = target - threadSize[index]
if cost < 0:
cost = 0
if f % 2 == 0:
groupA_cost += cost
else:
groupB_cost += cost

if m % 2 == 1:
return groupA_cost
else:
return min(groupA_cost, groupB_cost)

LinkedIn
MOD = 10**9 + 7

def comb3(n):
if n < 3:
return 0
return n * (n - 1) % MOD * (n - 2) % MOD * pow(6, MOD - 2, MOD) % MOD

def evenproduct(nums):
n = len(nums)
even = sum(1 for num in nums if num % 2 == 0)
odd = n - even

total_triplets = comb3(n)
odd_triplets = comb3(odd)

valid_triplets = (total_triplets - odd_triplets + MOD) % MOD

return valid_triplets


Even product// LinkedIn