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Test Clearance
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IBM// Data science
CODING & SQL are done▶️
All placement help available
@mrtrueliving_ix
Test accomplished ❤️
CODING & SQL are done
All placement help available
@mrtrueliving_ix
Test accomplished ❤️
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Juspay coding help available
End time :8 pm
Asap complete your coding round
CTS communication help
DM @mrtrueliving_ix▶️
End time :8 pm
Asap complete your coding round
CTS communication help
DM @mrtrueliving_ix
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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(®ex, "^0*10*$", REG_EXTENDED);
if (result) {
printf("Could not compile regex\n");
return 0;
}
result = regexec(®ex, binaryString, 0, NULL, 0);
regfree(®ex);
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) ✅
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
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Previous:
https://t.me/code_alphix/2970?single
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https://t.me/code_alphix/2964
💯 Test Clearance▶️
😭 Don't miss the opportunity 😭
DM @mrtrueliving_ix✅
@mrtrueliving_ix✅
Previous:
https://t.me/code_alphix/2970?single
https://t.me/code_alphix/2959?single
https://t.me/code_alphix/2964
💯 Test Clearance
DM @mrtrueliving_ix
@mrtrueliving_ix
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🍾1
Apply asap
Batch 23/24/25/26
Multiple roles
Limited application are accepted
DM for test clearance
https://whatsapp.com/channel/0029VahiS3p2v1IyoS891Y1g/713
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
💯 Remote access { Success}
Don't Miss the opportunity📣
Check prev proofs▶️
Rupeek sde
Accenture -7pm
Meesho sde
DM for clearance
@Mrtrueliving_ix
@Mrtrueliving_ix
💯 Remote access { Success}
Don't Miss the opportunity
Check prev proofs
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