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๐ŸšฉMain Group - @SuperExams
๐Ÿ“Job Updates - @FresherEarth

๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
โš ๏ธDaily Job Updates
โš ๏ธHackathon Updates & Solutions

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๐Ÿ“ŒWe Urgently Need Manual Testers - (QA Engineers FRESHERS) at Ventures Digital India

Role: Associate Manual Tester [QA Engineer]
Qualification: Bachelor's Degree
Industry: IT/Software
Experience: 0-3 Year

Required Skills:
- Basic knowledge of Manual Testing
- Familiarity with Test Case Creation and Defect Management
- Basic knowledge of JIRA Tools

Note- Hybrid Mode [2 Days work from office]

Email- Neha.Toke@venturesdigitalindia.com
ACCENTURE Interview Experience

1) Self Intro ?
2) Project - Major & Minor ?
3) Difficulty Faced in Project?
4) Least subject u like. Why it is least?
5) Hobbies I said in Self Intro
    From Hobbies ( chess) he Asked y u
    like that,  do u play by Moves Names
    or Randomly Last but not the Least.
6) Do u have any Questions ?
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 (intern) โœ…
#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 โœ…
๐Ÿ‘2
#include <iostream>
#include <vector>
#include <algorithm>

const int MOD = 1e9 + 7;

using namespace std;

long long solve(const vector<int>& a, int b) {
    int c = a.size();
    long long d = 0;
    for (int i = 0; i < c; i++) {
        d += a[i];
    }

    vector<long long> e(b + 1, 0);
    e[0] = 1;

    for (int i = 0; i < c; i++) {
        for (int j = b; j >= a[i]; j--) {
            e[j] = (e[j] + e[j - a[i]]) % MOD;
        }
    }
    long long f = 1;
    for (int i = 0; i < c; i++) {
        f = (f * 2) % MOD;
    }

    long long g = 0;
    for (int i = 0; i <= b; i++) {
        g = (g + e[i]) % MOD;
    }

    long long h = (f - 2 * g + MOD) % MOD;
    return h;
}


Sum of the elements โœ…
Oyo
def solve(n, m, arr):
    mod = 10**9 + 7
    fact = [1] * (n + 1)
    for i in range(2, n + 1):
        fact[i] = fact[i - 1] * i % mod

    arr.sort()
    if n == 1:
        return 1

    val = arr.pop()
    arr[-1] += val

    def mod_inverse(a, mod):
        return pow(a, mod - 2, mod)
   
    ans = fact[n]
    temp = 1
    for count in arr:
        temp = (temp * fact[count]) % mod
   
    temp = mod_inverse(temp, mod)
    return (ans * temp) % mod


minimum arrangements โœ…
Oyo