#include <bits/stdc++.h>
using namespace std;

long getDataDependenceSum(long n) {
    set<long> dependentDays; 

    for (long k = 1; k * k <= n; ++k) {
        dependentDays.insert(n / k); 
        dependentDays.insert(k);     
    }

    long sum = 0;
    for (auto day : dependentDays) {
        sum += day;
    }
    return sum;
}

Amazon ✅

#include <ibits/stdc++.h>
using namespace std;

int count(int num, int p) {
    int count = 0;
    while (num % p == 0 && num > 0) {
        count++;
        num /= p;
    }
    return count;
}

int solve(int n, vector<vector<int>>& v) {

    vector<vector<int>> dp2(n, vector<int>(n, INT_MAX));
    vector<vector<int>> dp5(n, vector<int>(n, INT_MAX));


    dp2[0][0] = count(v[0][0], 2);
    dp5[0][0] = count(v[0][0], 5);

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (i > 0) {
                dp2[i][j] = min(dp2[i][j], dp2[i - 1][j] + count(v[i][j], 2));
                dp5[i][j] = min(dp5[i][j], dp5[i - 1][j] + count(v[i][j], 5));
            }
            if (j > 0) {
                dp2[i][j] = min(dp2[i][j], dp2[i][j - 1] + count(v[i][j], 2));
                dp5[i][j] = min(dp5[i][j], dp5[i][j - 1] + count(v[i][j], 5));
            }
        }
    }


    return min(dp2[n - 1][n - 1], dp5[n - 1][n - 1]);
}

Treasure Room ✅
Netcore

Anyone got this mail Just React

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For the openings shared in the previous months.
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📌We Urgently Need Manual Testers - (QA Engineers FRESHERS) at Ventures Digital India

Role: Associate Manual Tester [QA Engineer]
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Required Skills:
- Basic knowledge of Manual Testing
- Familiarity with Test Case Creation and Defect Management
- Basic knowledge of JIRA Tools

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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 ?

📌Antino Is Hiring
Role: DevOps Engineer Intern

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MAQ Hiring !!
Role - Developer
Exp - 0 year

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📝 Qualification: Bachelor's degree

🎯 Experience: Freshers

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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 (intern) ✅

#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 ✅

https://leetcode.com/discuss/interview-question/645247/Toptal-or-OA-2020-or-Biggest-Integer-in-a-Matrix  task 3

Amex ✅

Batch 2023/2024

https://docs.google.com/forms/d/e/1FAIpQLSdQ-NzcoWBahQyqn_8fpL0DEWxmN1pu5F8RbRuPw2PY6qks6Q/viewform?pli=1

#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 ✅
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 ✅
Oyo

Company Name : Google
Batch : 2025 passouts
Role : Software Engineer - University Grad

Link : https://www.google.com/about/careers/applications/jobs/results/142576465748075206-software-engineer/