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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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#include <iostream>
#include <set>
#include <cmath>
using namespace std;
int ss(int N) {
    set<int> s;
    for (int p = 1; p * p <= N; ++p)
    {
    int r = p * p;
    for (int q = 1; q * q * q <= N; ++q)
    {
    int cube = q * q * q;
    int sum = r + cube;
    if (sum <= N)
    {
    s.insert(sum);
    }
    }
    }
    return s.size();
}
int main()
{
    int N;
    cin >> N;
    cout << ss(N) << endl;
    return 0;
}


Count Perfect Sums โœ…
Airtel
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๐Ÿ“ŒCompany Name: Orange mantra
Designation: Analyst

CTC: Up to 4 LPA
Qualification: Btech/MBA
Experience: Freshers or 6 months
Location: Gurgaon, Haryana

Employment type: Full Time
No of working days: 5

Skills Required
-Excellent Communication & Presentation skills
-Proficient in MS-office
-Data Analysis & reporting
-Research & Analytical skills.

*Note*: Immediate joiners preferred
Share your CV: chaudhary.pratima@orangemantra.in
long getMaxRequests(vector<int>& sc, vector<int>& ir, int k) {
    int n = sc.size();
    long total = 0;
    vector<int> extra(n, 0);

    for (int i = 0; i < n; i++) {
        int h = min(sc[i], ir[i]);
        total += h;

        if (sc[i] < ir[i]) {
            int dch = min(2 * sc[i], ir[i]);
            extra[i] = dch - h;
        }
    }

    sort(extra.rbegin(), extra.rend());

    for (int i = 0; i < k; i++) {
        total += extra[i];
    }

    return total;
}

Twillo โœ…
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
int getMaxOccurrences(string components, int minLength, int maxLength, int maxUnique) {
    const int mod = 1e9 + 7;
    map<long, int> freq;
    vector<int> count(26, 0);
    int l = 0;
    int ans = 0;
    int noOfLetter = 0;
    long hash_curr = 0;
    long pow = 1;
   
    for (int i = 0; i < minLength - 1; i++) {
        pow = (pow * 53) % mod;
    }
   
    for (int r = 0; r < components.size(); r++) {
        count[components[r] - 'a']++;
        if (count[components[r] - 'a'] == 1) {
            noOfLetter++;
        }
        if ((r - l + 1) > minLength) {
            hash_curr = ((hash_curr - (components[l] - 'a' + 1) * pow) % mod + mod) % mod;
            count[components[l] - 'a']--;
            if (count[components[l] - 'a'] == 0) {
                noOfLetter--;
            }
            l++;
        }
        hash_curr = ((hash_curr * 53) % mod + (components[r] - 'a' + 1)) % mod;
        if ((noOfLetter <= maxUnique) && (r - l + 1) >= minLength && (r - l + 1) <= maxLength) {
            ans = max(ans, ++freq[hash_curr]);
        }
    }
    return ans;
}

Twillo โœ…
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