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

๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
โš ๏ธDaily Job Updates
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Buy ads: https://telega.io/c/cs_algo
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Changepond recruiting Fresher_Software Developers and Testers_Registration Link given Below
Skills Required: JAVA / Net / Testing/ RPA / SQL/ PHP with Logical and analytical thinking towards programming.
Eligibility:
B.E. / B.Tech/ME/Mtech CSE / IT / ECE / EEE
B.Sc (Computer Science).BCA
MCA,MSC ( Computer Science )
Graduated in 2022 or 2023 Only with 70% and above in all academics
Good Communication
Job Location: SIPCOT, Siruseri, Chennai

Apply link : https://forms.office.com/Pages/ResponsePage.aspx?id=xXv5k9nOHE6DUIaS7Za9oqyRiK_P05VPnXJDDjk1bk1UMjRWVEtJM01LRERZUkNQVVJEWVhHUjlZWi4u
FresherOpportunityAlert

Macro Technology Solutions Hiring Freshers || .Net Trainee Developer for Trichy.

We are looking for candidates with a UG/PG Computer Science background (i.e., 2022/2023/2024) who have strong practical and theoretical knowledge in C/C++, OOPs, SQL, and RDBMS. The candidate should be based out of Trichy and available for immediate joining.

Title: .Net Trainee Developer
Location: Trichy
Duration: Full Time

Terms & Conditions:

The candidate should be from any UG/PG Computer science background (i.e., 2022/2023/2024 passed out).
The candidate should be based out of Trichy (i.e., home location).
The candidate should have consistently scored more than 75% in 10th, 12th, and graduation/post-graduation courses.
We are looking for immediate joiners only.

Interested can mail to hireme@macroglobal.co.uk
Freshers/Interns for the HR Internship for our organization.

Job Responsibilities-

โ€ขSourcing candidates through various channels, including job portals, social media platforms, and professional networks.
โ€ขScreen resumes and conduct initial interviews to assess candidate qualifications and suitability for open positions.
โ€ขCoordinate and schedule interviews between candidates and hiring managers, ensuring a seamless and efficient recruitment process.
โ€ขContribute to new employee orientation and paperwork.
โ€ขDrafting job descriptions and posting job openings on relevant platforms to
attract qualified candidates.
โ€ขCollaborate with team members on special projects and initiatives to enhance our recruitment strategies and processes.

Location: Gurugram, Haryana
No. of working days: 5 days

*Immediate joiners and candidates from Delhi NCR region or nearby areas preferred

If you find the above mentioned profile suitable or have any references for the same, kindly share the updated resumes at dhanashree.surve@cybercube.co.in

Regards,
Dhanashree Surve
HR Executive
CyberCube Services Pvt. Ltd.
Company Name : Collins
Role : Internship at Collins and Prat and Whitney
Batch : 2025 female passouts from mechanical and allied branches only (Aerospace, Mechatronics, Industrial Engineering Management & Automobile)
Prizes worth 1 Lakh and Samsung Tab too.
Link : https://bit.ly/CollinsInternship

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

bool wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> dict(wordDict.begin(), wordDict.end());
    vector<bool> dp(s.size() + 1, false);
    dp[0] = true;
   
    for (int i = 1; i <= s.size(); ++i) {
        for (int j = i - 1; j >= 0; --j) {
            if (dp[j]) {
                string word = s.substr(j, i - j);
                if (dict.find(word) != dict.end()) {
                    dp[i] = true;
                    break;
                }
            }
        }
    }
    return dp[s.size()];
}

int main() {
    string s = "penappleapple";
    vector<string> wordDict = {"apple","pen"};
    cout << (wordBreak(s, wordDict) ? "true" : "false") << endl;
    return 0;
}

Kitty in horror house
Salesforce โœ…
#include <iostream>
#include <vector>
#include <set>

using namespace std;

int getSmallestArea(vector<vector<int>>& grid) {
    int rows = grid.size();
    if (rows == 0) return 0;
    int cols = grid[0].size();
    if (cols == 0) return 0;

    set<int> rowsSet, colsSet;

    for (int i = 0; i < rows; ++i) {
        for (int j = 0; j < cols; ++j) {
            if (grid[i][j] == 1) {
                rowsSet.insert(i);
                colsSet.insert(j);
            }
        }
    }

    int width = colsSet.empty() ? 0 : *colsSet.rbegin() - *colsSet.begin() + 1;
    int height = rowsSet.empty() ? 0 : *rowsSet.rbegin() - *rowsSet.begin() + 1;

    return width * height;
}

Salesforce โœ…
Get smallest Area
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€
int closestToTarget(vector<int> oxygenLevels, int target) {     int n = oxygenLevels.size();     int minDiff = abs(oxygenLevels[0] - target); // Initialize the minimum difference with the first oxygen level     // Iterate through the oxygen levels to findโ€ฆ
#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int closestToTarget(vector<int>& oxygenLevels, int target) {
    int n = oxygenLevels.size();
    int minDiff = INT_MAX;

    for (int i = 0; i < n; ++i) {
        int andValue = oxygenLevels[i];

        for (int j = i; j < n; ++j) {
            andValue &= oxygenLevels[j];

            minDiff = min(minDiff, abs(andValue - target));
            if (andValue == 0) break;
        }
    }

    return minDiff;
}

Optimum oxygen
Salesforce โœ…
#include <unordered_map>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
using namespace std;

class designCache {
public:
    class Node {
    public:
        int key;
        int val;
        Node* prev;
        Node* next;

        Node(int key, int val) {
            this->key = key;
            this->val = val;
        }
    };

    Node* head = new Node(-1, -1);
    Node* tail = new Node(-1, -1);

    int cap;
    unordered_map<int, Node*> m;

    designCache(int capacity) {
        cap = capacity;
        head->next = tail;
        tail->prev = head;
    }

    void addNode(Node* newnode) {
        Node* temp = head->next;

        newnode->next = temp;
        newnode->prev = head;

        head->next = newnode;
        temp->prev = newnode;
    }

    void deleteNode(Node* delnode) {
        Node* prevv = delnode->prev;
        Node* nextt = delnode->next;

        prevv->next = nextt;
        nextt->prev = prevv;
    }

    int get(int key) {
        if (m.find(key) != m.end()) {
            Node* resNode = m[key];
            int ans = resNode->val;

            m.erase(key);
            deleteNode(resNode);
            addNode(resNode);

            m[key] = head->next;
            return ans;
        }
        return -1;
    }

    void put(int key, int value) {
        if (m.find(key) != m.end()) {
            Node* curr = m[key];
            m.erase(key);
            deleteNode(curr);
        }

        if (m.size() == cap) {
            m.erase(tail->prev->key);
            deleteNode(tail->prev);
        }

        addNode(new Node(key, value));
        m[key] = head->next;
    }
};

int main() {
    string attribs, attrib;
    getline(cin, attribs);
    stringstream ss(attribs);
    int capacity, key, val;
    cin >> capacity;
    designCache obj(capacity);
    while (getline(ss, attrib, ' ')) {
        if (attrib == "put") {
            cin >> key >> val;
            obj.put(key, val);
            cout << "null ";
        } else if (attrib == "get") {
            cin >> key;
            val = obj.get(key);
            cout << val << " ";
        }
    }
    return 0;
}.

Design a cache โœ…
static int maximize(int arr[], int n)
    {
        int prefixSum[] = new int[n];
       
        int totalSum = 0;
   
    
        int maxPrefixSum = 0;
       
        for (int i = 0; i < n; i++)
        {
            prefixSum[i] = maxPrefixSum ;
           
            maxPrefixSum += arr[i];
            totalSum += arr[i];
           
            maxPrefixSum = Math.max(maxPrefixSum, -totalSum);
        }
       
        int maxSum = Math.max(totalSum, maxPrefixSum);
       
        int suffixSum = 0;
        for (int i = n - 1; i >= 0; --i)
        {
            suffixSum -= arr[i];
   
 
            maxSum = Math.max(maxSum, suffixSum + prefixSum[i]);
        }
       
        return maxSum;
    }.  add this if size == 1:
return arr[0]

DE Shaw โœ