int solution(vector<int>arr){
    int ans=INT_MIN;
    int result=-1;
    vector<int>weight(arr.size(),0);
    for(int i=0;i<arr.size();i++){
        int source=i;
        int dest=arr[i];
        if(dest!=-1){
            weight[dest]+=source;
            if(ans<=weight[dest]){
                ans=max(ans,weight[dest]);
                result=dest;
            }
           
        }
    }
    if(ans!=INT_MIN)
        return result;
    return -1;
}

Maximum Weight Node ✅

Company Name: HackerEarth

🛄 Job Title: Problem Setter Intern
✍🏻 YOE: 2025, 2026 and 2027 grads

➡️ Apply: https://hackerearth.applytojob.com/apply/cEkOgMaelD/Technical-Engineer-Problem-Setter-Programming--Internship

Please do share in your college grps and in case you are applying please react on this post:) 😇❤️

Company: Rakuten Symphony
Location: N/A
Role: Associate Software Engineer
For Graduates: 2024, 2023

https://rakuten.skillate.com/jobs/54126

Requires -

Graduate & Diploma Engineer Trainee

Bharat FIH

Job Description :

He/She will be deployed in the department based on their interest and suitability.
He/She will work along with the team members and perform their tasks, which is assigned by his/her superior.
He or She should perform well for their position to confirm after successfully completed their training period.
He or She has to participate in the department meeting and present their project or task when is required.
He/She has to deliver their tasks on time.

Education & Skills :

Diploma Engineer Trainee (DET)

Diploma Graduates (EEE/ECE/Mech/Mechatronics/Robotics) who have passed out 2022 / 2023
Good Academic Records
No Standing Arrears

Skill Set :

Good Communication skills (Verbal & Written)
Eager to learn
Pleasing Personality
Good in System Knowledge (MS Office)

Graduate Engineer Trainee (GET) :

BE/B.Tech Graduates (EEE/ECE/Mech/Mechatronics/Robotics) who have passed out 2022 / 2023
Good Academic Records
No Standing Arrears

Skill Set :

Good Communication skills (Verbal & Written)
Eager to learn
Pleasing Personality
Good in System Knowledge (MS Office)

Note : Immediate Joiners will be preferred

Working Place: Sunguvarchatram Plant

You can share your application to careers@fih-foxconn.com (Subject Line : Applied for the position of GET or DET Batch 2022/2023)

FresherOpporunityAlert

Vembu Technologies is hiring Freshers 👨‍💼

🙋🏻‍♀️Product Success Engineer and QA Engineer Trainee

🎓B.E/B.Tech- CS/IT -2023 & 2024 batch passed out students

🙌Note: Minimum 70 % academic record required to attend the interview ( SSLC, HSC, UG) No standing Arears

🧳Salary-3.00 to 3.6 Lakhs/ CTC

👨‍💼Interview and selection process :-

Written Test (Aptitude, English and Technical)
Face 2 Face Interview

📧 Interested Candidates can email their resume with prabu@vembu.com

#include <iostream> 
#include <vector> 
#include <unordered_map> 
#include <queue> 
#include <algorithm> 
#include <unordered_set> 
using namespace std; 

vector<string> fep(unordered_map<string, vector<string>>& g, string s) { 
    vector<string> st = {s}; 
    deque<string> p; 
    while (!st.empty()) { 
        string u = st.back(); 
        if (!g[u].empty()) { 
            string v = g[u].front(); 
            g[u].erase(g[u].begin()); 
            st.push_back(v); 
        } else { 
            p.push_front(st.back()); 
            st.pop_back(); 
        } 
    } 
    return vector<string>(p.begin(), p.end()); 
} 

int main() { 
    int n, m; 
    cin >> n >> m; 
    unordered_map<string, vector<string>> g; 
    unordered_set<string> ads; 
    for (int i = 0; i < n; i++) { 
        string a, b; 
        cin >> a >> b; 
        g[a].push_back(b); 
        ads.insert(a); 
        ads.insert(b); 
    } 
    for (auto it = g.begin(); it != g.end(); ++it) { 
        sort(it->second.begin(), it->second.end()); 
    } 
    string s = "ABC"; 
    vector<string> p = fep(g, s); 
    for (const auto& x : p) { 
        cout << x << " "; 
    } 
    cout << endl; 
    return 0; 
}

Flipkart ✅

using namespace std;

vector<int> rc;
vector<vector<int>> adj;
vector<int> seq;

void dfs(int node) {
    if(adj[node].size() > 0) dfs(adj[node][0]);
    seq.push_back(rc[node]);
    if(adj[node].size() > 1) dfs(adj[node][1]);
}

int main() {
    int rCount;
    cin >> rCount;
    rc.resize(rCount);
    for(int i=0; i<rCount; i++) cin >> rc[i];
    int cRow, cCol;
    cin >> cRow >> cCol;
    adj.resize(rCount);
    for(int i=0; i<cRow; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
    }
    int rM;
    cin >> rM;
    dfs(0);
    cout << seq[rM-1] << endl;
    return 0;
}

Restaurant ✅
Flipkart

using namespace std;

void pop_front(vector<int> &v) {
    if (v.size() > 0) {
        v.erase(v.begin());
    }
}

vector<int> helper(vector<int> nums, int c_len) {
    vector<int> ans;
    int In = nums.size();
    for(int i = 0; i < In; i++) {
        while(ans.size() > 0 && ans.back() < nums[i] && ((In - i) > (c_len - ans.size()))) {
            ans.pop_back();
        }
        if(ans.size() < c_len) {
            ans.push_back(nums[i]);
        }
    }
    return ans;
}

vector<int> solve(vector<int> nums1, vector<int> nums2, int k) {
    int l1 = nums1.size();
    int l2 = nums2.size();
    vector<int> rs;
    for (int s1 = max(0, k - l2); s1 <= min(k, l1); s1++) {
        vector<int> p1, p2;
        p1 = helper(nums1, s1);
        p2 = helper(nums2, k - s1);
        vector<int> temp;
        for (int j = 0; j < k; j++) {
            vector<int> temp2 = max(p1, p2);
            int fr = temp2.front();
            if (p1 > p2) {
                pop_front(p1);
            } else {
                pop_front(p2);
            }
            temp.push_back(fr);
        }
        rs = max(rs, temp);
    }
    return rs;
}

int main() {
    int n;
    cin >> n;
    vector<int> arr1(n);
    for(int i = 0; i < n; i++) {
        cin >> arr1[i];
    }
    int m;
    cin >> m;
    vector<int> arr2(m);
    for(int i = 0; i < m; i++) {
        cin >> arr2[i];
    }
    int k;
    cin >> k;
    vector<int> v = solve(arr1, arr2, k);
    for(int i = 0; i < v.size(); i++) {
        cout << v[i] << " ";
    }
    return 0;
}

University
Flipkart ✅

Meesho is hiring for SDE 1 role.

Batch eligible: 2022 and 2023 grads only

Link: https://docs.google.com/forms/d/e/1FAIpQLSccczQRZsQPXR_GPaomStgK2Vem3MspsmiLBJWC0FDXMVWjuw/viewform

Note: this is the referral form and will take only 200 resumes and refer best out of that!!

Parent node ✅

int minimumChunksRequired(long long totalPackets,vector<vector<long long>>uploadedChunks){
    long long ans=0;
    long long prev=0;

    sort(uploadedChunks.begin(),uploadedChunks.end());

    for(int i=0;i<uploadedChunks.size();i++){
        if(uploadedChunks[i][0]==prev+1){
            prev=uploadedChunks[i][1];
            continue;
        }
        else{

            long diff=uploadedChunks[i][0]-prev-1;


            for(int k=0;k<64;k++){
                if(diff&(1LL<<k)){
                    ans++;
                }
            }

            prev=uploadedChunks[i][1];
        }
    }

    long diff=totalPackets-prev;

    for(int k=0;k<64;k++){
        if(diff&(1LL<<k)){
            ans++;
        }
    }

    return ans;

}

#include <iostream>
#include <sstream>
#include <vector>
#include <algorithm>
#include <cctype>

using namespace std;

string encodeString(const string &s) {
    stringstream ss(s);
    string word;
    vector<string> words;

    while (ss >> word) {
        reverse(word.begin(), word.end());
        words.push_back(word);
    }

    string reversedSentence;
    for (const auto &w : words) {
        reversedSentence += w + " ";
    }
    reversedSentence = reversedSentence.substr(0, reversedSentence.size() - 1);

    string encodedChars;
    for (size_t i = 0; i < reversedSentence.size(); ++i) {
        char c = reversedSentence[i];
        if (i % 2 == 0) {
            encodedChars += toupper(c);
        } else {
            encodedChars += tolower(c);
        }
    }

    for (size_t i = 0; i < encodedChars.size(); ++i) {
        if (encodedChars[i] == ' ') {
            encodedChars[i] = (i % 2 == 0) ? '#' : '*';
        } else if (encodedChars[i] == ',') {
            encodedChars[i] = '@';
        } else if (encodedChars[i] == '.') {
            encodedChars[i] = '@';
        } else if (encodedChars[i] == '-') {
            encodedChars[i] = ':';
        }
    }

    return encodedChars;
}

int main() {
    string inputString = "The mangy, scrawny stray dog hurriedly gobbled down the grain-free, organic dog food.";
    string outputString = encodeString(inputString);
    cout << outputString << endl;
    return 0;
}



// Mini orange 1st ✅

string shortestPalindrome(string s)
    {
        reverse(s.begin(), s.end());
        string rev = s;
        reverse(s.begin(), s.end());
        string temp = s + "@" + rev;
        vector<int> vr = fail_fun(temp);

        int maxm = vr[vr.size()-1];

        temp = "";
        for(int i = maxm; i < s.length(); i++)
        {
            temp = temp + s[i];
        }       
        reverse(temp.begin(), temp.end());
        return temp + s;
    }

    vector<int> fail_fun(string &s)
    {
        int n = s.length();
        vector<int> fail_ar(n, 0);

        for(int i = 1, k = 0; i < n; i++)
        {
            while(k > 0 && s[k] != s[i])
            {
                k = fail_ar[k-1];
            }

            if(s[k] == s[i])
            {
                fail_ar[i] = ++k;
            }
        }

        return fail_ar;
    }

Miniorange

const int MOD = 1e9 + 7;
const int MAXN = 1e6;
using namespace std;
vector<bool> sieve() {
    vector<bool> isPrime(MAXN + 1, true);
    isPrime[0] = isPrime[1] = false;
    for (int i = 2; i <= sqrt(MAXN); ++i) {
        if (isPrime[i]) {
            for (int j = i * i; j <= MAXN; j += i) {
                isPrime[j] = false;
            }
        }
    }
    return isPrime;
}

int solve(const string& s) {
    int n = s.size();
    vector<bool> isPrime = sieve();
    vector<int> dp(n + 1, 0);
    dp[0] = 1;

    for (int i = 1; i <= n; ++i) {
        int num = 0;
        for (int j = 1; j <= 6 && i - j >= 0; ++j) {
            num = num + (s[i - j] - '0') * pow(10, j - 1);
            if (s[i - j] != '0' && num <= MAXN && isPrime[num]) {
                dp[i] = (dp[i] + dp[i - j]) % MOD;
            }
        }
    }

    return dp[n];
}

String estee ✅

long getMaxCost(string s) {
    if (s.find('1') == string::npos) {
        return 0;
    }
    int ind = s.find('1');
    int count = 1;
    int n = s.length();
     long res = 0;
    int prev = ind;
   
    for (int i = ind + 1; i < n; ++i) {
        if (s[i] == '1') {
            if (prev != i - 1) {
                res += (i - prev) * count;
            }
            count += 1;
            prev = i;
        }
    }
   
    if (prev == n - 1) {
        prev = n;
    }
   
    return res + count * (n - prev);
}

Binary Circuit
Estee advisor✅