Split it up ✅

Pitney Bowes is hiring QA Interns

2025/2024/2023 passouts can apply

https://careers.pitneybowes.com/global/en/job/R18422/Intern-QA?rx_job=R18422&rx_medium=post&rx_paid=0&rx_r=none&rx_source=linkedinll&rx_ts=20240718T021606Z&rx_viewer=f31609e0b41311ee90450bb95a3380bef7666ebed64d4e95aade908439480d1d

intern at take you forward https://www.linkedin.com/posts/takeuforward_internship-drive-tomorrow-at-8-pm-if-you-activity-7220012171378798592-zXqN?utm_source=share&utm_medium=member_desktop

NoScrubs is hiring Full-Stack Engineer Intern (previous intern experience is must)

For 2024, 2025, 2026 grads

Location: Remote

Stipend - 40K per month

Apply - https://wellfound.com/jobs/3062383-full-stack-engineer-intern

https://forms.office.com/pages/responsepage.aspx?id=W8FT8jyv2EaRBTCyeq83uQqH6OaTWbdFvQLgqAudlD5URE1JVExNWE1QUjhDSjdOWUdUU0xFTEEwQy4u

iMocha Hiring Prompt Engineering Intern

Students pursuing or completed B.Sc/BCA/MCA/BE/B.Tech.
- Only freshers can apply.
- Proficiency in programming languages and technical skills.
- Excellent written and verbal communication skills.
- Ability to collaborate effectively within a team and across departments.
- Eagerness to learn and contribute to the development of AI-driven solutions.

Pay range: 7K-12K INR per month

Walmart | Sparkathon

Eligibility: 2025 and 2026 grads

Chance to get a software development internship at Walmart.

https://walmart.converge.tech/content/converge/en_in/sparkathon.html

Amazon is hiring for SDE l and SDE ll roles

SDE l : 2023/2022/2021 batches and below

SDE ll : 3+ year experience

Send resume : shivg7706@gmail.com

Subject Format of mail : Name | Company | Experience | Preferred Location | SDE l / SDE ll

Resume file name : YOURFULLNAME.pdf

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Urgent hiring Bulk Fresher and Interns- PAN -2024
Post Name - Multiple Positions and HR
non IT Freshers & Experience.

Interested Apply here:-https://lnkd.in/gerxxYci


Remote and WFH both options are available.
( Preferably Experience)
- Laptop and welcome kit will provided.
-5 Days working
Qualification - Any Degree and 12 th passout .
Salary - 3.5 LPA to 6 LPA

#include <iostream>
#include <vector>
#include <queue>
#include <climits>
using namespace std;
const vector<pair<int, int>> directions = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
const vector<char> dirChar = {'L', 'R', 'U', 'D'};
int minCostToReachDestination(int N, vector<vector<char>>& Dir, int Sx, int Sy, int Dx, int Dy) {
    priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<>> pq;
    vector<vector<int>> cost(N, vector<int>(N, INT_MAX));
    pq.push({0, {Sx, Sy}});
    cost[Sx][Sy] = 0;
   
    while (!pq.empty()) {
        auto [curCost, pos] = pq.top();
        pq.pop();
       
        int x = pos.first;
        int y = pos.second;
                if (x == Dx && y == Dy) {
            return curCost;
        }
            for (int i = 0; i < 4; ++i) {
            int nx = x + directions[i].first;
            int ny = y + directions[i].second;
           
            if (nx >= 0 && nx < N && ny >= 0 && ny < N) {
                int newCost = curCost + (Dir[x][y] == dirChar[i] ? 0 : 1);
                if (newCost < cost[nx][ny]) {
                    cost[nx][ny] = newCost;
                    pq.push({newCost, {nx, ny}});
                }
            }
        }
    }
   
    return -1;
}

Cost optimization
Google ✅

const int mod = 1e9 + 7;
void insert(vi &dp, int x){
    for(int i = 1000;i >= x; --i)
        dp[i] = (dp[i] + dp[i-x]) % mod;
}
void remove(vi &dp, int x){
    for(int i = x; i <= 1000; ++i)
        dp[i] = (dp[i] - dp[i-x] + mod) % mod;
}
int solve(){
    int n, s;
    cin >> n >> s;
    int a[n];
    inputarr(a, n);
    vector<int> dp(1001,0);
    dp[0] = 1;
    int r = 0;
    int ans = 1e9;
    for(int i = 0; i < n; i++){
        while(r < n && (dp[s] == 0)){
            insert(dp, a[r]);
            r++;
        }
        if(dp[s])
            ans = min(ans, r - i);
        remove(dp, a[i]);
    }
    if(ans == 1e9)
        ans = -1;
    cout << ans << endl;

    return 0;
}

Subset Queries ✅
Google

#include <bits/stdc++.h>

using namespace std;

const long long MOD = 1000000007;

long long calculateWays(long long N, long long K, const string& S, const vector<long long>& A);
bool isValidSubstr(const vector<long long>& charCount, const vector<long long>& A);
void updateCharCount(vector<long long>& charCount, char ch, int delta);

long long calculateWays(long long N, long long K, const string& S, const vector<long long>& A) {
    vector<vector<long long>> dp(N + 1, vector<long long>(K + 1, 0));
    dp[0][0] = 1;

    for (long long i = 1; i <= N; ++i) {
        vector<long long> charCount(26, 0);
        for (long long j = i; j >= 1; --j) {
            updateCharCount(charCount, S[j - 1], 1);
            if (isValidSubstr(charCount, A)) {
                for (long long k = 1; k <= K; ++k) {
                    dp[i][k] = (dp[i][k] + dp[j - 1][k - 1]) % MOD;
                }
            } else {
                break;
            }
        }
    }

    return dp[N][K];
}

bool isValidSubstr(const vector<long long>& charCount, const vector<long long>& A) {
    for (long long k = 0; k < 26; ++k) {
        if (charCount[k] > A[k]) {
            return false;
        }
    }
    return true;
}

void updateCharCount(vector<long long>& charCount, char ch, int delta) {
    charCount[ch - 'a'] += delta;
}

int main() {
    long long N, K;
    cin >> N;
    string S;
    cin >> S;
    vector<long long> A(26);
    for (long long i = 0; i < 26; ++i) {
        cin >> A[i];
    }
    cin >> K;

    long long result = calculateWays(N, K, S, A);
    cout << result << endl;
    return 0;
}

Maximum Partition
Google ✅

#include <bits/stdc++.h>

using namespace std;

const long long INF = 1e8;
const long long MOD = 1e9 + 7;

long long mod_pow(long long b, long long e, long long m) {
    long long r = 1;
    while (e > 0) {
        if (e % 2 == 1) {
            r = (r * b) % m;
        }
        b = (b * b) % m;
        e /= 2;
    }
    return r;
}

int main() {
    int n;
    cin >> n;
    vector<long long> arr(n);
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }

    long long max_v = LLONG_MIN;
    long long res = 0;

    for (int i = 1; i < n; ++i) {
        long long pow_v = mod_pow(INF, n - 1, MOD);
        long long val = (arr[i - 1] - arr[i]);
        long long cur_v = (val * pow_v);
        if (cur_v > max_v) {
            max_v = cur_v;
            res = cur_v;
        }
    }

    cout << res % MOD << endl;
    return 0;
}

Awesomeness factor ✅
Google