Rupeek is hiring for Android Engineer (2021, 2022 and 2023 grads)

JD: https://docs.google.com/document/u/0/d/1RYlGWEH31KPt5-YZ4G5PSMgrmaMywv-KQpQbPz1mBA4/mobilebasic

Apply Link: https://docs.google.com/forms/d/e/1FAIpQLSfz7XamMLri24X64d-B58m2H70YZl9aItL_o3Hv4N5r-PnVNg/viewform?usp=send_form

setlmint.com is hiring Full Stack Engineer

For 2024 grads

Check out this job at setlmint.com: https://www.linkedin.com/jobs/view/3957631749

#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main()
{
   int t;
   cin>>t;
   while(t--)
   {
    int m;
    cin>>m;
    int l;
    cin>>l;
    int a[l];
    for(int i=0;i<l;i++) cin>>a[i];
    int r;
    cin>>r;
    int b[r];
    for(int i=0;i<r;i++) cin>>b[i];

     int i=0;
     int j=0;int price=0;int cnt=0;
    while(i<l and j<r and price<=m)
    {
      if(a[i]>=a[j]) {price+=a[j];cnt++;j++;}
      else {price+=a[i];cnt++;i++;}
    }
    cout<<cnt<<endl;
   }

}

Sheldon And Penny ✅

public static long[] countSentences(String[] wordSet, String[] sentences) {
        long[] counts = new long[sentences.length];
        Map<String, Long> wordCountMap = new HashMap<>();

        for (String word : wordSet) {
            String sortedWord = sortString(word);
            wordCountMap.put(sortedWord, wordCountMap.getOrDefault(sortedWord, 0L) + 1);
        }

        for (int i = 0; i < sentences.length; i++) {
            String[] words = sentences[i].split(" ");
            long sentenceCount = 1;
            for (String word : words) {
                String sortedWord = sortString(word);
                sentenceCount *= wordCountMap.getOrDefault(sortedWord, 1L);
            }
            counts[i] = sentenceCount;
        }

        return counts;
    }

    private static String sortString(String s) {
        char[] charArray = s.toCharArray();
        Arrays.sort(charArray);
        return new String(charArray);
    }
}

How many Sentences ✅
BlackRock

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

int main() {
    int n, m, k;
    cin >> n >> m >> k;
   
    vector<vector<int>> arr(n, vector<int>(m));
    vector<vector<int>> dist(n, vector<int>(m, INT_MAX));
   
    for (auto &e : arr)
        for (auto &ee : e)
            cin >> ee;
   
    queue<pair<pair<int, int>, int>> q;
    vector<vector<bool>> vis(n, vector<bool>(m, false));
   
    vis[0][0] = true;
    dist[0][0] = 0;
    q.push({{0, 0}, 0});
   
    while (!q.empty()) {
        auto p = q.front();
        q.pop();
       
        int x = p.first.first;
        int y = p.first.second;
        int steps = p.second;
       
        // Right movement
        for (int i = 1; i <= k; i++) {
            int newY = y + i;
            if (newY >= m || arr[x][newY] == 1) break;
            if (vis[x][newY]) continue;
            q.push({{x, newY}, steps + 1});
            vis[x][newY] = true;
            dist[x][newY] = steps + 1;
        }
       
        // Left movement
        for (int i = 1; i <= k; i++) {
            int newY = y - i;
            if (newY < 0 || arr[x][newY] == 1) break;
            if (vis[x][newY]) continue;
            q.push({{x, newY}, steps + 1});
            vis[x][newY] = true;
            dist[x][newY] = steps + 1;
        }
       
        // Downward movement
        for (int i = 1; i <= k; i++) {
            int newX = x + i;
            if (newX >= n || arr[newX][y] == 1) break;
            if (vis[newX][y]) continue;
            q.push({{newX, y}, steps + 1});
            vis[newX][y] = true;
            dist[newX][y] = steps + 1;
        }
       
        // Upward movement
        for (int i = 1; i <= k; i++) {
            int newX = x - i;
            if (newX < 0 || arr[newX][y] == 1) break;
            if (vis[newX][y]) continue;
            q.push({{newX, y}, steps + 1});
            vis[newX][y] = true;
            dist[newX][y] = steps + 1;
        }
    }
   
    cout << (dist[n - 1][m - 1] == INT_MAX ? -1 : dist[n - 1][m - 1]) << endl;
}

Minimum moves ✅

Tomorrow will be upload Cisco Codes those Who will be give Exam

International opportunity alert 🚀

Cobblestone Energy is hiring Software Engineer

For 2024, 2023 grads

Location - Dubai

Apply - https://grnh.se/c8535a2c3us

Process Associates (P2P Process) at KGiS

Requirements:

Experience: Fresher
Position: Process Associates
Technical Skills: Core Accounting concepts, Bank and ledger reconciliation.
Shift: Rotational Shifts.
Location: Coimbatore
Notice period: Immediate


Interested candidates can come for a direct walk-in drive.

The Walk-in drive is scheduled as per the below details:
Date: (25-June and 27-June-2024)
Time: 10:00 AM to 01:00 PM

Address:
Opposite to holiday residency
KGISL campus
365, IT Tower 1, 
Thudiyalur Road, 
Saravanampatti, 
Coimbatore - 641035.


Kindly come directly to the office with documents and a resume.

Your point of contact:
Name: Priyanka S
Contact: +91 8754193339

#!/bin/python3

import math
import os
import random
import re
import sys



#
# Complete the 'findSubstring' function below.
#
# The function is expected to return a STRING.
# The function accepts following parameters:
#  1. STRING s
#  2. INTEGER k
#

def findSubstring(s, k):
    vowels = ["a", "e", "i", "o", "u"]
    cur = best = sum([c in vowels for c in s[:k]])
    ans = 0
    for i in range(k, len(s)):
        cur += s[i] in vowels
        cur -= s[i - k] in vowels
        if cur > best:
            best = cur
            ans = i - k + 1
    if best > 0:
        return s[ans:(ans+k)]
    else:
        return "Not found!"

if name == 'main':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    s = input()

    k = int(input().strip())

    result = findSubstring(s, k)

    fptr.write(result + '\n')

    fptr.close()

Vowel substring ✅

from collections import defaultdict

def f(k, costs, from_lst, to_lst):
  children = defaultdict(list)

  for i, u in enumerate(from_lst):
    children[u].append(to_lst[i])

  prefixes = defaultdict(int)
  prefixes[0] = 1

  def g(n, s):
    result = 0
    curr = (s + costs[n]) % k
    result += prefixes[curr]
    prefixes[curr] += 1
    for c in children[n]:
      result += g(c, curr)
    prefixes[curr] -= 1
    return result

  return g(0, 0)

Hack Tree ✅

Autodraft is hiring SDE

For 2023, 2022 grads

Expected CTC - 12LPA - 20LPA

Apply - https://wellfound.com/jobs/3040061-software-engineer

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll sum=0,cnt1=0;
ll dfs1(ll node,vector<ll>adj[],vector<ll>& vis) {
    vis[node]=1;
    for(auto &it:adj[node]) 
    {
        if(!vis[it])  dfs1(it, adj, vis);
    }
    sum+=node;
    cnt1++;
}
ll dfs(ll node,vector<ll>adj[],vector<ll>&vis,stack<ll>& st) {
    vis[node]=1;
    for(auto &it:adj[node]) 
    {
        if(!vis[it])  dfs(it, adj, vis, st);
    }
    st.push(node);
}
ll solve(ll N,vector<ll> Edge) 
{
    vector<ll>adj[N];
    for(ll i=0;i<N;i++) 
    {
        ll u=i;
        ll v=Edge[i];
        if(v!=-1) adj[u].push_back(v);
    }
    vector<ll>vis(N+1);
    stack<ll>st;
    for(ll i=0;i<N;i++) 
    {
        if(!vis[i])  dfs(i,adj,vis,st);
    }
    vector<ll>adj1[N];
    for(ll i=0;i<N;i++) 
    {
        vis[i]=0;
        for(auto &it:adj[i]) {
            adj1[it].push_back(i);
        }
    }
    ll ans=0,cnt=0;
    while(st.size()) 
    {
        ll node=st.top();
        st.pop();
        if(!vis[node]) 
        {
            sum=0;
            cnt++;
            cnt1=0;
            dfs1(node, adj1, vis);
        }
        if(cnt1>1)  ans=max(ans, sum);
    }
    if(cnt==N) return -1;
    return ans;
}

signed main() {
    ll t; cin >> t;
    while(t--) 
    {
        ll n; cin>>n;
        vector<ll>a(n);
        for(ll i=0;i<n;i++) cin>>a[i];
        cout<<solve(n,a)<<endl;
    }
    return 0;
}

Largest sum Cycle
Juspay ✅

Company Name: Flipkart
Role: SDE-1 & UI-1
Batch: 2024 Female passouts

CTC Overview: 32.57 LPA (18 Lakh Base pay, 10% variable, ESOPs worth 5 lakhs; Joining Bonus: 3 lakhs & Retention Bonus 3 Lakhs - detailed comp breakup will be shared with candidates shortlisted for online test

https://docs.google.com/forms/d/e/1FAIpQLSdITyHEhZ3W0Rdtd7fuRfIdlk8UyZa2yAIimfb2_w9XpLT9NQ/viewform?usp=sf_link

Circuit Board ✅