from collections import deque
def can_reach_all_with_p(teammates, P):
    n = len(teammates)
    graph = [[] for _ in range(n)]
    for i in range(n):
        for j in range(n):
            if i != j:
                dist = abs(teammates[i][0] - teammates[j][0]) + abs(teammates[i][1] - teammates[j][1])
                if dist <= P:
                    graph[i].append(j)
    def bfs(start):
        visited = [False] * n
        q = deque([start])
        visited[start] = True
        count = 1
        while q:
            node = q.popleft()
            for neighbor in graph[node]:
                if not visited[neighbor]:
                    visited[neighbor] = True
                    q.append(neighbor)
                    count += 1
        return count == n
    for i in range(n):
        if bfs(i):
            return True
    return False

def min_initial_points(teammates):
    left, right = 0, 2 * 10**9  
    while left < right:
        mid = (left + right) // 2
        if can_reach_all_with_p(teammates, mid):
            right = mid
        else:
            left = mid + 1
    return left

Reach teammate
Groww ✅

📌Bank of America Hiring Apprentice

Qualification:
Final year Graduates from the Class of 2025 ONLY

- Must Have Major Specialization in Computer Science & Information Technology ONLY
- Must have scored 60% in the last semester OR CGPA of 6 on a scale of 10 in the last semester
- No Active Backlogs in any of the current or prior semesters
- Students should be willing to join any of the roles/skills/segment as per company requirement
- Students should be willing to work in any shifts/night shifts as per company requirement
- Students should be willing to work in any locations namely – Mumbai, Chennai, Gurugram, Gandhinagar (GIFT), Hyderabad as per company requirement

💻Apply: https://careers.bankofamerica.com/en-us/job-detail/24035855/apprentice-multiple-locations

🚀 Naukri Campus Young Turks Skill Assessment Test 🚀

Your chance to showcase your skills in demand by employers by taking two tests.

Eligibility : Students currently pursuing UG courses and 2024 undergraduates looking for their first job ( BA, B.Com, B.Sc, B.Tech, BBA, BCA & more)

Round 1 : Basic Aptitude Test

Round 2 : Skills Test in fields like coding & six other areas

Rewards 🎉: Earn merit certificates from top brands to enhance your CV,  even win cash prizes, goodies up to Rs. 20,00,000! 💶

Enroll Here 🔗 : https://feji.us/zh7zag

Adidas is hiring Junior Data and Platform Engineer

For 2022, 2023 grads
Location: Gurugram

https://jobs.adidas-group.com/adidas/job/Gurgaon-Junior-Data-&-Platform-Engineer-HR/1113244801/?feedId=301201

Company Name : Qualcomm
Role : Software Engineer (Test)
Batch : 2023 passouts - ECE/CS

Mail your resume to : aayushar@qti.qualcomm.com

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int solve(const vector<int>& arr) {
    int n = arr.size();
        int a = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] == 1) a++;
    }
        if (a == n) {
        return n - 1;
    }
    vector<int> f(n);
    for (int i = 0; i < n; i++) {
        f[i] = (arr[i] == 1) ? -1 : 1;
    }
    int m = f[0], v = f[0];
    for (int i = 1; i < n; i++) {
        v = max(f[i], v + f[i]);
        m = max(m, v);
    }
        int e = a + m;
    return e;
}

flipping frenzy game✅

TCS Codevita is back with its 12 edition!!

Batch eligible: 2025, 2026, 2027 and 2028 grads

Apply: https://codevita.tcsapps.com/

Company Name: PhonePe
Role: Software Engineer - Backend
YOE: 0-2 years

Apply: https://boards.greenhouse.io/embed/job_app?token=5890043003&gh_src=961e65dc3us

int warpedSpell(int secret) {
    if(secret <= 1001) return 1001;
    if(secret >= 9999999) return 9999999;
    string secretStr = to_string(secret);
    int n = secretStr.length();
    string warpedStr = secretStr;

    for (int i = 0; i < (n + 1) / 2; ++i) {
        warpedStr[n - i - 1] = warpedStr[i];
    }

    if (warpedStr > secretStr) {
        return stoi(warpedStr);
    }

    for (int i = (n - 1) / 2; i >= 0; --i) {
        if (warpedStr[i] < '9') {
            warpedStr[i]++;
            break;
        }
        warpedStr[i] = '0';
    }

    for (int i = 0; i < (n + 1) / 2; ++i) {
        warpedStr[n - i - 1] = warpedStr[i];
    }

    if (warpedStr[0] == '0') {
        warpedStr = "1" + warpedStr;
        warpedStr[n] = '1';
    }

    return stoi(warpedStr);
}

Thoughtwork ✅

import re

def orderNumberFinder(message):
    pattern = r'\b\d{3}[-. ]?\d{4}[-. ]?[A-Z]\b'
    if re.search(pattern, message):
        return 1
    else:
        return 0

Thoughtwork ✅

#include <climits>
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
int t[31];
int sol(int i, unordered_map<int, int>& h)
{
    if (i >= 30)
        return 0;
    else if (t[i] != -1)
        return t[i];
    int min_value = INT_MAX;
    for (int j = 1; j <= 6; j++) {
        int k = i + j;
        if (h.count(k) > 0) {
            if (h[k] < k)
                continue;
            k = h[k];
        }

        min_value = min(min_value, sol(k, h) + 1);
    }
    t[i] = min_value;
    return t[i];
}

int min_throw(int n, vector<int> arr)
{
    for (int i = 0; i < 31; i++) {
        t[i] = -1;
    }
    unordered_map<int, int> h;
    for (int i = 0; i < 2 * n; i += 2) {
        h[arr[i]] = arr[i + 1];
    }

    return sol(1, h);
}

Winning Strategy ✅

double solve(int n,int k,int row,int col,vector<vector<vector<double>>>&dp)
    {
        if(k==0) return 1;
     
        int dx[8]={2,1,-1,-2,-2,-1,1,2};
        int dy[8]={1,2,2,1,-1,-2,-2,-1};
        double ans=0;
        if(dp[k][row][col]!=-1) return dp[k][row][col];
        for(int i=0;i<8;i++)
        {
            int nrow=row+dx[i];
            int ncol=col+dy[i];
            if(nrow>=0 and ncol>=0 and nrow<n and ncol<n)  ans+=solve(n,k-1,nrow,ncol,dp);
        }
        return dp[k][row][col]=ans/8;
       
    }
    double knightProbability(int n, int k, int row, int column) {
        vector<vector<vector<double>>>dp(k+1,vector<vector<double>>(n+1,vector<double>(n+1,-1)));
        return solve(n,k,row,column,dp);
    }

Still There ✅

class Solution {
    public int solution(int[] p, int[] s) {
        int n = p.length;
        int totalS = 0, totalP = 0;

        for (int i = 0; i < n; i++) {
            totalS += s[i];
            totalP += p[i];
        }

        Arrays.sort(s);
        int cars = 0;
        for (int i = n - 1; i >= 0; i--) {
            totalP -= s[i];
            cars++;
            if (totalP <= 0) break;
        }

        return cars;
    }
}

📌Tetra Looking for Freshers

Linux - Red Hat Certified Engineers

Mode of Work : Remote

Interested candidates may mail resumes to recruitment@tetrain.com

👉Red Hat Certification mandatory !!

Rubrik OA ✅

def oracles(n, q, ops, qs):
    a = []
    for b, x in ops:
        if b == 1:
            a.append(x)
        else:
            a = a * (x + 1)
    return [a[i-1] if i <= len(a) else 0 for i in qs]

n, q = map(int, input().split())
ops = [map(int, input().split()) for _ in range(n)]
qs = list(map(int, input().split()))

print(*oracles(n, q, ops, qs))

The Oracles ✅

MAX_N = 100000

e = [0] * MAX_N

def canSplit(n, k, b):
    maxSum = 0
    s = 0
    br = 0

    for i in range(n):
        if s + e[i] > k or br == b - 1:
            br += 1
            s = e[i]
        else:
            s += e[i]
        
        maxSum = max(maxSum, s)
        
        if br >= b:
            return False
    
    return maxSum <= k

def minBranches(n, k):
    l, r = 1, n
    while l < r:
        mid = l + (r - l) // 2
        if canSplit(n, k, mid):
            r = mid
        else:
            l = mid + 1
    
    return l if canSplit(n, k, l) else -1

def main():
    import sys
    input = sys.stdin.read
    data = input().split()
    
    n, m, k = map(int, data[:3])
    global e
    e = list(map(int, data[3:3+n]))
    
    print(minBranches(n, k))

if __name__ == "__main__":
    main()

Save the universe✅

#define ff first
#define ss second
struct minq{
    stack<pair<int,int>>s,t;
    void push(int x){
        int mn=(s.size()>0)?min(s.top().ss,x):x;
        s.push({x,mn});
    }   
    void pop(){
        if(t.empty()){
            while(!s.empty()){
                auto x=s.top();
                int mn=(t.size()>0)?min(x.ff,t.top().ss):x.ff;
                t.push({x.ff,mn});
                s.pop();
            }
        }
        t.pop();
    }
    int getmin(){
        int mn=INT_MAX;
        if(s.size())mn=min(mn,s.top().ss);
        if(t.size())mn=min(mn,t.top().ss);
        return mn;
    }
};

bool ok(string &s,vector<int>&a,int mid){
    minq v;
    for(int i=0;i<mid;i++){
        v.push(a[s[i]-'a']);
    }
    bool alpha=v.getmin()>=mid;
    for(int i=mid;i<s.size();i++){
        v.pop();
        v.push(a[s[i]-'a']);
        alpha|=(v.getmin()>=mid);
    }
    return alpha;
}
int getLongest(string &s,vector<int>&a){
    int st=0,en=s.size()-1;
    while(en-st>1){
        int mid=(en+st+1)/2;
        if(ok(s,a,mid)){
            st=mid;
        }
        else en=mid;
    }
    if(ok(s,a,en))return en;
    else return st;
}


vector<int>solve(vector<string>universe,vector<vector<int>>s){
    vector<int>ans;
    int n=universe.size();
    for(int i=0;i<n;i++){
        ans.push_back(getLongest(universe[i],s[i]));
    }
    return ans;
}

Anti Aging Serum ✅