#include <iostream>
#include <vector>
#include <string>
using namespace std;
int countSubstrings(string input_str) {
    string d[9] = {"ab", "cde", "fgh", "ijk", "lmn", "opq", "rst", "uvw", "xyz"};
    vector<int> mp(26, 0);
    for (int i = 0; i < 9; ++i) {
        for (char c : d[i]) {
            mp[c - 'a'] = i + 1;
        }
    }

    int ans = 0;
    int n = input_str.size();
    for (int i = 0; i < n; ++i) {
        int s = 0;
        for (int j = i; j < n; ++j) {
            s += mp[input_str[j] - 'a'];
            if (s % (j - i + 1) == 0) {
                ++ans;
            }
        }
    }

    return ans;
}

Countsubstring ✅

Kindly share your CV with hr@althisolutions.com

def is_valid_email(email):
    if not email.endswith('@hackerrank.com'):
        return False
    
    user = email[:-15]
    
    if len(user) < 1 or len(user) > 11:
        return False
    
    letters = 0
    underscore = 0
    digits = 0
    
    for char in user:
        if 'a' <= char <= 'z':
            if underscore or digits:
                return False
            letters += 1
        elif char == '_':
            if underscore or digits:
                return False
            underscore += 1
        elif '0' <= char <= '9':
            digits += 1
        else:
            return False
    
    if letters < 1 or letters > 6:
        return False
    if underscore > 1:
        return False
    if digits > 4:
        return False

    return True

query = int(input())
result = ['False'] * query

for i in range(query):
    someString = input()
    if is_valid_email(someString):
        result[i] = 'True'

with open('output.txt', 'w') as fileOut:
    fileOut.write('\n'.join(result))

Valid Email Addresses ✅

int help(vl arr, vl arr2, int n)
{
    vector<pair<ll, ll>> ans;
    for (int i = 0; i < n; i++)
    {
        ans.pb({arr[i], arr2[i]});
    }
    sort(all(ans));
    debug(ans);
    ll ans1 = *max_element(all(arr));
    ll finalans = ans1;
    ll sum = 0;
    for (int i = n - 1; i >= 0; i--)
    {
        ans1 = ans[i].ff;
        finalans = min(finalans, max(ans1, sum));
        sum += ans[i].ss;
    }
    finalans = min(finalans, sum);
    return finalans;
}

Efficient Application Development ✅

long getMaximumEfficiency(int n,int k,vl capacity,vl numserver){
    sort(numserver.begin(),numserver.end());
    sort(capacity.begin(),capacity.end());
    ll sum=0;
    int x=0;
    int y=n-1;
    for(int i=k-1;i>=0;i--){
        if(numserver[i]==1){
            break;
        }
        sum+=capacity[y]-capacity[x];
        x++;
        y--;
    }
    return sum;
}

Server upgrade ✅

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 ✅