Coding | EXAMS | IBM ACCENTURE | VIRTUSA | IBM | AMAZON | TCS | EPAM | WILEY EDGE | TECH MAHINDRA | JPMORGAN | HCL | WIPRO

ACCOLITE DIGITAL CODING + APTITUDE SOLUTIONS 🔥 Increasing Subarray Code Weird Code Uploaded in all languages:- Python, C, C++, Java ALL TEST CASES PASSED ✅ All Aptitude solutions uploaded Note:- ALL Students will get the same questions.😅 Telegram:- https://t.me/Coding_000

Guys Now I am disable the sharing options..So pls share the Coding mcq to ur friends Share our channel keep support 👨‍💻 @Coding_000 ❤️

WILEY EDGE Exam Help Available Msg: @ILOVEU_143 Note: It's Paid 💰💰

Accolite Offcampus Drive Answers here @Coding_000 ❤️ Scroll above MCQ & Coding ans there✅ @Examhelp_143👨‍💻 Share to ur frds help them to get placed😊👨‍💻☺️ Application link - @Offcampus_000 😍

Code forces solutions 👆👆

Codeforces B C++

Codeforces A Flip Flop C✅

Any exams..?

Qspiders Help available.. Contact -@ILOVEU_143 ✅

Off campus updates join this 👉 @Offcampus_000 ❤️

My request to our group family don't trust anyone simply and don't lose money 🙏🥹 @Examhelp_143 ❤️

Any one want projects -mini or Major 😊  Unique project💥💥 Domain AI/ML contact -@ILOVEU_143 Note -paid 🤑 share✅ share ✅@Coding_000

✅💥❤️

#include <bits/stdc++.h> using namespace std; vector<string> rows; char alphabet; string findPrecedingAlphabets() {     int r = -1, c = -1;     for (int i = 0; i < rows.size(); i++)     {         int idx = rows[i].find(alphabet);         if (idx != string::npos)         {             r = i;             c = idx;             break;         }     }     if (r == -1)         return "alphabet not found";     string a = "";     if (r > 0)         a += rows[r - 1][c];     if (c > 0)         a += rows[r][c - 1];     return a; } int main() {     int n;     cin >> n;     for (int i = 0; i < n; i++)     {         string row;         cin >> row;         rows.push_back(row);     }     cin >> alphabet;     cout << findPrecedingAlphabets() << endl;     return 0; }

Share our channel for...ibm 6pm code..✅

timer code

def get_min_max_time(time):     # Convert time string to a list of integers     time = [int(t) for t in time.split(":")]     # Calculate the minimum and maximum time     min_time = [time[0], max(time[1] // 10 * 10, time[1] % 10), time[2]]     max_time = [time[0], min(time[1] // 10 * 10 + 5, 59), time[2]]     # Convert the minimum and maximum time back to strings     min_time = ":".join(str(t).zfill(2) for t in min_time)     max_time = ":".join(str(t).zfill(2) for t in max_time)     return min_time, max_time # Test the function time = "10: 5:48" print(get_min_max_time(time)) # Output: ('10:05:48', '10:55:48') Time code