allcoding1

#include<bits/stdc++.h> using namespace std; const int MOD = 1e9 + 7; vector<vector<vector<string>>> dp; vector<string> solve(string& alice, string& bob, int i, int j) {     if(i == alice.size() || j == bob.size()) {         return {""};     }     if(dp[i][j].size() != 0) {         return dp[i][j];     }     if(alice[i] == bob[j]) {         vector<string> tmp = solve(alice, bob, i+1, j+1);         for(string& s : tmp) {             s = alice[i] + s;         }         return dp[i][j] = tmp;     }     vector<string> left = solve(alice, bob, i+1, j);     vector<string> right = solve(alice, bob, i, j+1);     if(left[0].size() > right[0].size()) {         return dp[i][j] = left;     }     if(left[0].size() < right[0].size()) {         return dp[i][j] = right;     }     left.insert(left.end(), right.begin(), right.end());     sort(left.begin(), left.end());     left.erase(unique(left.begin(), left.end()), left.end());     return dp[i][j] = left; } int main() {     int T;     cin >> T;     while(T--) {         string alice, bob;         cin >> alice >> bob;         dp = vector<vector<vector<string>>>(alice.size(), vector<vector<string>>(bob.size()));         vector<string> trips = solve(alice, bob, 0, 0);         for(string& trip : trips) {             cout << trip << endl;         }     }     return 0; }

#include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public:     int maxSumOptimalArrangement(vector<int>& coins) {         vector<int> positive;         vector<int> negative;         for (int coin : coins) {             if (coin >= 0)                 positive.push_back(coin);             else                 negative.push_back(coin);         }         sort(positive.rbegin(), positive.rend());                 sort(negative.begin(), negative.end());         int totalSum = 0;         for (size_t i = 0; i < max(positive.size(), negative.size()); ++i) {             if (i < positive.size())                 totalSum += positive[i];             if (i < negative.size())                 totalSum -= negative[i];         }         return totalSum;     } }; ZS campus beat code Circuit Board Telegram:- @allcoding1

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available

Company Name: Qualcomm Job Title: Software Engineer Intern 2024 and 2025 Apply Now:- https://careers.qualcomm.com/careers/job?domain=qualcomm.com&pid=446693743471&query=intern&location=India&domain=qualcomm.com&sort_by=relevance&job_index=0 Telegram:- @allcoding1

#include<stdio.h> #include<stdlib.h> #define MOD 1000000007 int compare(const void *a, const void *b) {     return (*(int*)a - *(int*)b); } int main() {     int N;     scanf("%d", &N);     if (N <= 1) {         printf("NO HOURS\n");         return 0;     }     int *A = (int*)malloc(N * sizeof(int));     for (int i = 0; i < N; i++) {         scanf("%d", &A[i]);     }     qsort(A, N, sizeof(int), compare);     int count = 0;     for (int i = 0; i < N - 1; i++) {         for (int j = i + 1; j < N; j++) {             if ((A[i] + A[j]) % 60 == 0) {                 count = (count + 1) % MOD;             }         }     }     if (count > 0) {         printf("%d\n", count);     } else {         printf("NO HOURS\n");     }     free(A);     return 0; } Hours Count Telegram:-

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available Proofs:- https://t.me/+mP2YTsMihZozNWQ1

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available Proofs:- https://t.me/+mP2YTsMihZozNWQ1

proofs👇 https://t.me/+mP2YTsMihZozNWQ1

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 All courses (100 rupees) Contact:- @meterials_available

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

Exam Ans:- @allcoding1 Jobs post:- @allcoding1_official

Salesperson ID's - IBM

Lexicographical smallest substring - IBM

def max_call_executives(n, start_times, end_times):     timeline = [0] * (24 * 60 + 1)         for i in range(n):         start = int(start_times[i][:2]) * 60 + int(start_times[i][2:])         end = int(end_times[i][:2]) * 60 + int(end_times[i][2:])         timeline[start] += 1         timeline[end] -= 1         max_executives = 0     current_executives = 0     for i in range(len(timeline)):         current_executives += timeline[i]         max_executives = max(max_executives, current_executives)         return max_executives Call Centre

import heapq def distance(x, y):     return x*x + y*y def nearest_houses(P, T, queries):     distances = []     heapq.heapify(distances)     for query in queries:         if query[0] == 1:             x, y = query[1], query[2]             heapq.heappush(distances, distance(x, y))         else:             nearest = heapq.nsmallest(T, distances)[-1]             print(nearest) Nearest House

📌IT learning courses 📌All programing courses 📌Abdul bari courses 📌Ashok IT Tutorials + Books + Courses + Trainings + Workshops + Educational Resources 🔹Data science 🔹Python 🔹Artificial Intelligence 🔹AWS Certified 🔹Cloud 🔹BIG DATA 🔹Data Analytics 🔹BI 🔹Google Cloud Platform 🔹IT Training 🔹MBA 🔹Machine Learning 🔹Deep Learning 🔹Ethical Hacking 🔹SPSS 🔹Statistics 🔹Data Base 🔹Learning language resources English , 🇫🇷 100 rupees Contact:- @meterials_available

#include <iostream> #include <vector> using namespace std; const int MOD = 1e9 + 7; int F(int i, int k, int n, vector<int> &level, vector<int> &dp) {     if (i == n) return 1;     if (dp[i] != -1) return dp[i];     int ans = 0, odds = 0;     vector<int> hash(n + 10, 0);     for (int j = i; j < n; j++) {         if (++hash[level[j]] % 2 == 0) odds -= 1;         else odds += 1;         if (odds <= k) {             ans = (ans + F(j + 1, k, n, level, dp)) % MOD;         }     }     return dp[i] = ans; } int countValidPartitions(vector<int> level, int k) {     int n = level.size();     vector<int> dp(n, -1);     return F(0, k, n, level, dp); } DE Shaw