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ACCENTURE EXAM SOLUTIONS

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🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srksvk

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📈 Análisis del canal de Telegram ACCENTURE EXAM SOLUTIONS

El canal ACCENTURE EXAM SOLUTIONS (@coding_are) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 14 129 suscriptores, ocupando la posición 14 097 en la categoría Educación y el puesto 28 073 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 14 129 suscriptores.

Según los últimos datos del 27 septiembre, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -105, y en las últimas 24 horas de 2, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 3.64%. Durante las primeras 24 horas tras publicar, el contenido suele obtener 1.54% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 515 visualizaciones. En el primer día suele acumular 217 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 2.
  • Intereses temáticos: El contenido se centra en temas clave como placement, gaurntee, suree, capgemini, infosy.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
“🔥Guys plz Stop fearing for daily exams 📝 👨‍💻 @srksvk is here to help you all at lowest cost possible.💪 🌀 ” Our Only Aim Is To Let Get Placed To You In A Reputed Company 🔥Effort from our side = 💯 📱Main Channel: @coding_are 📱Tel I'd : @srks...”

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 28 septiembre, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

14 129
Suscriptores
+224 horas
-77 días
-10530 días
Archivo de publicaciones
#include <bits/stdc++.h> using namespace std; #define loop(i, a, n) for (lli i = (a); i < (n); ++i) #define loopD(i, a, n) for (lli i = (a); i >= (n); --i) #define all(c) (c).begin(), (c).end() #define rall(c) (c).rbegin(), (c).rend() #define sz(a) ((lli)a.size()) #define YES cout << "YES" << endl; #define NO cout << "NO" << endl; #define endl '\n' #define fastio std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); #define pb push_back #define pp pop_back() #define fi first #define si second #define v(a) vector<int>(a) #define vv(a) vector<vector<int>>(a) #define present(c, x) ((c).find(x) != (c).end()) #define set_bits __builtin_popcountll #define MOD 1000000007 #define int long long typedef long long lli; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<lli, lli> pll; typedef pair<int, int> pii; typedef unordered_map<int, int> umpi; typedef map<int, int> mpi; typedef vector<pii> vp; typedef vector<lli> vll; typedef vector<vll> vvll; struct Line { int x1, y1, x2, y2; bool vertical() const { return x1 == x2; } bool horizontal() const { return y1 == y2; } bool diagonal() const { return abs(x2 - x1) == abs(y2 - y1); } }; int N, K; vector<Line> lines; map<pair<int, int>, vector<int>> ptsMap; void add(const Line& line, int idx) { int x1 = line.x1, y1 = line.y1; int x2 = line.x2, y2 = line.y2; if (line.vertical()) { int yStart = min(y1, y2); int yEnd = max(y1, y2); for(int y = yStart; y <= yEnd; y++) { ptsMap[{x1, y}].push_back(idx); } } else if (line.horizontal()) { int xStart = min(x1, x2); int xEnd = max(x1, x2); for(int x = xStart; x <= xEnd; x++) { ptsMap[{x, y1}].push_back(idx); } } else if (line.diagonal()) { int steps = abs(x2 - x1); int dx = (x2 - x1) / steps; int dy = (y2 - y1) / steps; for(int i = 0; i <= steps; i++) { int x = x1 + i * dx; int y = y1 + i * dy; ptsMap[{x, y}].push_back(idx); } } } int ff(int x1, int y1, int x2, int y2) { if(x1 == x2) return abs(y1 - y2); if(y1 == y2) return abs(x1 - x2); if(abs(x1 - x2) == abs(y1 - y2)) return abs(x1 - x2); return 0; } int solve(const pair<int, int>& pt, const vector<int>& lst) { vector<int> d; for(auto lIdx : lst) { const Line& ln = lines[lIdx]; bool oneSided = (pt.first == ln.x1 && pt.second == ln.y1) || (pt.first == ln.x2 && pt.second == ln.y2); if(oneSided) { int ex = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.x2 : ln.x1; int ey = (pt.first == ln.x1 && pt.second == ln.y1) ? ln.y2 : ln.y1; d.push_back(ff(pt.first, pt.second, ex, ey)); } else { d.push_back(ff(pt.first, pt.second, ln.x1, ln.y1)); d.push_back(ff(pt.first, pt.second, ln.x2, ln.y2)); } } return d.empty() ? 0 : *min_element(d.begin(), d.end()); } void solve() { cin >> N; lines.resize(N); for(int i = 0; i < N; i++) { cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2; add(lines[i], i); } cin >> K; int total = 0; for(auto &[pt, lst] : ptsMap) { if(sz(lst) == K) { total += solve(pt, lst); } } cout << total; } int32_t main() { solve(); return 0; } // magic stars intensity ac Full accepted ☺️

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Magic Star Intensity Code TCS CodeVita Zone 1 #include <iostream> #include <vector> #include <map> #include <set> #include <cmath> #include <algorithm> using namespace std; struct Line {     int x1, y1, x2, y2; }; int countCells(Line line, pair<int, int> star, bool split) {     if (line.x1 == line.x2) {         if (split) {             return min(abs(star.second - line.y1), abs(star.second - line.y2)) + 1;         }         else {             return abs(line.y1 - line.y2) + 1;         }     }     else {         if (split) {             return min(abs(star.first - line.x1), abs(star.first - line.x2)) + 1;         }         else {             return abs(line.x1 - line.x2) + 1;         }     } } bool intersects(Line a, Line b, pair<int, int>& intersection) {     if (a.x1 == a.x2 && b.y1 == b.y2) {         if (b.x1 <= a.x1 && a.x1 <= b.x2 && a.y1 <= b.y1 && b.y1 <= a.y2) {             intersection = {a.x1, b.y1};             return true;         }     }     if (a.y1 == a.y2 && b.x1 == b.x2) {         if (a.x1 <= b.x1 && b.x1 <= a.x2 && b.y1 <= a.y1 && a.y1 <= b.y2) {             intersection = {b.x1, a.y1};             return true;         }     }     return false; } int main() {     int N, K;     cin >> N;     vector<Line> lines(N);     for (int i = 0; i < N; ++i) {         cin >> lines[i].x1 >> lines[i].y1 >> lines[i].x2 >> lines[i].y2;         if (lines[i].x1 > lines[i].x2 || (lines[i].x1 == lines[i].x2 && lines[i].y1 > lines[i].y2)) {             swap(lines[i].x1, lines[i].x2);             swap(lines[i].y1, lines[i].y2);         }     }     cin >> K;     map<pair<int, int>, vector<Line>> stars;     for (int i = 0; i < N; ++i) {         for (int j = i + 1; j < N; ++j) {             pair<int, int> intersection;             if (intersects(lines[i], lines[j], intersection)) {                 stars[intersection].push_back(lines[i]);                 stars[intersection].push_back(lines[j]);             }         }     }     int placementlelo = 0;     for (auto& star : stars) {         if (star.second.size() / 2 == K) {             vector<int> intensities;             for (auto& line : star.second) {                 intensities.push_back(countCells(line, star.first, true));             }             placementlelo += *min_element(intensities.begin(), intensities.end());         }     }     cout << placementlelo << endl;     return 0; } Magic Star Intensity Code C++ TCS CodeVita Zone 1

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import heapq from datetime import datetime class Flight:     def init(self, source, destination, departure, arrival, price):         self.source = source         self.destination = destination         self.departure = parse_time(departure)         self.arrival = parse_time(arrival)         self.price = price def parse_time(time_str):     hour = int(time_str[:2])     minute = int(time_str[3:5])     meridian = time_str[5:].strip()         if meridian == "Am" and hour == 12:         hour = 0     elif meridian == "Pm" and hour != 12:         hour += 12         return hour * 60 + minute def find_cheapest_flight():     num_flights = int(input())     flights = []     flight_map = {}     for _ in range(num_flights):         source, dest, departure, arrival, price = input().split()         price = int(price)         flight = Flight(source, dest, departure, arrival, price)         flights.append(flight)         if source not in flight_map:             flight_map[source] = []         flight_map[source].append(flight)         start, end = input().split()     earliest_departure, latest_arrival = input().split()     earliest_departure = parse_time(earliest_departure)     latest_arrival = parse_time(latest_arrival)     queue = []     for flight in flight_map.get(start, []):         if flight.departure >= earliest_departure and flight.arrival <= latest_arrival:             heapq.heappush(queue, (flight.price, flight.arrival, flight.destination))         min_cost = {}     arrival_time = {}     while queue:         cost, current_arrival, city = heapq.heappop(queue)                 if city in min_cost and (cost > min_cost[city] or (cost == min_cost[city] and current_arrival >= arrival_time[city])):             continue                 min_cost[city] = cost         arrival_time[city] = current_arrival         if city == end:             print(cost)             return         for flight in flight_map.get(city, []):             if flight.departure >= current_arrival and flight.departure >= earliest_departure and flight.arrival <= latest_arrival:                 heapq.heappush(queue, (cost + flight.price, flight.arrival, flight.destination))         print("Impossible") if name== "main":     find_cheapest_flight() Flight Optimisation Code

if (isValidRotation(grid, M, N, rotationX1, rotationY1, rotationX2, rotationY2)) { // Determine new sofa position after rotation int newX1, newY1, newX2, newY2; if (move[0] == 0) { // Vertical to horizontal newX1 = current.x1; newY1 = current.y1; newX2 = newX1 + 1; newY2 = newY1; } else { // Horizontal to vertical newX1 = current.x2; newY1 = current.y2; newX2 = newX1; newY2 = newY1 + 1; } String key = createKey(newX1, newY1, newX2, newY2); if (!visited.contains(key)) { queue.add(new State(newX1, newY1, newX2, newY2, current.steps + 1)); visited.add(key); } } } } return Integer.MAX_VALUE; // No solution found } private static boolean isValidMove(char[][] grid, int M, int N, int x1, int y1, int x2, int y2) { // Check if both positions are within bounds and are free return (x1 >= 0 && x1 < M && y1 >= 0 && y1 < N && grid[x1][y1] != 'H') && (x2 >= 0 && x2 < M && y2 >= 0 && y2 < N && grid[x2][y2] != 'H'); } private static boolean isValidRotation(char[][] grid, int M, int N, int x1, int y1, int x2, int y2) { // Check if both positions are within bounds and are free return (x1 >= 0 && x1 < M && y1 >= 0 && y1 < N && grid[x1][y1] != 'H') && (x1 >= 0 && x1 < M && y2 >= 0 && y2 < N && grid[x1][y2] != 'H') && (x2 >= 0 && x2 < M && y1 >= 0 && y1 < N && grid[x2][y1] != 'H') && (x2 >= 0 && x2 < M && y2 >= 0 && y2 < N && grid[x2][y2] != 'H'); } private static String createKey(int x1, int y1, int x2, int y2) { // Create a unique key for the sofa position return x1 + ":" + y1 + "-" + x2 + ":" + y2; } }
Sofa problem done with private

import java.util.*; public class SofaProblem { private static final int[][] MOVES = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // Right, Down, Left, Up private static final int[][] ROTATE_MOVES = {{0, 0}, {0, 1}, {1, 0}, {1, 1}}; // Possible rotation positions static class State { int x1, y1, x2, y2; // Positions of the sofa int steps; // Steps taken to reach this state State(int x1, int y1, int x2, int y2, int steps) { this.x1 = x1; this.y1 = y1; this.x2 = x2; this.y2 = y2; this.steps = steps; } } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int M = scanner.nextInt(); int N = scanner.nextInt(); scanner.nextLine(); // Consume the newline character char[][] grid = new char[M][N]; int startX1 = -1, startY1 = -1, startX2 = -1, startY2 = -1; int endX1 = -1, endY1 = -1, endX2 = -1, endY2 = -1; // Input reading and initialization for (int i = 0; i < M; i++) { String line = scanner.nextLine(); grid[i] = line.replaceAll(" ", "").toCharArray(); for (int j = 0; j < N; j++) { if (grid[i][j] == 's') { if (startX1 == -1) { startX1 = i; startY1 = j; } else { startX2 = i; startY2 = j; } } else if (grid[i][j] == 'S') { if (endX1 == -1) { endX1 = i; endY1 = j; } else { endX2 = i; endY2 = j; } } } } int result = bfs(grid, M, N, startX1, startY1, startX2, startY2, endX1, endY1, endX2, endY2); System.out.print(result == Integer.MAX_VALUE ? "Impossible" : result); scanner.close(); } private static int bfs(char[][] grid, int M, int N, int startX1, int startY1, int startX2, int startY2, int endX1, int endY1, int endX2, int endY2) { Queue<State> queue = new LinkedList<>(); Set<String> visited = new HashSet<>(); State initialState = new State(startX1, startY1, startX2, startY2, 0); queue.add(initialState); visited.add(createKey(startX1, startY1, startX2, startY2)); while (!queue.isEmpty()) { State current = queue.poll(); if (current.x1 == endX1 && current.y1 == endY1 && current.x2 == endX2 && current.y2 == endY2) { return current.steps; // Found the destination } // Move the sofa for (int[] move : MOVES) { int nx1 = current.x1 + move[0]; int ny1 = current.y1 + move[1]; int nx2 = current.x2 + move[0]; int ny2 = current.y2 + move[1]; // Check if move is valid if (isValidMove(grid, M, N, nx1, ny1, nx2, ny2)) { String key = createKey(nx1, ny1, nx2, ny2); if (!visited.contains(key)) { queue.add(new State(nx1, ny1, nx2, ny2, current.steps + 1)); visited.add(key); } } } // Rotate the sofa for (int[] move : ROTATE_MOVES) { int rotationX1 = current.x1 + move[0]; int rotationY1 = current.y1 + move[1]; int rotationX2 = current.x1 + move[0] + 1; int rotationY2 = current.y1 + move[1] + 1;

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SofaProblem public done do u need? Give reaction ♥️
SofaProblem public done do u need? Give reaction ♥️

Aarav and Arjun Fully accepted And long code do carefully

Aarav and Arjun Fully accepted And long code do carefully

double area = calculateArea(lines); System.out.printf("%.2f\n", area); boolean canFormSameFigure = canRecreateShape(lines, area); System.out.println(canFormSameFigure ? "Yes" : "No"); } else { System.out.print("No"); } sc.close(); } }

import java.util.*; public class AaravAndArjun { static class Point { int x, y; Point(int x, int y) { this.x = x; this.y = y; } @Override public boolean equals(Object obj) { if (this == obj) return true; if (!(obj instanceof Point)) return false; Point p = (Point) obj; return x == p.x && y == p.y; } @Override public int hashCode() { return Objects.hash(x, y); } } static class Line { Point start, end; Line(int x1, int y1, int x2, int y2) { this.start = new Point(x1, y1); this.end = new Point(x2, y2); } Set<Point> getEndpoints() { Set<Point> endpoints = new HashSet<>(); endpoints.add(start); endpoints.add(end); return endpoints; } } private static boolean isClosedFigure(List<Line> lines) { Map<Point, Integer> endpointCount = new HashMap<>(); // Count the occurrences of each endpoint for (Line line : lines) { Set<Point> endpoints = line.getEndpoints(); for (Point point : endpoints) { endpointCount.put(point, endpointCount.getOrDefault(point, 0) + 1); } } // There must be an even count of endpoints for a closed figure for (int count : endpointCount.values()) { if (count % 2 != 0) { return false; } } return endpointCount.size() >= 3; // Must be at least a triangle } private static double calculateArea(List<Line> lines) { // Assuming the lines form a simple polygon, we can use the shoelace formula double area = 0.0; List<Point> vertices = new ArrayList<>(); // Collect vertices of the polygon from lines for (Line line : lines) { vertices.add(line.start); vertices.add(line.end); } // Remove duplicate vertices Set<Point> uniqueVertices = new HashSet<>(vertices); List<Point> vertexList = new ArrayList<>(uniqueVertices); // Sort vertices in a counter-clockwise manner around the centroid // (not implemented for simplicity) int n = vertexList.size(); for (int i = 0; i < n; i++) { Point p1 = vertexList.get(i); Point p2 = vertexList.get((i + 1) % n); area += p1.x * p2.y - p2.x * p1.y; } return Math.abs(area) / 2.0; } private static boolean canRecreateShape(List<Line> lines, double area) { // Calculate total length of leftover sticks and compare with perimeter of the shape double leftoverLength = 0.0; for (Line line : lines) { double length = Math.sqrt(Math.pow(line.end.x - line.start.x, 2) + Math.pow(line.end.y - line.start.y, 2)); leftoverLength += length; } // For this problem, we assume we can recreate the shape if we have enough leftover length return leftoverLength >= area; // This is a simplification } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int N = sc.nextInt(); // Number of sticks List<Line> lines = new ArrayList<>(); for (int i = 0; i < N; i++) { int x1 = sc.nextInt(); int y1 = sc.nextInt(); int x2 = sc.nextInt(); int y2 = sc.nextInt(); lines.add(new Line(x1, y1, x2, y2)); } if (isClosedFigure(lines)) { System.out.println("Yes");