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ACCENTURE EXAM SOLUTIONS

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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ACCENTURE EXAM SOLUTIONS (@coding_are) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 14 125 obunachidan iborat bo'lib, Taʼlim toifasida 14 097-o'rinni va Hindiston mintaqasida 28 073-o'rinni egallagan.

📊 Auditoriya ko‘rsatkichlari va dinamika

невідомо sanasidan buyon loyiha tez o‘sib, 14 125 obunachiga ega bo‘ldi.

28 Sentabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -105 ga, so‘nggi 24 soatda esa -4 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.

  • Tasdiqlash holati: Tasdiqlanmagan
  • Jalb etish (ER): Auditoriya o‘rtacha 3.68% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 1.57% ini tashkil etuvchi reaksiyalarni to‘playdi.
  • Post qamrovi: Har bir post o‘rtacha 520 marta ko‘riladi; birinchi sutkada odatda 222 ta ko‘rish yig‘iladi.
  • Reaksiyalar va o‘zaro ta’sir: Auditoriya faol: har bir postga o‘rtacha 2 ta reaksiya keladi.
  • Tematik yo‘nalishlar: Kontent placement, gaurntee, suree, capgemini, infosy kabi asosiy mavzularga jamlangan.

📝 Tavsif va kontent siyosati

Muallif resursni shaxsiy fikrni ifoda etish maydoni sifatida ta’riflaydi:
“🔥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...”

Yuqori yangilanish chastotasi (oxirgi ma’lumot 28 Sentabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.

14 125
Obunachilar
-424 soatlar
-127 kun
-10530 kun
Postlar arxiv
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");

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Passed
Passed

AaravAndArjun.... this code want Give reaction ♥️

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import java.util.HashMap; import java.util.Map; import java.util.Scanner; public class MinimumMoves { public static int minimumMoves(String[] instruct) { String[] directions = {"up", "down", "left", "right"}; int s = instruct.length; int[][][] dp = new int[s + 1][4][4]; for (int i = 0; i <= s; i++) { for (int j = 0; j < 4; j++) { for (int k = 0; k < 4; k++) { dp[i][j][k] = Integer.MAX_VALUE; } } } for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { dp[0][i][j] = 0; } } for (int k = 1; k <= s; k++) { int instrIdx = -1; for (int i = 0; i < 4; i++) { if (directions[i].equals(instruct[k - 1])) { instrIdx = i; break; } } for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { if (dp[k - 1][i][j] != Integer.MAX_VALUE) { if (instrIdx == i || instrIdx == j) { dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][i][j]); } else { dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][instrIdx][j] + 1); dp[k][i][j] = Math.min(dp[k][i][j], dp[k - 1][i][instrIdx] + 1); } } } } } int minimumMoves = Integer.MAX_VALUE; for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { minimumMoves = Math.min(minimumMoves, dp[s][i][j]); } } return minimumMoves; } public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int m = Integer.parseInt(scanner.nextLine().trim()); String[] instruct = new String[m]; for (int i = 0; i < m; i++) { instruct[i] = scanner.nextLine().trim(); } int result = minimumMoves(instruct); System.out.print(result); scanner.close(); } } Dance rev Java 8

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#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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