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ACCENTURE EXAM SOLUTIONS

ACCENTURE EXAM SOLUTIONS

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πŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srksvk

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πŸ“ˆ Analytical overview of Telegram channel ACCENTURE EXAM SOLUTIONS

Channel ACCENTURE EXAM SOLUTIONS (@coding_are) in the English language segment is an active participant. Currently, the community unites 14 125 subscribers, ranking 14 097 in the Education category and 28 073 in the India region.

πŸ“Š Audience metrics and dynamics

Since its creation on Π½Π΅Π²Ρ–Π΄ΠΎΠΌΠΎ, the project has demonstrated rapid growth, gathering an audience of 14 125 subscribers.

According to the latest data from 28 September, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -105 over the last 30 days and by -4 over the last 24 hours, overall reach remains high.

  • Verification status: Not verified
  • Engagement rate (ER): The average audience engagement rate is 3.68%. Within the first 24 hours after publication, content typically collects 1.57% reactions from the total number of subscribers.
  • Post reach: On average, each post receives 520 views. Within the first day, a publication typically gains 222 views.
  • Reactions and interaction: The audience actively supports content: the average number of reactions per post is 2.
  • Thematic interests: Content is focused on key topics such as placement, gaurntee, suree, capgemini, infosy.

πŸ“ Description and content policy

The author describes the resource as a platform for expressing subjective opinions:
β€œπŸ”₯Guys plz Stop fearing for daily exams πŸ“ πŸ‘¨β€πŸ’» @srksvk is here to help you all at lowest cost possible.πŸ’ͺ πŸŒ€ ” Our Only Aim Is To Let Get Placed To You In A Reputed Company πŸ”₯Effort from our side = πŸ’― πŸ“±Main Channel: @coding_are πŸ“±Tel I'd : @srks...”

Thanks to the high frequency of updates (latest data received on 28 September, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.

14 125
Subscribers
-424 hours
-127 days
-10530 days
Posts Archive
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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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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Changes variable

Full accepted ☺️

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