if currentLevelSize == 1: bottlenecks.append(queue[0]) # Process all tasks at current time step for i in range(currentLevelSize): task = queue.popleft() # Update dependencies for all dependent tasks for dependent in graph[task]: inDegree[dependent] -= 1 if inDegree[dependent] == 0: queue.append(dependent) return bottlenecks1a:T68f,#include #include #include #include using namespace std; class Solution { public: vector findBottlenecks(int n, vector> edges) { // Build adjacency list and calculate in-degrees vector> graph(n); vector inDegree(n, 0); for (vector edge : edges) { int u = edge[0], v = edge[1]; graph[u].push_back(v); inDegree[v]++; } // Initialize queue with tasks that have no dependencies queue queue; for (int i = 0; i < n; i++) { if (inDegree[i] == 0) { queue.push(i); } } vector bottlenecks; // Simulate execution time step by time step while (!queue.empty()) { int currentLevelSize = queue.size(); // If exactly one task is ready at this time step, it's a bottleneck if (currentLevelSize == 1) { bottlenecks.push_back(queue.front()); } // Process all tasks at current time step for (int i = 0; i < currentLevelSize; i++) { int task = queue.front(); queue.pop(); // Update dependencies for all dependent tasks for (int dependent : graph[task]) { inDegree[dependent]--; if (inDegree[de

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