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Version Reviewing: 1
"Alternative to proprietary mobile development"
Summary:
- dramatically faster than other apps - built once, optimized everywhere - easy navigation
Version Reviewing: 1
"lightning fast"
Summary:
- smooth transitions for teams with ionic - simple to use, user friendly
Version Reviewing: 1
"Spend more time on what matters"
Summary:
- Ship stable apps faster than other app builders - Automate app delivery
Version Reviewing: 1.3
"security visibility"
Summary:
- Uses BPF to define network-layer and application-layer security policies for microservices - Supports different frameworks, such as Kubernates, Docker, etc.
Version Reviewing: 2018.3
"Easy to write in C, C++"
Summary:
- Lower barrier entry to write in C/C++ - Only able to write in C language, not compatible with others - version control and organization of IDE for simpler use
Version Reviewing: 1
"Helix Products by Perforce"
Summary:
- Variety of Helix products, from APM to Static Analysis tools - Useful for bigger companies, rather than start-ups - difficult to setup, costly, but designed for engineering teams bigger in size
Version Reviewing: 3.0
"Easy tools, Full access"
Summary:
- Models APIs with accuracy - visualize while designing - Share with team easily
Version Reviewing: 3.1
"Updated capabilities for intelligent, compute-intensive apps"
Summary:
- Device manager plug-in support - Secure Kubernetes on your own platform
Version Reviewing: 6.0
"End to end visibility, control, and automation"
Summary:
- Plan, optimize, manage - simplify your cloud journey through planning migration
Version Reviewing: 1
"Better Model, Better Customer Service"
Summary:
- AI platform makes seamless efforts for customer service - Anywhere APIs - Gig and AI meet in the middle
What does this code do?
void guess_my_function(int arr[], int left, int middle, int right){ int low = middle - left + 1; int high = right - middle; int L[] = new int[low]; int R[] = new int[high]; int i = 0, j = 0; for (i = 0; i < low; i++) L[i] = arr[left + i]; for (j = 0; j < high; j++) R[j] = arr[middle + 1 + j]; int k = left; i = 0; j = 0; while (i < low && j < high){ if (L[i] <= R[j]){ arr[k] = L[i]; i++; } else { arr[k] = R[j]; j++; } k++; } while (i < low){ arr[k] = L[i]; i++; k++; } while (j < high){ arr[k] = R[j]; j++; k++; } }
Programming Language: Java