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Candle bunch

Cross number puzzle

Resource power

def Resource(A, B, C):     same_type_systems = (A // 3) + (B // 3) + (C // 3)     remaining_A = A % 3     remaining_B = B % 3     remaining_C = C % 3     different_type_systems = min(A, B, C)     remaining_A -= different_type_systems     remaining_B -= different_type_systems     remaining_C -= different_type_systems     remaining_A = max(0, remaining_A)     remaining_B = max(0, remaining_B)     remaining_C = max(0, remaining_C)     total_systems = same_type_systems + different_type_systems     leftover_resources = remaining_A + remaining_B + remaining_C     total_systems += leftover_resources // 3         return total_systems Resource power ✅

class Solution {     public int wateringPlants(int[] plants, int capacity) {         int ans = 0;         int vol = capacity;         for (int i = 0; i < plants.length; i++) {             int cur = plants[i];             boolean isEnoughWater = cur <= vol;             ans += (isEnoughWater ? 0 : 2) * i + 1;             vol = (isEnoughWater ? vol : capacity) - cur;         }         return ans;     } } Watering plants ✅

Cisco Ideathon All Answers✅ 1. # 2. def 3. [1], [2], [1,3] 4. [4,3,2,1] 5. [1,3,5] 6. List comprehension is a concise way to create a new list based on an existing iterable, while a for loop can be used for various operations 7. Both b and c 8. True, False 9. VLAN - All of the above 10. IP - Address 11. Internet Backbone 12. Aggregation, Access 13. To put the interface into forwarding state immediately, bypassing the listening and learning states. 14. Partial mesh topology 15. The next hop self command allows an IBGP ... 16. IP Phone and Access Points 17. Events 18. It allows the attacker to intercept and manipulate network traffic. 19. The DHCP server that responds first 20. It indicates the best part to reach the destination network. 21. Supported Versions 22. LOCAL_PREF 23. SSID 24. BPDU Filter 25. Switch 26. NA 27. Use NAT Protocol Translation 28. Non broadcast multi access 29. discarding 30. NAT64 is used to translate IPv6 addresses to IPv4 addresses for communication between IPv6-only and IPv4-only hosts. 31. Encryption and MIC with AES and GCMP 32. Packet 5 arrived ... 33. All of them 34. High amount of output 35. IGP 36. 255 37. Activity Tracking and Access Approval 38. Area id 39. Command 40. Ssl and Http 41. Network tap 42. Design to filter 43. Interceptinf and decrypt 44. Engery 45. 12 46. 4.54 47. 3 48. C 49. QOS 50. 37 51. It enables router to forward to IPv6 packets. 52. BGP sync ensured that a router ... its IGP 53. Inverse Query 54. Server virtualisation 55. Administrative distance is a measure ... route.

#include <iostream> #include <vector> #include <algorithm> #include <climits> using namespace std; int solve(int i, vector<int>& lst, int ad, const vector<int>& machineCount, const vector<int>& finalMachineCount, int shiftingCost) {     int n = machineCount.size();     if (i == n) {         int res = ad;         for (int k = 0; k < 3; ++k) {             res += abs(lst[k] - finalMachineCount[k]);         }         return res;     }         int res = INT_MAX;     res = min(res, solve(i + 1, lst, ad, machineCount, finalMachineCount, shiftingCost));         for (int j = 0; j < 3; ++j) {         lst[j] += machineCount[i];         if (lst[j] - machineCount[i] != 0) {             res = min(res, solve(i + 1, lst, ad + shiftingCost, machineCount, finalMachineCount, shiftingCost));         } else {             res = min(res, solve(i + 1, lst, ad, machineCount, finalMachineCount, shiftingCost));         }         lst[j] -= machineCount[i];     }         return res; } int getMinimumCost(const vector<int>& machineCount, const vector<int>& finalMachineCount, int shiftingCost) {     vector<int> lst(3, 0);     return solve(0, lst, 0, machineCount, finalMachineCount, shiftingCost); } int main() {     vector<int> machineCount = { /* your values here */ };     vector<int> finalMachineCount = { /* your values here */ };     int shiftingCost = /* your value here */;         cout << getMinimumCost(machineCount, finalMachineCount, shiftingCost) << endl;     return 0; }