🎯Johnson Controls Off Campus Drive 2024 Hiring As Graduate Engineering Trainee | INR 4-6 LPA
Job Role Graduate Engineer Trainee
Qualification B.E/B.Tech
Batch 2023
Experience Freshers
Salary Rs 4-6 LPA
Job Location Pune
Apply Now:- https://www.allcoding1.com/2024/05/johnson-controls-off-campus-drive-2024.html
Telegram:- @allcoding1
Job Role Graduate Engineer Trainee
Qualification B.E/B.Tech
Batch 2023
Experience Freshers
Salary Rs 4-6 LPA
Job Location Pune
Apply Now:- https://www.allcoding1.com/2024/05/johnson-controls-off-campus-drive-2024.html
Telegram:- @allcoding1
❤1🥰1
Cognizant Careers Off Campus Drive 2024 | Programmer Analyst Trainee | Apply Now
Company Name Cognizant
Post Name Programmer Analyst Trainee
Qualification BE, B Tech, BSC, BCA, ME, M Tech & MCA
Branch Any Branches
Batch 2019, 2020, 2021, 2022, 2023 & 2024
Salary Up to ₹3.5 LPA
Experience Fresher
Job Location Bangalore, Chennai & Kolkata
Last Date Asap
Required Qualification:-
BE, B Tech, BSC, BCA, ME, M Tech & MCA
Apply Now:- https://www.allcoding1.com/2024/05/cognizant-off-campus-drive-2024.html
Telegram:- @allcoding1
Company Name Cognizant
Post Name Programmer Analyst Trainee
Qualification BE, B Tech, BSC, BCA, ME, M Tech & MCA
Branch Any Branches
Batch 2019, 2020, 2021, 2022, 2023 & 2024
Salary Up to ₹3.5 LPA
Experience Fresher
Job Location Bangalore, Chennai & Kolkata
Last Date Asap
Required Qualification:-
BE, B Tech, BSC, BCA, ME, M Tech & MCA
Apply Now:- https://www.allcoding1.com/2024/05/cognizant-off-campus-drive-2024.html
Telegram:- @allcoding1
👍10
#include <iostream>
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {
// Base case: if we have reached the end of bags
if (index > idxy) {
return 0;
}
// Calculate the cost of picking from the left and the right
int cost_left = weights[index] * X + (left_consecutive ? El : 0) +
minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);
int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +
minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);
// Return the minimum of both choices
return min(cost_left, cost_right);
}
int main() {
int N, X, Y, El, Er;
cin >> N >> X >> Y >> El >> Er;
vector<int> weights(N);
for (int i = 0; i < N; ++i) {
cin >> weights[i];
}
cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;
return 0;
}
Amazon Hackon exam
#include <vector>
#include <climits>
#include <algorithm>
using namespace std;
int minimum_energy_cost_recursive(int index,int idxy, int left_consecutive, int right_consecutive, int N, int X, int Y, int El, int Er, vector<int>& weights) {
// Base case: if we have reached the end of bags
if (index > idxy) {
return 0;
}
// Calculate the cost of picking from the left and the right
int cost_left = weights[index] * X + (left_consecutive ? El : 0) +
minimum_energy_cost_recursive(index + 1,idxy, 1, 0, N, X, Y, El, Er, weights);
int cost_right = weights[idxy] * Y + (right_consecutive ? Er : 0) +
minimum_energy_cost_recursive(index ,idxy-1, 0, 1, N, X, Y, El, Er, weights);
// Return the minimum of both choices
return min(cost_left, cost_right);
}
int main() {
int N, X, Y, El, Er;
cin >> N >> X >> Y >> El >> Er;
vector<int> weights(N);
for (int i = 0; i < N; ++i) {
cin >> weights[i];
}
cout << minimum_energy_cost_recursive(0,N-1, 0, 0, N, X, Y, El, Er, weights) << endl;
return 0;
}
Amazon Hackon exam
👍6🤯1
def count_arrays(N, K, X):
MOD = 1000000007
if N == 1:
return 1 if X == 1 else 0
a = [0] * (N + 1)
b = [0] * (N + 1)
a[1] = 1
b[1] = 0
for i in range(2, N + 1):
a[i] = (a[i - 1] * (K - 2) + b[i - 1] * (K - 1)) % MOD
b[i] = a[i - 1] % MOD
return b[N]
N = int(input())
K = int(input())
X = int(input())
print(count_arrays(N, K, X))
Number Of Array code
Thoughtwork exam
Python3
Telegram:- @allcoding1
MOD = 1000000007
if N == 1:
return 1 if X == 1 else 0
a = [0] * (N + 1)
b = [0] * (N + 1)
a[1] = 1
b[1] = 0
for i in range(2, N + 1):
a[i] = (a[i - 1] * (K - 2) + b[i - 1] * (K - 1)) % MOD
b[i] = a[i - 1] % MOD
return b[N]
N = int(input())
K = int(input())
X = int(input())
print(count_arrays(N, K, X))
Number Of Array code
Thoughtwork exam
Python3
Telegram:- @allcoding1
👍5👎2🔥1