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L&T Technology Services - MBA Intern Hiring โ 2024 or 2025 pass out batch.
Please register using the below link: https://forms.office.com/pages/responsepage.aspx?id=eDMbMYqOXkujP-gKPYumClSsrm1sEv1BmOXg5_0vsfxUOUY2WUU1MUpRWlBROUtMSFFaSDRSOFZHRi4u
** Please note that applicants that do not fill the above form will not be considered.
Registration End Date: 6:00PM, 31st July 2024
Type: Full time Offline Internship
Location: Mysore
Specialization: B. Tech + Mba in Sales & Marketing
Year of Pass out: Batch of 2024 or 2025
Stipend: As per Industry Standards
Job Brief:
- A Presales Intern acts as a bridge between the sales and technical teams, providing valuable insights and solutions to potential clients.
- This role involves a strategic approach to understanding client needs and aligning product or service offerings accordingly.
Requirement and Skills:
- Excellent communication and presentation skills.
- Strong analytical and problem-solving abilities.
- Ability to work collaboratively in a fast-paced and dynamic environment.
Please register using the below link: https://forms.office.com/pages/responsepage.aspx?id=eDMbMYqOXkujP-gKPYumClSsrm1sEv1BmOXg5_0vsfxUOUY2WUU1MUpRWlBROUtMSFFaSDRSOFZHRi4u
** Please note that applicants that do not fill the above form will not be considered.
Registration End Date: 6:00PM, 31st July 2024
Type: Full time Offline Internship
Location: Mysore
Specialization: B. Tech + Mba in Sales & Marketing
Year of Pass out: Batch of 2024 or 2025
Stipend: As per Industry Standards
Job Brief:
- A Presales Intern acts as a bridge between the sales and technical teams, providing valuable insights and solutions to potential clients.
- This role involves a strategic approach to understanding client needs and aligning product or service offerings accordingly.
Requirement and Skills:
- Excellent communication and presentation skills.
- Strong analytical and problem-solving abilities.
- Ability to work collaboratively in a fast-paced and dynamic environment.
๐ฑ1
#include <iostream>
#include <vector>
using namespace std;
// Function to find the bitwise OR of sums of all sub-sequences
int bitwiseOrOfSums(const vector<int>& arr) {
int result = 0;
for (int num : arr) {
result |= num;
}
return result;
}
int main() {
int T;
cin >> T;
while (T--) {
int N;
cin >> N;
vector<int> arr(N);
for (int i = 0; i < N; ++i) {
cin >> arr[i];
}
cout << bitwiseOrOfSums(arr) << endl;
}
return 0;
}
Unify โ
#include <iostream>
#include <vector>
using namespace std;
int main() {
int N, Q;
cin >> N >> Q;
vector<int> A(N);
for (int i = 0; i < N; ++i) {
cin >> A[i];
}
vector<int> results(Q);
for (int q = 0; q < Q; ++q) {
int L, R, X;
cin >> L >> R >> X;
--L;
--R;
long long sumXor = 0;
for (int i = L; i <= R; ++i) {
sumXor += (A[i] ^ X);
}
results[q] = sumXor;
}
for (const auto& res : results) {
cout << res << endl;
}
return 0;
}
The XOR Operation โ
Unify
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int maxReward(const vector<int>& reward_1, const vector<int>& reward_2, int k) {
int n = reward_1.size();
vector<pair<int, int>> diffs(n);
for (int i = 0; i < n; ++i) {
diffs[i] = {reward_1[i] - reward_2[i], i};
}
sort(diffs.rbegin(), diffs.rend());
int aa = 0;
for (int i = 0; i < k; ++i) {
aa += reward_1[diffs[i].second];
}
for (int i = k; i < n; ++i) {
aa += reward_2[diffs[i].second];
}
return aa;
}
Atlassian โ
๐2
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
#include <climits>
using namespace std;
int solve(int center_nodes, int center_edges, vector<int>& center_from, vector<int>& center_to, vector<int>& status) {
vector<vector<int>> graph(center_nodes);
for (int i = 0; i < center_edges; ++i) {
int u = center_from[i] - 1;
int v = center_to[i] - 1;
graph[u].push_back(v);
graph[v].push_back(u);
}
queue<pair<int, int>> q;
vector<int> min_time(center_nodes, INT_MAX);
for (int i = 0; i < center_nodes; ++i) {
if (status[i] == 3) {
q.push({i, 0});
min_time[i] = 0;
}
}
while (!q.empty()) {
int node = q.front().first;
int time = q.front().second;
q.pop();
for (int neighbor : graph[node]) {
if (min_time[neighbor] > time + 1) {
min_time[neighbor] = time + 1;
q.push({neighbor, time + 1});
}
}
}
int max_time = 0;
for (int i = 0; i < center_nodes; ++i) {
if (status[i] == 1) {
if (min_time[i] == INT_MAX) {
return -1;
}
max_time = max(max_time, min_time[i]);
}
}
return max_time;
}
Atlassian โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <climits>
using namespace std;
vector<int>solve(vector<int>& cost, vector<string>& aa) {
int n = cost.size();
vector<int> result(n, -1);
vector<int> costA, costB, costBoth;
for (int i = 0; i < n; ++i) {
if (aa[i] == "10") {
costA.push_back(cost[i]);
} else if (aa[i] == "01") {
costB.push_back(cost[i]);
} else if (aa[i] == "11") {
costBoth.push_back(cost[i]);
}
}
sort(costA.begin(), costA.end());
sort(costB.begin(), costB.end());
sort(costBoth.begin(), costBoth.end());
vector<int> prefixA(costA.size() + 1, 0);
vector<int> prefixB(costB.size() + 1, 0);
vector<int> prefixBoth(costBoth.size() + 1, 0);
for (int i = 0; i < costA.size(); ++i) prefixA[i + 1] = prefixA[i] + costA[i];
for (int i = 0; i < costB.size(); ++i) prefixB[i + 1] = prefixB[i] + costB[i];
for (int i = 0; i < costBoth.size(); ++i) prefixBoth[i + 1] = prefixBoth[i] + costBoth[i];
for (int k = 1; k <= n; ++k) {
int minCost = INT_MAX;
for (int i = 0; i <= min(k, (int)costBoth.size()); ++i) {
if (k - i <= (int)costA.size() && k - i <= (int)costB.size()) {
int currentCost = prefixBoth[i] + prefixA[k - i] + prefixB[k - i];
minCost = min(minCost, currentCost);
}
}
if (minCost != INT_MAX) {
result[k - 1] = minCost;
}
}
return result;
}
Atlassian โ
๐1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
int bfs(int start, vector<vector<int>>& adj, vector<bool>& visited) {
queue<int> q;
q.push(start);
visited[start] = true;
int size = 0;
while (!q.empty()) {
int node = q.front();
q.pop();
size++;
for (int neighbor : adj[node]) {
if (!visited[neighbor]) {
visited[neighbor] = true;
q.push(neighbor);
}
}
}
return size;
}
int solve(int K) {
vector<int> primes;
int num = 2;
while (primes.size() < K) {
bool is_prime = true;
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) {
is_prime = false;
break;
}
}
if (is_prime) {
primes.push_back(num);
}
num++;
}
return primes[K-1];
}
int main() {
int N, M;
cin >> N >> M;
vector<vector<int>> adj(N + 1);
for (int i = 0; i < M; i++) {
int U, V;
cin >> U >> V;
adj[U].push_back(V);
adj[V].push_back(U);
}
vector<bool> visited(N + 1, false);
int max_landmarks = 0;
for (int i = 1; i <= N; i++) {
if (!visited[i]) {
int component_size = bfs(i, adj, visited);
max_landmarks = max(max_landmarks, component_size);
}
}
int ticket_price = solve(max_landmarks);
cout << max_landmarks << " " << ticket_price << endl;
return 0;
}
Texas instrument โ
๐1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <queue>
#include <vector>
#include <unordered_map>
#include <functional>
#include <algorithm>
using namespace std;
struct Compare {
bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
if (a.second == b.second) return a.first > b.first;
return a.second < b.second;
}
};
int main() {
int N;
cin >> N;
priority_queue<pair<int, int>, vector<pair<int, int>>, Compare> pq;
unordered_map<int, int> elementPriority;
unordered_map<int, bool> removedElements;
for (int i = 0; i < N; ++i) {
int x, y;
cin >> x >> y;
pq.push({x, y});
elementPriority[x] = y;
removedElements[x] = false;
}
int K;
cin >> K;
for (int i = 0; i < K && !pq.empty(); ++i) {
int removedElement = pq.top().first;
pq.pop();
removedElements[removedElement] = true;
}
vector<pair<int, int>> remainingElements;
for (const auto& entry : elementPriority) {
if (!removedElements[entry.first]) {
remainingElements.push_back({entry.first, entry.second});
}
}
sort(remainingElements.begin(), remainingElements.end(), [](const pair<int, int>& a, const pair<int, int>& b) {
if (a.second == b.second) return a.first < b.first;
return a.second > b.second;
});
for (const auto& el : remainingElements) {
cout << el.first << endl;
}
return 0;
}
Texas instrument โ
#include <iostream>
#include <string>
#include <cctype>
using namespace std;
char encodeChar(char c) {
if (isalpha(c)) {
char base = islower(c) ? 'a' : 'A';
char newChar = base + (c - base + 3) % 26;
return toupper(newChar);
}
return c;
}
string solve(const string& s) {
string result = s;
for (size_t i = 1; i < s.length(); i += 2) {
result[i] = encodeChar(s[i]);
}
return result;
}
Encoding โ
#include <string>
#include <cctype>
using namespace std;
char encodeChar(char c) {
if (isalpha(c)) {
char base = islower(c) ? 'a' : 'A';
char newChar = base + (c - base + 3) % 26;
return toupper(newChar);
}
return c;
}
string solve(const string& s) {
string result = s;
for (size_t i = 1; i < s.length(); i += 2) {
result[i] = encodeChar(s[i]);
}
return result;
}
Encoding โ
๐1