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Piyush Ranjan on LinkedIn: #hiring #insurtech | 26 comments
โจ Hiring alert! โจ
We're looking for exceptional Backend (Python+Django) and Frontend (React) Engineers to join our engineering team. Join us and get theโฆ | 26 comments on LinkedIn
We're looking for exceptional Backend (Python+Django) and Frontend (React) Engineers to join our engineering team. Join us and get theโฆ | 26 comments on LinkedIn
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Business Analytics ( Freshers to 20 Months Exp)
Candidates Can apply :-
TECHNICAL QUALIFICATION BACKGROUND MANDATE
Freshers can apply who have Complete Certification or Training in SAS, SQL Skills.
OR
Experience with SAS, SQL Technologies.
Skills Required
1. SAS
2. SQL
3. BFSI Domain :- Risk, Card, Insurance, Finance, Fintech ETC.
Work Location - WFH
Please share profiles at jasneet@supportstar.in || jasneet.sscs@gmail.com OR call@7303249754
Open for Reference
Candidates Can apply :-
TECHNICAL QUALIFICATION BACKGROUND MANDATE
Freshers can apply who have Complete Certification or Training in SAS, SQL Skills.
OR
Experience with SAS, SQL Technologies.
Skills Required
1. SAS
2. SQL
3. BFSI Domain :- Risk, Card, Insurance, Finance, Fintech ETC.
Work Location - WFH
Please share profiles at jasneet@supportstar.in || jasneet.sscs@gmail.com OR call@7303249754
Open for Reference
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We are hiring B.Tech (CSE/IT) Freshers.
Grazitti Interactive is looking forward to connecting with Placement Cells and TPOs for B.Tech Campus Recruitment.
Please share the required details as per the below-mentioned criteria with kavita.rani@grazitti.com
Eligibility Criteria for the students:
ยท Year of graduation:2024/2025
ยท 60%+ overall education
ยท Location โ Panchkula/Mohali
ยท No standing Backlogs
ยท Flexible for any location
. B.Tech CSE/IT only
Give someone a chance to be a part of a Great place to work!
We will connect back for further process soon!
Grazitti Interactive is looking forward to connecting with Placement Cells and TPOs for B.Tech Campus Recruitment.
Please share the required details as per the below-mentioned criteria with kavita.rani@grazitti.com
Eligibility Criteria for the students:
ยท Year of graduation:2024/2025
ยท 60%+ overall education
ยท Location โ Panchkula/Mohali
ยท No standing Backlogs
ยท Flexible for any location
. B.Tech CSE/IT only
Give someone a chance to be a part of a Great place to work!
We will connect back for further process soon!
def getMaxGoodSubarrayLength(limit, financialMetrics):
n = len(financialMetrics)
max_length = 0
left = 0
right = 0
while right < n:
while right < n and financialMetrics[right] > limit / (right - left + 1):
right += 1
max_length = max(max_length, right - left)
left = right
right += 1
return max_length if max_length > 0 else -1
Tik tok metrics Analysis
Tik Tok โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
SELECT a.iban, COUNT(*) AS transactions, SUM(t.amount) AS total FROM accounts a JOIN transactions t ON a.id = t.account_id WHERE EXTRACT(MONTH FROM t.dt) = 9 AND EXTRACT(YEAR FROM t.dt) = 2023 GROUP BY a.iban ORDER BY โฆ
MOD = 10**9 + 7
def count_ways(s):
n = len(s)
if n == 0:
return 0
dp = [0] * 10
new_dp = [0] * 10
if s[0] == '?':
for digit in range(10):
dp[digit] = 1
else:
dp[int(s[0])] = 1
for i in range(1, n):
new_dp = [0] * 10
if s[i] == '?':
for current in range(10):
for prev in range(10):
if current != prev:
new_dp[current] = (new_dp[current] + dp[prev]) % MOD
else:
current_digit = int(s[i])
for prev in range(10):
if current_digit != prev:
new_dp[current_digit] = (new_dp[current_digit] + dp[prev]) % MOD
dp = new_dp
result = sum(dp) % MOD
return result
IWD Coding Challenge
ZSโ
#include <bits/stdc++.h>
using namespace std;
int solve(const string& num) {
char mini = '9';
char maxi = '0';
for (char digit : num) {
if (digit > maxi) maxi = digit;
if (digit < mini) mini = digit;
}
int sum = 0;
for (char digit : num) {
if (digit != maxi && digit != mini) {
sum += digit - '0';
}
}
return sum;
}
int main() {
int N;
cin >> N;
vector<int> v(N);
for (int i = 0; i < N; ++i) {
cin >> v[i];
}
int total = 0;
for (int num : v) {
total += solve(to_string(num));
}
cout << total << endl;
return 0;
}
Browserstack โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> calculateRanks(const vector<int>& leaderboard, const vector<int>& attempts) {
vector<int> ranks;
vector<int> uniqueScores = leaderboard;
uniqueScores.erase(unique(uniqueScores.begin(), uniqueScores.end()), uniqueScores.end());
int n = uniqueScores.size();
for (int score : attempts) {
while (n > 0 && score >= uniqueScores[n-1]) {
n--;
}
ranks.push_back(n + 1);
}
return ranks;
}
int main() {
int s;
cin >> s;
vector<int> leaderboard(s);
for (int i = 0; i < s; ++i) {
cin >> leaderboard[i];
}
int x;
cin >> x;
vector<int> attempts(x);
for (int i = 0; i < x; ++i) {
cin >> attempts[i];
}
vector<int> ranks = calculateRanks(leaderboard, attempts);
for (int rank : ranks) {
cout << rank << " ";
}
cout << endl;
return 0;
}
Browserstack โ
int minCacheSize(int n, const vector<string>& reqs, int k){
auto getHits = [&](int sz) {
unordered_set<string> cache;
list<string> order;
int hits = 0;
for (const string& item : reqs) {
if (cache.find(item) != cache.end()) {
hits++;
order.erase(find(order.begin(), order.end(), item));
order.push_front(item);
}
else {
if (cache.size() == sz) {
cache.erase(order.back());
order.pop_back();
}
cache.insert(item);
order.push_front(item);
}
}
return hits;
};
int l = 1, r = n, res = -1;
while (l <= r) {
int m = l + (r - l) / 2;
if (getHits(m) >= k) {
res = m;
r = m - 1;
} else {
l = m + 1;
}
}
return res;
}
ion mincachesizeโ
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Paisabazaar.com is hiring Software Engineer
For 2021, 2022, 2023, 2024 grads
Location: Gurugram
https://forms.gle/pB5UsHKejjb4MV3e6
For 2021, 2022, 2023, 2024 grads
Location: Gurugram
https://forms.gle/pB5UsHKejjb4MV3e6
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
Node *left, *right;
Node(int x) : data(x), left(nullptr), right(nullptr) {}
};
int solve(Node* root, int &maxi) {
if (root == NULL) return INT_MAX;
if (root->left == NULL && root->right == NULL) return root->data;
int left = solve(root->left, maxi);
int right = solve(root->right, maxi);
if (left != INT_MAX) maxi = max(maxi, root->data - left);
if (right != INT_MAX) maxi = max(maxi, root->data - right);
return min(root->data, min(left, right));
}
int maxDiff(Node* root) {
if (root == NULL) return 0;
int maxi = INT_MIN;
solve(root, maxi);
return maxi;
}
Node* buildTree(const vector<int>& levelOrder) {
if (levelOrder.empty() || levelOrder[0] == -1) return nullptr;
Node* root = new Node(levelOrder[0]);
queue<Node*> q;
q.push(root);
int index = 1;
while (!q.empty() && index < levelOrder.size()) {
Node* node = q.front();
q.pop();
if (index < levelOrder.size() && levelOrder[index] != -1) {
node->left = new Node(levelOrder[index]);
q.push(node->left);
}
index++;
if (index < levelOrder.size() && levelOrder[index] != -1) {
node->right = new Node(levelOrder[index]);
q.push(node->right);
}
index++;
}
return root;
}
int main() {
int height;
cin >> height;
vector<int> levelOrder;
int value;
while (cin >> value) {
levelOrder.push_back(value);
}
if (height == 0 levelOrder.empty() (levelOrder.size() == 1 && levelOrder[0] == -1)) {
cout << 0 << endl;
return 0;
}
Node* root = buildTree(levelOrder);
int result = maxDiff(root);
cout << result << endl;
return 0;
}
nodes and their ancestor Samsungโ
#include <bits/stdc++.h>
using namespace std;
struct TreeNode {
int val;
TreeNode *left, *right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
TreeNode* buildTree(const vector<string>& levelOrder) {
if (levelOrder.empty() || levelOrder[0] == "-1") return nullptr;
TreeNode* root = new TreeNode(stoi(levelOrder[0]));
queue<TreeNode*> q;
q.push(root);
int index = 1;
while (!q.empty() && index < levelOrder.size()) {
TreeNode* node = q.front();
q.pop();
if (index < levelOrder.size() && levelOrder[index] != "-1") {
node->left = new TreeNode(stoi(levelOrder[index]));
q.push(node->left);
}
index++;
if (index < levelOrder.size() && levelOrder[index] != "-1") {
node->right = new TreeNode(stoi(levelOrder[index]));
q.push(node->right);
}
index++;
}
return root;
}
int sumOfNodesWithoutSiblings(TreeNode* root) {
if (!root) return -1;
queue<TreeNode*> q;
q.push(root);
int sum = 0;
while (!q.empty()) {
TreeNode* node = q.front();
q.pop();
if ((node->left && !node->right) || (!node->left && node->right)) {
sum += node->left ? node->left->val : node->right->val;
}
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
return sum == 0 ? -1 : sum;
}
int main() {
int height;
cin >> height;
vector<string> levelOrder;
string value;
while (cin >> value) {
levelOrder.push_back(value);
}
if (height == 0 levelOrder.empty() (levelOrder.size() == 1 && levelOrder[0] == "-1")) {
cout << -1 << endl;
return 0;
}
TreeNode* root = buildTree(levelOrder);
int result = sumOfNodesWithoutSiblings(root);
cout << result << endl;
return 0;
}
no sibling node
Samsungโ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <bits/stdc++.h>
using namespace std;
class UnionFind {
vector<int> parent, rank;
public:
UnionFind(int n) {
parent.resize(n);
iota(parent.begin(), parent.end(), 0);
rank.resize(n, 0);
}
int find(int u) {
if (u != parent[u])
parent[u] = find(parent[u]);
return parent[u];
}
void unite(int u, int v) {
int rootU = find(u);
int rootV = find(v);
if (rootU != rootV) {
if (rank[rootU] > rank[rootV])
parent[rootV] = rootU;
else if (rank[rootU] < rank[rootV])
parent[rootU] = rootV;
else {
parent[rootV] = rootU;
rank[rootU]++;
}
}
}
};
int solve(int n, int c_service, int c_link, vector<pair<int, int>>& pairs) {
UnionFind uf(n);
for (auto& p : pairs) {
uf.unite(p.first - 1, p.second - 1);
}
vector<bool> conn(n, false);
int cnt = 0;
for (int i = 0; i < n; i++) {
int root = uf.find(i);
if (!conn[root]) {
conn[root] = true;
cnt++;
}
}
int op1 = n * c_service;
int op2 = cnt * c_service + (n - cnt) * c_link;
return min(op1,op2);
}
int main() {
int n, c_service, c_link, numPairs;
cin >> n >> numPairs >> c_service >> c_link;
vector<pair<int, int>> pairs(numPairs);
for (int i = 0; i < numPairs; i++) {
cin >> pairs[i].first >> pairs[i].second;
}
int minCost = solve(n, c_service, c_link, pairs);
cout << minCost << endl;
return 0;
}
IBMโ