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We looking for guys he have experience 5 years working in kuwait
Role: Python Developer (Fresher)
Location: kuwait
Note: - Immediate
Python Development
Machine Learning
Natural Language Processing
Generative AI
Data Analysis
Salary and commission from our project
contact +965 9979 7898 (WhatsApp)
Role: Python Developer (Fresher)
Location: kuwait
Note: - Immediate
Python Development
Machine Learning
Natural Language Processing
Generative AI
Data Analysis
Salary and commission from our project
contact +965 9979 7898 (WhatsApp)
๐คฎ2
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๐DataChannel is hiring for Front End Developer (Intern)
Experience: 0 - 1 year's
Expected Stipend: 2-5 LPA
Apply here: https://www.datachannel.co/career/front-end-developer-intern
๐Acies is hiring for Intern-Quantitative Developer
Experience: 0 - 1 year's
Expected Stipend: 2-4 LPA
Apply here: https://www.linkedin.com/jobs/view/4053339964/?alternateChannel=search
Experience: 0 - 1 year's
Expected Stipend: 2-5 LPA
Apply here: https://www.datachannel.co/career/front-end-developer-intern
๐Acies is hiring for Intern-Quantitative Developer
Experience: 0 - 1 year's
Expected Stipend: 2-4 LPA
Apply here: https://www.linkedin.com/jobs/view/4053339964/?alternateChannel=search
www.datachannel.co
Job Opening - Front End Developer (Intern) | DataChannel
DataChannel is a cloud based data integration and reverse ETL platform to automate the data collection, preparation and management process of all your marketing, CRM, financial & operational data.
๐1
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We are seeking 1-2 research interns to work on image and video synthesis at NVIDIA, with a special focus on generative image and video enhancement. Internships are expected to begin either in late January or summer 2025. Potential areas of research include, but are not limited to: 1. Generative image/video enhancement with perceptual guidance 2. Foundational image/video enhancement models emphasizing controllability, creativity, and identity preservation 3. Efficient image/video generation while maintaining diversity and texture details 4. More related topics in visual synthesis.
We are particularly interested in PhD candidates with strong publication records in visual synthesis. If you are interested, please send your resume to wxiongur@gmail.com
We are particularly interested in PhD candidates with strong publication records in visual synthesis. If you are interested, please send your resume to wxiongur@gmail.com
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
New Network Protocol โ
#include <vector>
#include <string>
#include <algorithm>
#include <climits>
using namespace std;
int ss(const string& substring, int sameTime) {
vector<int> freq(26, 0);
for (char ch : substring) freq[ch - 'a']++;
int count = 0;
for (int f : freq) {
if (f > 1) count += (f * (f - 1)) / 2;
}
return count * sameTime;
}
int getMinimumTime(string s, int sameTime, int partitionTime) {
int n = s.size();
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 0; j < i; ++j) {
string substring = s.substr(j, i - j);
int extraTime = ss(substring, sameTime);
dp[i] = min(dp[i], dp[j] + extraTime + (j > 0 ? partitionTime : 0));
}
}
return dp[n];
}
New Network Protocol โ
๐1
#include <iostream>
#include <vector>
using namespace std;
int main() {
string S;
cin >> S;
int n = S.length();
vector<vector<int>> pref(n, vector<int>(26, 0));
// Prefix count array to store count of each character in all prefixes
for (int i = 0; i < n; i++) {
for (int j = 0; j < 26; j++) {
pref[i][j] = (i == 0 ? 0 : pref[i - 1][j]);
}
pref[i][S[i] - 'a']++;
}
int q;
cin >> q;
while (q--) {
int x;
char c;
cin >> x >> c;
x--; // Decrement x to change it to 0-based indexing
long long ans = 0;
for (int i = 0; i < n - x; i++) {
ans += pref[i + x][c - 'a'] - (i == 0 ? 0 : pref[i - 1][c - 'a']);
}
cout << ans << endl;
}
return 0;
}
Substring queries โ
#include <vector>
using namespace std;
int main() {
string S;
cin >> S;
int n = S.length();
vector<vector<int>> pref(n, vector<int>(26, 0));
// Prefix count array to store count of each character in all prefixes
for (int i = 0; i < n; i++) {
for (int j = 0; j < 26; j++) {
pref[i][j] = (i == 0 ? 0 : pref[i - 1][j]);
}
pref[i][S[i] - 'a']++;
}
int q;
cin >> q;
while (q--) {
int x;
char c;
cin >> x >> c;
x--; // Decrement x to change it to 0-based indexing
long long ans = 0;
for (int i = 0; i < n - x; i++) {
ans += pref[i + x][c - 'a'] - (i == 0 ? 0 : pref[i - 1][c - 'a']);
}
cout << ans << endl;
}
return 0;
}
Substring queries โ
โค1๐1
class Solution {
public:
int minOperations(vector<int>& nums, int x, int y) {
int l = 0, r = *max_element(nums.begin(), nums.end());
auto check = [&](int t) {
long long cnt = 0;
for (int v : nums) {
if (v > 1LL * t * y) {
cnt += (v - 1LL * t * y + x - y - 1) / (x - y);
}
}
return cnt <= t;
};
while (l < r) {
int mid = (l + r) >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l;
}
};
Job execution โ
๐คฉ1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <string>
#include <unordered_map>
using namespace std;
class UnionFind {
public:
UnionFind(int size) {
parent.resize(size);
for (int i = 0; i < size; ++i) {
parent[i] = i;
}
}
int find(int u) {
if (parent[u] != u) {
parent[u] = find(parent[u]);
}
return parent[u];
}
void unionSets(int u, int v) {
parent[find(u)] = find(v);
}
private:
vector<int> parent;
};
bool can_transform(int n, int d, const string& mantra, const string& curse) {
if (mantra.length() != curse.length()) {
return false;
}
UnionFind uf(n);
for (int i = 0; i < n; ++i) {
if (i + d < n) {
uf.unionSets(i, i + d);
}
if (i + d + 1 < n) {
uf.unionSets(i, i + d + 1);
}
}
unordered_map<int, unordered_map<char, int>> mantraGroups;
unordered_map<int, unordered_map<char, int>> curseGroups;
for (int i = 0; i < n; ++i) {
int root = uf.find(i);
mantraGroups[root][mantra[i]]++;
curseGroups[root][curse[i]]++;
}
for (const auto& entry : mantraGroups) {
int root = entry.first;
if (mantraGroups[root] != curseGroups[root]) {
return false;
}
}
return true;
}
int main() {
int n, d;
string mantra, curse;
cin >> n >> d >> mantra >> curse;
if (can_transform(n, d, mantra, curse)) {
cout << "yes" << endl;
} else {
cout << "no" << endl;
}
return 0;
}
Rubrik โ
Vishwamitra and Raktbij
๐3๐คฎ1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
#include <unordered_map>
#include <queue>
#include <climits>
using namespace std;
class Solution {
public:
vector<vector<int>> constructGridLayout(int n, vector<vector<int>>& edges) {
vector<vector<int>> adj(n);
for(int i = 0; i < edges.size(); i++){
adj[edges[i][0]].push_back(edges[i][1]);
adj[edges[i][1]].push_back(edges[i][0]);
}
vector<int> dig(n,0);
for(int i = 0; i < n; i++){
for(auto val : adj[i]) dig[val]++;
}
int root = 0;
for(int i = 0; i < n; i ++){
if(dig[root] > dig[i]) root = i;
}
// cout<<root;
vector<int> vis(n,0);
vis[root] = 1;
vector<int> row1;
int x = adj[root][0];
if(adj[root].size() == 2){
if(dig[x] > dig[adj[root][1]]) x = adj[root][1];
}
row1.push_back(root);
row1.push_back(x);
vis[x] = 1;
while(dig[root] != dig[x]){
int d = 5;
int k = INT_MAX;
for(auto val : adj[x]){
if(vis[val] == 0){
if(dig[val] < d) {
k = val;
d = dig[val];
}
}
}
if(k == INT_MAX) break;
vis[k] = 1;
row1.push_back(k);
x = k;
}
// for(auto val : row1) cout<<val<<" ";
// cout<<endl;
// for(auto val : vis) cout<<val<<" ";
int c = row1.size();
int r = n / c;
vector<vector<int>> ans(r, vector<int>(c,0));
for(int i = 0; i < c; i++) ans[0][i] = row1[i];
for(int i = 1; i < r; i++){
for(int j = 0; j < c; j++){
for(auto val : adj[ans[i-1][j]]){
if(vis[val] == 0){
vis[val] = 1;
ans[i][j] = val;
break;
}
}
}
}
return ans;
}
};
Backup Network โ
๐คฎ1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
const long long MOD = 1e9 + 7;
long long basicForm(long long A) {
long long B = 1;
for (long long i = 2; i * i <= A; i++) {
if (A % i == 0) {
B *= i;
while (A % i == 0) {
A /= i;
}
}
}
if (A > 1) {
B *= A;
}
return B;
}
long long countDivisors(long long n) {
long long divisors = 1;
for (long long i = 2; i * i <= n; i++) {
if (n % i == 0) {
int count = 0;
while (n % i == 0) {
n /= i;
count++;
}
divisors *= (count + 1);
}
}
if (n > 1) {
divisors *= 2;
}
return divisors;
}
long long findD(long long A) {
long long B = basicForm(A);
long long C = 1;
for (long long i = 1; i * i <= B; i++) {
if (B % i == 0) {
C = (C * i) % MOD;
if (i != B / i) {
C = (C * (B / i)) % MOD;
}
}
}
long long D = countDivisors(C);
return D;
}
int main() {
long long A;
cin >> A;
cout << findD(A) << endl;
return 0;
}
converting numbersโ
๐4๐1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <unordered_set>
#include <vector>
using namespace std;
vector<bool> solution(int n, vector<vector<int>>queens, vector<vector<int>>queries) {
unordered_set<int> a, b, cs, d;
for (const auto &queen : queens) {
int r = queen[0], c = queen[1];
a.insert(r);
b.insert(c);
cs.insert(r - c);
d.insert(r + c);
}
vector<bool> r;
for (const auto &query : queries) {
int r_q = query[0], c_q = query[1];
bool u = (a.count(r_q) > 0 || b.count(c_q) > 0 || cs.count(r_q - c_q) > 0 || d.count(r_q + c_q) > 0);
r.push_back(u);
}
return r;
}
๐3๐คฎ2
def p(packet):
bitmask = 0
for ch in packet:
bitmask |= (1 << (ord(ch) - ord('a')))
return bitmask
def ss(packets):
n = len(packets)
bitmasks = [p(p) for p in packets]
lengths = [len(p) for p in packets]
m = 0
for i in range(n):
for j in range(i + 1, n):
if (bitmasks[i] & bitmasks[j]) == 0:
m = max(m, lengths[i] * lengths[j])
return m
num = int(input())
packets = [input() for _ in range(num)]
result = ss(packets)
print(result)
Wipro โ
๐1๐คฉ1
def ss(cipherCode):
digits = list(str(abs(cipherCode)))
digits.sort()
if digits[0] == '0':
for i in range(1, len(digits)):
if digits[i] != '0':
digits[0], digits[i] = digits[i], digits[0]
break
d = int(''.join(digits))
if cipherCode < 0:
d = -d
print(d)
cipherCode = int(input())
ss(cipherCode)
Wipro โ
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AspenTech Hiring !!
Role - C++ Developer
Exp - Fresher
https://aspentech.wd5.myworkdayjobs.com/aspentech/job/Bengaluru-India/C---Developers_R6599
Role - C++ Developer
Exp - Fresher
https://aspentech.wd5.myworkdayjobs.com/aspentech/job/Bengaluru-India/C---Developers_R6599
Myworkdayjobs
C++ Developers
The driving force behind our success has always been the people of AspenTech. What drives us, is our aspiration, our desire and ambition to keep pushing the envelope, overcoming any hurdle, challenging the status quo to continually find a better way. Youโฆ