Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
๐IQVIA is hiring for Software Developer
Experience: 0 - 1 years
Expected Salary: 5 - 12 LPA
Apply here: https://jobs.iqvia.com/en/job/-/-/24443/70955092656?source=LinkedIn_Slots
Experience: 0 - 1 years
Expected Salary: 5 - 12 LPA
Apply here: https://jobs.iqvia.com/en/job/-/-/24443/70955092656?source=LinkedIn_Slots
๐1
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
Software Engineer Internship 2025 (4 months) - India-based, Gurgaon office - Careers at Agoda
https://careersatagoda.com/job/6323041-software-engineer-internship-2025-4-months-india-based-gurgaon-office/
https://careersatagoda.com/job/6323041-software-engineer-internship-2025-4-months-india-based-gurgaon-office/
๐1
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
I am looking for a motivated Fresher to join in Bangalore. This role will involve providing HR support and managing coordination tasks within the HR department.
Role - HR Apprentice
Location - Bangalore
Start Date - Immediate
Qualification MBA in HR (Fresher)
Preferred Candidate Female
Essential Skills - Proficient in MS Excel, good communication, and coordination skills
Email resume - reetikaraj@fedfina.com
Role - HR Apprentice
Location - Bangalore
Start Date - Immediate
Qualification MBA in HR (Fresher)
Preferred Candidate Female
Essential Skills - Proficient in MS Excel, good communication, and coordination skills
Email resume - reetikaraj@fedfina.com
โค2๐2
LinkedIn SDE Intern Opening that I shared a few weeks ago
Test 1 is already done, Have you given it ?
Test 2 is today, Have you got the link?
All the best โฅ๏ธ
Test 1 is already done, Have you given it ?
Test 2 is today, Have you got the link?
All the best โฅ๏ธ
๐5โค2๐คฉ1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <bits/stdc++.h>
using namespace std;
/*
* Complete the 'countGoodSubsequences' function below.
*
* The function is expected to return an INTEGER.
* The function accepts STRING word as parameter.
*/
const int M = 1000000007;
int add(int x, int y) {
if ((x += y) >= M) {
x -= M;
}
return x;
}
int mul(long long x, long long y) {
return x * y % M;
}
int rev(int x) {
int r = 1;
for (int y = M - 2; y; y >>= 1) {
if (y & 1) {
r = mul(r, x);
}
x = mul(x, x);
}
return r;
}
int countGoodSubsequences(string s) {
unordered_map<char, int> have;
for (char c : s) {
++have[c];
}
const int n = s.length();
vector<int> r(n + 1);
for (int i = 1; i <= n; ++i) {
r[i] = rev(i);
}
vector<int> dp(n + 1);
int m = 0;
for (const auto& p : have) {
m = max(m, p.second);
for (int i = 1, t = 1; i <= p.second; ++i) {
t = mul(mul(t, p.second - i + 1), r[i]);
dp[i] = add(mul(dp[i], t + 1), t);
}
}
int ans = 0;
for (int i = 1; i <= m; ++i) {
ans = add(ans, dp[i]);
}
return ans;
}
LinkedIn โ
๐1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
using namespace std;
const int MOD = 1e9 + 7;
long long comb3(long long n) {
if (n < 3) return 0;
return ((n * (n - 1) % MOD) * (n - 2) % MOD) / 6 % MOD;
}
int countEvenProductTriplets(const vector<int>& arr) {
int n = arr.size();
int odd_count = 0, even_count = 0;
for (int num : arr) {
if (num % 2 == 0) {
even_count++;
} else {
odd_count++;
}
}
long long total_triplets = comb3(n);
long long odd_triplets = comb3(odd_count);
long long result = (total_triplets - odd_triplets + MOD) % MOD;
return (int)result;
}
LinkedIn โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
public static long getMinimumCost(int edgeDeviceCost, int inputPeripheralCost, int bundleCost, int x, int y) {
long minCost = Long.MAX_VALUE;
for (int bundles = 0; bundles <= Math.max(x, y); bundles++) {
int remainingEdgeDevices = Math.max(0, x - bundles);
int remainingPeripherals = Math.max(0, y - bundles);
long cost = (long) bundles * bundleCost
+ (long) remainingEdgeDevices * edgeDeviceCost
+ (long) remainingPeripherals * inputPeripheralCost;
minCost = Math.min(minCost, cost);
}
return minCost;
}
LinkedIn โ
typedef long long ll;
const int MOD = 1e9 + 7;
ll comb3(ll n) {
if(n < 3) return 0;
ll res = n;
res = res * (n - 1) % MOD;
res = res * (n - 2) % MOD;
ll inv6 = 166666668;
res = res * inv6 % MOD;
return res;
}
int evenproduct(vector<int> &nums) {
ll n = nums.size();
ll even = 0, odd = 0;
for(auto num : nums){
if(num % 2 == 0) even++;
else odd++;
}
ll total = comb3(n);
ll all_odd = comb3(odd);
ll valid = (total - all_odd + MOD) % MOD;
return (int)valid;
}
LinkedIn โ
Credit : ishaan
const int MOD = 1e9 + 7;
ll comb3(ll n) {
if(n < 3) return 0;
ll res = n;
res = res * (n - 1) % MOD;
res = res * (n - 2) % MOD;
ll inv6 = 166666668;
res = res * inv6 % MOD;
return res;
}
int evenproduct(vector<int> &nums) {
ll n = nums.size();
ll even = 0, odd = 0;
for(auto num : nums){
if(num % 2 == 0) even++;
else odd++;
}
ll total = comb3(n);
ll all_odd = comb3(odd);
ll valid = (total - all_odd + MOD) % MOD;
return (int)valid;
}
LinkedIn โ
Credit : ishaan
typedef long long ll;
ll getMinOperations(vector<int> &change, vector<int> &arr)
{
int n = change.size(), m = arr.size();
vector<int> hasSetOperation(m + 1, 0);
for (int i = 0; i < n; ++i)
{
if (change[i] > 0 && change[i] <= m)
{
hasSetOperation[change[i]] = 1;
}
}
for (int j = 1; j <= m; ++j)
{
if (hasSetOperation[j] == 0)
{
return -1;
}
}
ll total_decrements = 0;
for (auto val : arr)
{
total_decrements += (ll)val;
}
ll total_sets = (ll)m;
ll total_operations = total_decrements + total_sets;
if (total_operations <= (ll)n)
{
return total_operations;
}
else
{
return -1;
}
}
Minimum operation
LinkedInโ
ll getMinOperations(vector<int> &change, vector<int> &arr)
{
int n = change.size(), m = arr.size();
vector<int> hasSetOperation(m + 1, 0);
for (int i = 0; i < n; ++i)
{
if (change[i] > 0 && change[i] <= m)
{
hasSetOperation[change[i]] = 1;
}
}
for (int j = 1; j <= m; ++j)
{
if (hasSetOperation[j] == 0)
{
return -1;
}
}
ll total_decrements = 0;
for (auto val : arr)
{
total_decrements += (ll)val;
}
ll total_sets = (ll)m;
ll total_operations = total_decrements + total_sets;
if (total_operations <= (ll)n)
{
return total_operations;
}
else
{
return -1;
}
}
Minimum operation
LinkedInโ
๐1
static final int MOD = (int)1e9 + 7;
public static int countGoodSubsequences(String word) {
int[] freq = new int[26];
int maxFreq = 0;
for (char c : word.toCharArray()) {
freq[c - 'a']++;
maxFreq = Math.max(maxFreq, freq[c - 'a']);
}
int n = word.length();
long[] factorial = new long[maxFreq + 1];
long[] invFactorial = new long[maxFreq + 1];
factorial[0] = invFactorial[0] = 1;
for (int i = 1; i <= maxFreq; i++) {
factorial[i] = factorial[i - 1] * i % MOD;
}
invFactorial[maxFreq] = modInverse(factorial[maxFreq], MOD);
for (int i = maxFreq - 1; i >= 1; i--) {
invFactorial[i] = invFactorial[i + 1] * (i + 1) % MOD;
}
long totalGoodSequences = 0;
for (int k = 1; k <= maxFreq; k++) {
long total_k = 1;
boolean hasValidChar = false;
for (int i = 0; i < 26; i++) {
if (freq[i] >= k) {
hasValidChar = true;
long c = comb(freq[i], k, factorial, invFactorial);
total_k = total_k * (c + 1) % MOD;
}
}
if (hasValidChar) {
total_k = (total_k - 1 + MOD) % MOD;
totalGoodSequences = (totalGoodSequences + total_k) % MOD;
}
}
return (int) totalGoodSequences;
}
static long modInverse(long a, int mod) {
return modPow(a, mod - 2, mod);
}
static long modPow(long a, long b, int mod) {
long result = 1;
a %= mod;
while (b > 0) {
if ((b & 1) == 1)
result = result * a % mod;
a = a * a % mod;
b >>= 1;
}
return result;
}
static long comb(int n, int k, long[] factorial, long[] invFactorial) {
if (k > n) return 0;
return factorial[n] * invFactorial[k] % MOD * invFactorial[n - k] % MOD;
}
Good sequence
LinkedIn โ
โค1๐1
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
typedef long long ll;
long findMinIncrease(vector<int> threadSize) {
int n = threadSize.size();
if(n < 3) return 0;
vector<ll> required_increase(n, 0);
for(int i=1; i<n-1; ++i){
ll needed = max((ll)threadSize[i-1], (ll)threadSize[i+1]) + 1;
if(threadSize[i] < needed){
required_increase[i] = needed - (ll)threadSize[i];
}
else{
required_increase[i] = 0;
}
}
pair<int, ll> prev0 = {0, 0};
pair<int, ll> prev1 = {-1, 0};
for(int i=0; i<n; ++i){
pair<int, ll> current0 = {0, 0};
pair<int, ll> current1 = {0, 0};
if(i == 0 || i == n-1){
if(prev0.first > prev1.first){
current0 = prev0;
}
else if(prev0.first < prev1.first){
current0 = prev1;
}
else{
current0.first = prev0.first;
current0.second = min(prev0.second, prev1.second);
}
prev0 = current0;
prev1 = {-1, 0};
continue;
}
if(prev0.first > prev1.first){
current0.first = prev0.first;
current0.second = prev0.second;
}
else if(prev0.first < prev1.first){
current0.first = prev1.first;
current0.second = prev1.second;
}
else{
current0.first = prev0.first;
current0.second = min(prev0.second, prev1.second);
}
if(prev0.first != -1){
current1.first = prev0.first + 1;
current1.second = prev0.second + required_increase[i];
}
else{
current1.first = -1;
current1.second = 0;
}
pair<int, ll> new_prev0 = current0;
pair<int, ll> new_prev1;
if(current1.first != -1){
new_prev1 = current1;
}
else{
new_prev1 = {-1, 0};
}
prev0 = new_prev0;
prev1 = new_prev1;
}
pair<int, ll> final_ans;
if(prev0.first > prev1.first){
final_ans = prev0;
}
else if(prev0.first < prev1.first){
final_ans = prev1;
}
else{
final_ans.first = prev0.first;
final_ans.second = min(prev0.second, prev1.second);
}
return final_ans.second;
}
LinkedIn โ
long findMinIncrease(vector<int> threadSize) {
int n = threadSize.size();
if(n < 3) return 0;
vector<ll> required_increase(n, 0);
for(int i=1; i<n-1; ++i){
ll needed = max((ll)threadSize[i-1], (ll)threadSize[i+1]) + 1;
if(threadSize[i] < needed){
required_increase[i] = needed - (ll)threadSize[i];
}
else{
required_increase[i] = 0;
}
}
pair<int, ll> prev0 = {0, 0};
pair<int, ll> prev1 = {-1, 0};
for(int i=0; i<n; ++i){
pair<int, ll> current0 = {0, 0};
pair<int, ll> current1 = {0, 0};
if(i == 0 || i == n-1){
if(prev0.first > prev1.first){
current0 = prev0;
}
else if(prev0.first < prev1.first){
current0 = prev1;
}
else{
current0.first = prev0.first;
current0.second = min(prev0.second, prev1.second);
}
prev0 = current0;
prev1 = {-1, 0};
continue;
}
if(prev0.first > prev1.first){
current0.first = prev0.first;
current0.second = prev0.second;
}
else if(prev0.first < prev1.first){
current0.first = prev1.first;
current0.second = prev1.second;
}
else{
current0.first = prev0.first;
current0.second = min(prev0.second, prev1.second);
}
if(prev0.first != -1){
current1.first = prev0.first + 1;
current1.second = prev0.second + required_increase[i];
}
else{
current1.first = -1;
current1.second = 0;
}
pair<int, ll> new_prev0 = current0;
pair<int, ll> new_prev1;
if(current1.first != -1){
new_prev1 = current1;
}
else{
new_prev1 = {-1, 0};
}
prev0 = new_prev0;
prev1 = new_prev1;
}
pair<int, ll> final_ans;
if(prev0.first > prev1.first){
final_ans = prev0;
}
else if(prev0.first < prev1.first){
final_ans = prev1;
}
else{
final_ans.first = prev0.first;
final_ans.second = min(prev0.second, prev1.second);
}
return final_ans.second;
}
LinkedIn โ
๐1
#define INF INT_MAX
int minimumStressLevel(int N, vector<vector<int>>& abw, int X, int Y) {
vector<vector<pair<int, int>>> graph(N + 1);
for (auto& road : abw) {
int a = road[0], b = road[1], w = road[2];
graph[a].push_back({b, w});
graph[b].push_back({a, w});
}
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
vector<int> dist(N + 1, INF);
pq.push({0, X});
dist[X] = 0;
while (!pq.empty()) {
int u = pq.top().second;
int stress = pq.top().first;
pq.pop();
if (u == Y) return stress;
for (auto& neighbor : graph[u]) {
int v = neighbor.first;
int w = neighbor.second;
int new_stress = max(stress, w);
if (new_stress < dist[v]) {
dist[v] = new_stress;
pq.push({new_stress, v});
}
}
}
return -1;
}
ROBIN HOODโ
Probo (Intern)
int minimumStressLevel(int N, vector<vector<int>>& abw, int X, int Y) {
vector<vector<pair<int, int>>> graph(N + 1);
for (auto& road : abw) {
int a = road[0], b = road[1], w = road[2];
graph[a].push_back({b, w});
graph[b].push_back({a, w});
}
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
vector<int> dist(N + 1, INF);
pq.push({0, X});
dist[X] = 0;
while (!pq.empty()) {
int u = pq.top().second;
int stress = pq.top().first;
pq.pop();
if (u == Y) return stress;
for (auto& neighbor : graph[u]) {
int v = neighbor.first;
int w = neighbor.second;
int new_stress = max(stress, w);
if (new_stress < dist[v]) {
dist[v] = new_stress;
pq.push({new_stress, v});
}
}
}
return -1;
}
ROBIN HOODโ
Probo (Intern)
๐1
#include <bits/stdc++.h>
using namespace std;
string getPerfectExecutionString(string& category) {
vector<int> freq(26, 0);
for (char& x : category) {
freq[x - 'a']++;
}
string temp;
string ans;
int p = 0;
for (int i = 0; i < 26; i++) {
while (!temp.empty() && temp.back() - 'a' <= i) {
ans.push_back(temp.back());
temp.pop_back();
}
while (freq[i]) {
if (category[p] - 'a' == i) {
ans.push_back(category[p]);
} else {
temp.push_back(category[p]);
}
freq[category[p++] - 'a']--;
}
}
while (!temp.empty()) {
ans.push_back(temp.back());
temp.pop_back();
}
return ans;
}
Get perfect executing string โ
DE Shaw