๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <iostream>
#include <vector>
using namespace std;
int countWays(int n, int k) {
vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));
vector<vector<int>> cumSum(n + 1, vector<int>(k + 1, 0));
for (int i = 1; i <= k; ++i) {
dp[1][i] = 1;
cumSum[1][i] = cumSum[1][i - 1] + dp[1][i];
}
for (int len = 2; len <= n; ++len) {
for (int j = 1; j <= k; ++j) {
dp[len][j] = cumSum[len - 1][j / 2];
cumSum[len][j] = cumSum[len][j - 1] + dp[len][j];
}
}
return cumSum[n][k];
}
Commvault โ
#include <vector>
using namespace std;
int countWays(int n, int k) {
vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));
vector<vector<int>> cumSum(n + 1, vector<int>(k + 1, 0));
for (int i = 1; i <= k; ++i) {
dp[1][i] = 1;
cumSum[1][i] = cumSum[1][i - 1] + dp[1][i];
}
for (int len = 2; len <= n; ++len) {
for (int j = 1; j <= k; ++j) {
dp[len][j] = cumSum[len - 1][j / 2];
cumSum[len][j] = cumSum[len][j - 1] + dp[len][j];
}
}
return cumSum[n][k];
}
Commvault โ
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Good for 1st and 2nd yr students
If you want to start Open Source Contribution, Here are 5 beginner friendly repos where you can start by solving issues:
- CodeTriage: A tool to help developers find open issues in GitHub repositories that need attention.
https://lnkd.in/gdxsNnds
- Tirith: A policy framework that scans declarative Infrastructure as Code (IaC) configurations.
https://lnkd.in/g9RW6pE5
- RealWorld: It is a full-stack example app that demonstrates various frameworks and libraries by building the same application, a medium clone.
https://lnkd.in/gxz6sXfS
- PublicLab: It is a community where people can come together to explore environmental concerns using DIY techniques.
https://lnkd.in/gNzvXN5N
- Oppia: It is an online learning platform that enables users to create and share interactive educational activities.
https://lnkd.in/gDgZds_n
If you want to start Open Source Contribution, Here are 5 beginner friendly repos where you can start by solving issues:
- CodeTriage: A tool to help developers find open issues in GitHub repositories that need attention.
https://lnkd.in/gdxsNnds
- Tirith: A policy framework that scans declarative Infrastructure as Code (IaC) configurations.
https://lnkd.in/g9RW6pE5
- RealWorld: It is a full-stack example app that demonstrates various frameworks and libraries by building the same application, a medium clone.
https://lnkd.in/gxz6sXfS
- PublicLab: It is a community where people can come together to explore environmental concerns using DIY techniques.
https://lnkd.in/gNzvXN5N
- Oppia: It is an online learning platform that enables users to create and share interactive educational activities.
https://lnkd.in/gDgZds_n
lnkd.in
LinkedIn
This link will take you to a page thatโs not on LinkedIn
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
TalentCorp Hiring Graduate Trainee โค๏ธโค๏ธโค๏ธ
Qualifications: BAMS, Any Graduate, B.Arch , B.Tech/B.E. , BCA , B.B.A/ B.M.S , ITI Certification , B.Sc in , B.Com , B.El.Ed , B.Ed , B.Des. , Diploma , B.A.
Experience: 0 โ 3 years
Job Type: Full Time
Location: Pune
Skills/Requirements:
1. Ability to quickly learn and adjust to new tasks and environments.
2. Strong verbal and written communication to effectively interact with colleagues and supervisors.
3. Ability to assess information, solve problems, and make data-driven decisions.
4. Working collaboratively with others, contributing to team goals, and supporting peers.
Date of Interview: 10th August, 2024
Time : 8.30 AM โ 12.30 PM
Venue: TalentCorp Solutions Pvt Ltd Ground Floor, Soham Complex, Opposite Gajanan Hospital, Ranjangaon (Pune).
Qualifications: BAMS, Any Graduate, B.Arch , B.Tech/B.E. , BCA , B.B.A/ B.M.S , ITI Certification , B.Sc in , B.Com , B.El.Ed , B.Ed , B.Des. , Diploma , B.A.
Experience: 0 โ 3 years
Job Type: Full Time
Location: Pune
Skills/Requirements:
1. Ability to quickly learn and adjust to new tasks and environments.
2. Strong verbal and written communication to effectively interact with colleagues and supervisors.
3. Ability to assess information, solve problems, and make data-driven decisions.
4. Working collaboratively with others, contributing to team goals, and supporting peers.
Date of Interview: 10th August, 2024
Time : 8.30 AM โ 12.30 PM
Venue: TalentCorp Solutions Pvt Ltd Ground Floor, Soham Complex, Opposite Gajanan Hospital, Ranjangaon (Pune).
Forwarded from OffCampus Jobs | OnCampus Jobs | Daily Jobs Updates | Lastest Jobs | All Jobs | CSE Jobs | Fresher Jobs โฅ (Dushyant)
Electronics Art is hiring ML Engineer
For 2021, 2022, 2023 grads
Location: Hyderabad
https://ea.gr8people.com/jobs/184256/ml-engineer?sid=4&s=08
For 2021, 2022, 2023 grads
Location: Hyderabad
https://ea.gr8people.com/jobs/184256/ml-engineer?sid=4&s=08
Electronic Arts
Find Jobs - Careers
Find Jobs Careers
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Photo
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000000;
const int SEG_TREE_SIZE = 2097152;
vector<pair<int, int>> upd[MAXN + 10];
long long rao[SEG_TREE_SIZE];
long long sum[SEG_TREE_SIZE];
void update(int t, int l, int r, int x, int y) {
if (x < l || r < x) return;
rao[t] += y;
sum[t] += 1LL * x * y;
if (l == r) return;
update(2 * t, l, (l + r) / 2, x, y);
update(2 * t + 1, (l + r) / 2 + 1, r, x, y);
}
void get(int t, int l, int r, long long k, long long &B) {
if (rao[t] <= k) {
B += sum[t];
return;
}
if (k <= 0) return;
if (l == r) {
B += 1LL * k * l;
return;
}
get(2 * t, l, (l + r) / 2, k, B);
get(2 * t + 1, (l + r) / 2 + 1, r, k - rao[2 * t], B);
}
long getMinCost(int n, int k, vector<vector<int>> plans) {
memset(rao, 0, sizeof(rao));
memset(sum, 0, sizeof(sum));
long long res = 0;
for (const auto& plan : plans) {
int l = plan[0];
int r = plan[1];
int c = plan[2];
int p = plan[3];
upd[l].push_back({c, p});
upd[r + 1].push_back({-c, p});
}
for (int i = 1; i <= n; ++i) {
for (const auto& operation : upd[i]) {
update(1, 1, MAXN, operation.second, operation.first);
}
long long B = 0;
get(1, 1, MAXN, min(1LL * k, rao[1]), B);
res += B;
}
return res;
}
CLOUD DATA TRANSFERโ
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll solve(vector<ll>&a,vector<ll>&b,ll n,ll m)
{
ll ans=0;
ll s1=0,s2=0;
ll i=0,j=0;
while(i<n and j<m)
{
if(a[i]<b[j]) s1+=a[i++];
else if(a[i]>b[j]) s2+=b[j++];
else
{
ans+=max(s1,s2)+a[i];
i++;
j++;
s1=0;
s2=0;
}
}
while(i<n) s1+=a[i++];
while(j<m) s2+=b[j++];
return max(s1,s2)+ans;
}
signed main()
{
ll n; cin>>n;
vector<ll>a(n);
for(ll i=0;i<n;i++) cin>>a[i];
ll m; cin>>m;
vector<ll>b(m);
for(ll i=0;i<n;i++) cin>>b[i];
cout<<solve(a,b,n,m);
}
Cave coin collection โ
Intuit
#define ll long long
using namespace std;
ll solve(vector<ll>&a,vector<ll>&b,ll n,ll m)
{
ll ans=0;
ll s1=0,s2=0;
ll i=0,j=0;
while(i<n and j<m)
{
if(a[i]<b[j]) s1+=a[i++];
else if(a[i]>b[j]) s2+=b[j++];
else
{
ans+=max(s1,s2)+a[i];
i++;
j++;
s1=0;
s2=0;
}
}
while(i<n) s1+=a[i++];
while(j<m) s2+=b[j++];
return max(s1,s2)+ans;
}
signed main()
{
ll n; cin>>n;
vector<ll>a(n);
for(ll i=0;i<n;i++) cin>>a[i];
ll m; cin>>m;
vector<ll>b(m);
for(ll i=0;i<n;i++) cin>>b[i];
cout<<solve(a,b,n,m);
}
Cave coin collection โ
Intuit
#include <bits/stdc++.h>The Chemist
#include <bits/stdc++.h>
using namespace std;
int main() {
int k;
cin >> k;
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++) cin >> arr[i];
sort(arr.begin(), arr.end());
int min_diff = 1e9;
int best_sum = -1;
for (int i = 0; i < n; i++) {
int first = arr[i];
int lb = lower_bound(arr.begin() + i + 1, arr.end(), first + k) - arr.begin();
for (int j = lb - 1; j <= lb; j++) {
if (j > i && j < n) {
int second = arr[j];
int diff = abs(k - (second - first));
int sum = first + second;
if (diff < min_diff || (diff == min_diff && sum > best_sum)) {
min_diff = diff;
best_sum = sum;
}
}
}
}
cout << best_sum << endl;
return 0;
}
Uber โ
๐ฅ1
#include <bits/stdc++.h>>
using namespace std;
const int MOD = 10000;
int countArrays(string &S, int C, int K) {
int N = S.length();
vector<int> dp(N + 1, 0);
dp[0] = 1;
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= K && j <= i; j++) {
string num_str = S.substr(i - j, j);
long long num = stoll(num_str);
if (num <= C) {
dp[i] = (dp[i] + dp[i - j]) % MOD;
}
}
}
return dp[N];
}
int main() {
int N, C, K;
cin >> N >> C >> K;
string S;
cin >> S;
cout << countArrays(S, C, K) << endl;
return 0;
}
Without whitespaces
Uber โ
๐ฅ1๐1
#include <bits/stdc++.h>
using namespace std;
bool chkValid(const vector<string>& eq, unordered_map<char, int>& l2d) {
vector<long long> nums;
for (const string& p : eq) {
long long num = 0;
for (char c : p) {
num = num * 10 + l2d[c];
}
nums.push_back(num);
}
return nums[0] + nums[1] == nums[2];
}
int dfs(vector<string>& eq, unordered_map<char, int>& l2d, unordered_set<int>& tD, vector<char>& let, int pos) {
if (pos == let.size()) {
return chkValid(eq, l2d) ? 1 : 0;
}
char l = let[pos];
int count = 0;
for (int d = 0; d < 10; ++d) {
if (tD.count(d)) continue;
if (d == 0 && find_if(eq.begin(), eq.end(), [&](string& p) { return p[0] == l; }) != eq.end()) continue;
tD.insert(d);
l2d[l] = d;
count += dfs(eq, l2d, tD, let, pos + 1);
tD.erase(d);
l2d.erase(l);
}
return count;
}
int cntCryptarithmSolutions(vector<string> eq) {
unordered_set<char> allLet;
for (const string& p : eq) {
for (char c : p) {
allLet.insert(c);
}
}
vector<char> let(allLet.begin(), allLet.end());
unordered_map<char, int> l2d;
unordered_set<int> tD;
return dfs(eq, l2d, tD, let, 0);
}
Crypt
Trilogy โ
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
#include <bits/stdc++.h> using namespace std; bool chkValid(const vector<string>& eq, unordered_map<char, int>& l2d) { vector<long long> nums; for (const string& p : eq) { long long num = 0; for (char c : p) { num = numโฆ
โ
bool placeBlock(vector<vector<char>>& field, int row, int& finalRow, int& finalCol) {
int col = 0;
while (col < field[0].size() && field[row][col] == '.') ++col;
while (row < field.size() - 1 && field[row + 1][col] == '.') ++row;
if (col >= 0 && row >= 0) {
field[row][col] = '#';
finalRow = row;
finalCol = col;
return true;
}
return false;
}
void removeBlock(vector<vector<char>>& field, int row, int col) {
field[row][col] = '.';
}
void solve(vector<vector<char>>& field, vector<int>& ans, int cnt) {
bool check = true;
for (int i = 0; i < field.size(); ++i) {
if (field[i][0] != '#') {
check = false;
break;
}
}
if (check) {
ans[0] = min(ans[0], cnt);
ans[1] = max(ans[1], cnt);
return;
}
for (int i = 0; i < field.size(); ++i) {
int finalRow = -1, finalCol = -1;
if (placeBlock(field, i, finalRow, finalCol)) {
solve(field, ans, cnt + 1);
removeBlock(field, finalRow, finalCol);
}
}
}
vector<int> solution(vector<vector<char>> field) {
vector<int> ans = {INT_MAX, INT_MIN};
solve(field, ans, 0);
return ans;
}
Block placing
Trilogy โ
int col = 0;
while (col < field[0].size() && field[row][col] == '.') ++col;
while (row < field.size() - 1 && field[row + 1][col] == '.') ++row;
if (col >= 0 && row >= 0) {
field[row][col] = '#';
finalRow = row;
finalCol = col;
return true;
}
return false;
}
void removeBlock(vector<vector<char>>& field, int row, int col) {
field[row][col] = '.';
}
void solve(vector<vector<char>>& field, vector<int>& ans, int cnt) {
bool check = true;
for (int i = 0; i < field.size(); ++i) {
if (field[i][0] != '#') {
check = false;
break;
}
}
if (check) {
ans[0] = min(ans[0], cnt);
ans[1] = max(ans[1], cnt);
return;
}
for (int i = 0; i < field.size(); ++i) {
int finalRow = -1, finalCol = -1;
if (placeBlock(field, i, finalRow, finalCol)) {
solve(field, ans, cnt + 1);
removeBlock(field, finalRow, finalCol);
}
}
}
vector<int> solution(vector<vector<char>> field) {
vector<int> ans = {INT_MAX, INT_MIN};
solve(field, ans, 0);
return ans;
}
Block placing
Trilogy โ
๐2
๐๐ฆ ๐๐น๐ด๐ผ ๐ป ๐ ใ๐๐ผ๐บ๐ฝ๐ฒ๐๐ถ๐๐ถ๐๐ฒ ๐ฃ๐ฟ๐ผ๐ด๐ฟ๐ฎ๐บ๐บ๐ถ๐ป๐ดใ
Maze Craze โ
#include <iostream>
#include <vector>
using namespace std;
const int MOD = 1e9 + 7;
int solve(int N, int M, vector<vector<int>>& nums) {
vector<vector<long long>> dp(N, vector<long long>(M, 0));
dp[0][0] = nums[0][0];
for (int j = 1; j < M; ++j) {
dp[0][j] = (dp[0][j - 1] * nums[0][j]) % MOD;
}
for (int i = 1; i < N; ++i) {
dp[i][0] = (dp[i - 1][0] * nums[i][0]) % MOD;
}
for (int i = 1; i < N; ++i) {
for (int j = 1; j < M; ++j) {
dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) * nums[i][j] % MOD;
}
}
return dp[N - 1][M - 1];
}
int main() {
int N, M;
cin >> N >> M;
vector<vector<int>> nums(N, vector<int>(M));
for (int i = 0; i < N; ++i) {
for (int j = 0; j < M; ++j) {
cin >> nums[i][j];
}
}
cout << solve(N, M, nums) << endl;
return 0;
}
Maze Crazeโ
๐1
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Take Home Assignment based Hiring
Sample Invoices for reference:
https://drive.google.com/drive/folders/1Q7sU_eiisOvtCAFCRSesyMDkVsMtNF7h?usp=drive_link
Sample Invoices for reference:
https://drive.google.com/drive/folders/1Q7sU_eiisOvtCAFCRSesyMDkVsMtNF7h?usp=drive_link