Accenture Work From Home
π―Position: HR Partner Analyst
Qualification: Any Graduate
Experience: 0-3 Years
Location: Remote (Bangalore Office)
Salary: 3 LPA β 5 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π―Position: HR Partner Analyst
Qualification: Any Graduate
Experience: 0-3 Years
Location: Remote (Bangalore Office)
Salary: 3 LPA β 5 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π3
π― IBM Off Campus Drive Hiring 2024 | Sofware Developer | 6-20 LPA
Job Role : Software Developer
Qualification : BE/ B.Tech/ ME/ M.Tech
Experience : 0-3 years
Package : 6-20 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
Job Role : Software Developer
Qualification : BE/ B.Tech/ ME/ M.Tech
Experience : 0-3 years
Package : 6-20 LPA
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π2β€1
Forwarded from allcoding1_official
Guys β₯οΈ
@allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's not me don't lose your money
#scammer
Please report
Please share with your friends and telegram Group's
@allcoding, @Allcodingoffical1, @offical1allcoding @Allcodingoffical, @codingoffical and @Allcodingofficalmain it's not me don't lose your money
#scammer
Please report
Please share with your friends and telegram Group's
π5
Forwarded from allcoding1_official
#scammer
Please report π
@offical1allcoding , Vds , @Allcodingoffical
The changing names
Guys β₯οΈ be careful don't pay money
Please report π
@offical1allcoding , Vds , @Allcodingoffical
The changing names
Guys β₯οΈ be careful don't pay money
π1
π―Phone pe hiring
Job role:- Advisor,ONDC
Qualification:- Any
Experience:- 0-2years
Location:- Bangalore
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
Job role:- Advisor,ONDC
Qualification:- Any
Experience:- 0-2years
Location:- Bangalore
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π3
π―Phone pe hiring
Job role:- Advisor,ONDC
Qualification:- Any
Experience:- 0-2years
Location:- Bangalore
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
Job role:- Advisor,ONDC
Qualification:- Any
Experience:- 0-2years
Location:- Bangalore
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π3
π―Google Internship 2024 | Software Student Training in Engineering Program | Apply Now
Job Role : Software Student Training in Engineering Program
Qualification : B.E/B.Tech/B.Sc
Experience : Freshers
Last Date : 19 January 2024
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
Job Role : Software Student Training in Engineering Program
Qualification : B.E/B.Tech/B.Sc
Experience : Freshers
Last Date : 19 January 2024
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π4β€1
KPMG Off Campus Drive Hiring 2024 | Front End Developer | 4 LPA+
Job Title : Front End Developer
Qualification : B.E / B.Tech
Batch : Any Batch
Package : 4 LPA+
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
Job Title : Front End Developer
Qualification : B.E / B.Tech
Batch : Any Batch
Package : 4 LPA+
Apply Now:- www.allcoding1.com
Telegram:- @allcoding1
π6
class Solution {
public:
void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)
{
vis[i][j]=1;
queue<pair<int,int>>q;
q.push({i,j});
while(!q.empty())
{
int row=q.front().first;
int col=q.front().second;
q.pop();
int delrow[4]={1,0,-1,0};
int delcol[4]={0,1,0,-1};
for(int k=0;k<=3;k++){
int nrow=row+delrow[k];
int ncol=col+delcol[k];
if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])
{
vis[nrow][ncol]=1;
q.push({nrow,ncol});
}
}
}
}
int numIslands(vector<vector<char>>& grid) {
int n=grid.size();
int m=grid[0].size();
vector<vector<int>>vis(n,vector<int>(m,0));
int cnt=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!vis[i][j] and grid[i][j]=='1')
{
cnt++;
bfs(vis,grid,i,j,n,m);
}
}
}
return cnt;
}
};
Zeta β
Telegram:- @allcoding1
public:
void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)
{
vis[i][j]=1;
queue<pair<int,int>>q;
q.push({i,j});
while(!q.empty())
{
int row=q.front().first;
int col=q.front().second;
q.pop();
int delrow[4]={1,0,-1,0};
int delcol[4]={0,1,0,-1};
for(int k=0;k<=3;k++){
int nrow=row+delrow[k];
int ncol=col+delcol[k];
if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])
{
vis[nrow][ncol]=1;
q.push({nrow,ncol});
}
}
}
}
int numIslands(vector<vector<char>>& grid) {
int n=grid.size();
int m=grid[0].size();
vector<vector<int>>vis(n,vector<int>(m,0));
int cnt=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!vis[i][j] and grid[i][j]=='1')
{
cnt++;
bfs(vis,grid,i,j,n,m);
}
}
}
return cnt;
}
};
Zeta β
Telegram:- @allcoding1
π5
class Solution {
public:
void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)
{
vis[i][j]=1;
queue<pair<int,int>>q;
q.push({i,j});
while(!q.empty())
{
int row=q.front().first;
int col=q.front().second;
q.pop();
int delrow[4]={1,0,-1,0};
int delcol[4]={0,1,0,-1};
for(int k=0;k<=3;k++){
int nrow=row+delrow[k];
int ncol=col+delcol[k];
if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])
{
vis[nrow][ncol]=1;
q.push({nrow,ncol});
}
}
}
}
int numIslands(vector<vector<char>>& grid) {
int n=grid.size();
int m=grid[0].size();
vector<vector<int>>vis(n,vector<int>(m,0));
int cnt=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!vis[i][j] and grid[i][j]=='1')
{
cnt++;
bfs(vis,grid,i,j,n,m);
}
}
}
return cnt;
}
};
Zeta β
Telegram:- @allcoding1
public:
void bfs(vector<vector<int>>&vis,vector<vector<char>>&grid,int i,int j,int n,int m)
{
vis[i][j]=1;
queue<pair<int,int>>q;
q.push({i,j});
while(!q.empty())
{
int row=q.front().first;
int col=q.front().second;
q.pop();
int delrow[4]={1,0,-1,0};
int delcol[4]={0,1,0,-1};
for(int k=0;k<=3;k++){
int nrow=row+delrow[k];
int ncol=col+delcol[k];
if(nrow>=0 and nrow<n and ncol>=0 and ncol<m and grid[nrow][ncol]=='1' and !vis[nrow][ncol])
{
vis[nrow][ncol]=1;
q.push({nrow,ncol});
}
}
}
}
int numIslands(vector<vector<char>>& grid) {
int n=grid.size();
int m=grid[0].size();
vector<vector<int>>vis(n,vector<int>(m,0));
int cnt=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(!vis[i][j] and grid[i][j]=='1')
{
cnt++;
bfs(vis,grid,i,j,n,m);
}
}
}
return cnt;
}
};
Zeta β
Telegram:- @allcoding1
#include <bits/stdc++.h>
using namespace std;
vector<int> solution(vector<int> a, int n, int k) {
vector<int> v;
deque<int> dq;
for (int i = 0; i < n; i++) {
while (!dq.empty() && dq.front() <= i - k)
dq.pop_front();
while (!dq.empty() && a[dq.back()] <= a[i])
dq.pop_back();
dq.push_back(i);
if (i >= k - 1)
v.push_back(a[dq.front()]);
}
return v;
}
int main() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++)
cin >> a[i];
vector<int> result = solution(a, n, k);
for (int i = 0; i < result.size(); i++)
cout << result[i] << " ";
return 0;
}.
//cricket match β
Zeta
Telegram:- @allcoding1
using namespace std;
vector<int> solution(vector<int> a, int n, int k) {
vector<int> v;
deque<int> dq;
for (int i = 0; i < n; i++) {
while (!dq.empty() && dq.front() <= i - k)
dq.pop_front();
while (!dq.empty() && a[dq.back()] <= a[i])
dq.pop_back();
dq.push_back(i);
if (i >= k - 1)
v.push_back(a[dq.front()]);
}
return v;
}
int main() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++)
cin >> a[i];
vector<int> result = solution(a, n, k);
for (int i = 0; i < result.size(); i++)
cout << result[i] << " ";
return 0;
}.
//cricket match β
Zeta
Telegram:- @allcoding1
π2
string make_string_S_to_T(string S) {
string T=βprogrammingβ;
bool possible = false;
int M = T.length();
int N = S.length();
for (int i = 0; i <= M; i++) {
int prefix_length = i;
int suffix_length = M - i;
string prefix = S.substr(0, prefix_length);
string suffix = S.substr(N - suffix_length, suffix_length);
if (prefix + suffix == T) {
possible = true;
break;
}
}
if (possible)
return "YES";
else
return "NO";
}
Deleting substring β
Zeta
Telegram:- @allcoding1
string T=βprogrammingβ;
bool possible = false;
int M = T.length();
int N = S.length();
for (int i = 0; i <= M; i++) {
int prefix_length = i;
int suffix_length = M - i;
string prefix = S.substr(0, prefix_length);
string suffix = S.substr(N - suffix_length, suffix_length);
if (prefix + suffix == T) {
possible = true;
break;
}
}
if (possible)
return "YES";
else
return "NO";
}
Deleting substring β
Zeta
Telegram:- @allcoding1
π1