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Cognizant Interview Experience - 2021
Completed Cognizant Genc Interview today lasted for about 45mins:
1. Asked about my projects in depth
2. Asked about ML algorithms and cloud (Since i had mentioned)
3. Gave a banking scenario and asked me to implement in c++
4. MySQL and DBMS concepts
5. Oops
6. What changes would you do implement in your past projects
7. Basic HR questions
8. Would you work for Cognizant
9. Tell me about your hobbies
10. Do you have any questions
Join @placementupdatess on telegram for more useful material and off campus updates 🔥🔥🔥
Wipro Recruitment 2021 For Associate
Location: Hyderabad
Qualification: Any Graduate
Work Experience: Freshers
Apply - https://bit.ly/3ht6uXe
Free Online Courses with Certification - https://bit.ly/3yvGdNt
#include <bits/stdc++.h>
using namespace std;
void merge(vector<vector<int>>&arr,int l,int m,int r)
{
vector<vector<int>>left,right;
for(int i=l;i<=m;i++)
{
left.push_back(arr[i]);
}
for(int i=m+1;i<=r;i++)
{
right.push_back(arr[i]);
}
int n1=left.size();
int n2=right.size();
int i=0,j=0,pointer=l;
while(i<n1 and j<n2)
{
vector<int> v1=left[i],v2=right[j];
if(v1[1]>v2[1])
{
arr[pointer]=v1;
i++;
}
else if(v2[1]>v1[1])
{
arr[pointer]=v2;
j++;
}
else
{
if(v1[2]>=v2[2])
{
arr[pointer]=v1;
i++;
}
else
{
arr[pointer]=v2;
j++;
}
}
pointer++;
}
while(i<n1)
{
vector<int> v1=left[i];
arr[pointer]=v1;
pointer++;
i++;
}
while(j<n2)
{
vector<int> v2=right[j];
arr[pointer]=v2;
pointer++;
j++;
}
}
void merge_sort(vector<vector<int>>&arr,int l,int r)
{
if(l>=r)return;
int m=l+(r-l)/2;
merge_sort(arr,l,m);
merge_sort(arr,m+1,r);
merge(arr,l,m,r);
}
int main()
{
unordered_map<int,int>freq,ind;
int n;
cin>>n;
unordered_set<int>arr;
for(int i=0;i<n;i++)
{
int ele;
cin>>ele;
freq[ele]++;
ind[ele]=i;
arr.insert(ele);
}
vector<vector<int>>nums;
for(int i:arr)
{
nums.push_back({i,freq[i],ind[i]});
}
merge_sort(nums,0,nums.size()-1);
for(auto i:nums)
{
for(int j=0;j<i[1];j++)cout<<i[0]<<' ';
}
return 0;
}
Sort array
@placementupdatess : TELEGRAM JION
https://surveys.infosysapps.com/r/a/infy_se_2021_july
*Infosys Off Campus Drive for B.Tech 2021 batch.*
*Last date to apply: 20th July 2021*
Cooling period : 6 months
⚠️ Reminder - who didn't apply yet
Kumaran - t.me/campusdrive/984
Zoho - t.me/campusdrive/987
Wipro - t.me/campusdrive/994
NPCI - t.me/campusdrive/996
ZenQ - t.me/campusdrive/1001
✅ @placementupdatess
Dear Students,
Greetings from NRI Institute of Technology, Guntur !!
Due to current pandemic situation, we couldn’t complete the recruitment process by Columbus, to those students who have registered earlier. Sorry for the delay in our recruitment process.
We would like to start recruitment process for those registered candidates who are currently unplaced/available for COLUMBUS’s recruitment process.
In this connection we are requesting the available and interested candidates to register for the COLUBUS recruitment drive using the link given below.
Registration link: https://forms.gle/sKRU7xgwEbLEH9YM8
Deadline for registration is: 07-07-2021(Today) before 9:30 AM.
Regards
TPO, NRIIT-Guntur.
telegram t.me/placementupdatess
Maxxton India Technologies hiring for below roles:
Angular 2+ exp :- 4positions
Java 2+ exp :- 8 positions
PO 4+exp :- 2 positions
QA 2+ exp :- 6 positions
React developer 2+ exp :- 1position
Support role 1+exp :- 1position
SQL developer 2+exp - 2positions
This all are referral positions
mail Resume at s.soni@maxxton.com (eg: <position name> | Exp: <your experience>)
GREEDY FLORISTS ALL CASES PASSED
#include<iostream>
#include<functional>
#include<algorithm>
using namespace std;
int c[100];
int main()
{
int n, k;
long long ans = 0;
cin >> n >> k;
for (int i = 0; i < n; i++)
cin >> c[i];
sort(c, c + n, greater<int>());
int cnt = 0;
for (int i = 0; i < n; i++)
{
ans += c[i] * (cnt / k + 1);
cnt++;
}
cout << ans << endl;
return 0;
}
.......................................................
Telegram:1:
@placementupdatess
#include<bits/stdc++.h>
using namespace std;
bool isSubsetSum(vector<int> &arr, int n, int sum){
bool subset[2][sum + 1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= sum; j++) {
if (j == 0)
subset[i % 2][j] = true;
else if (i == 0)
subset[i % 2][j] = false;
else if (arr[i - 1] <= j)
subset[i % 2][j] = subset[(i + 1) % 2]
[j - arr[i - 1]] || subset[(i + 1) % 2][j];
else
subset[i % 2][j] = subset[(i + 1) % 2][j];
}
}
return subset[n % 2][sum];
}
int main(){
int t;
cin>>t;
map<int, set<int> > m1;
for(int i = 1; i <= 10000; i++){
string temp = to_string(i);
for(auto ch: temp)
m1[ch-'0'].insert(i);
}
while(t--){
int n, k, i;
cin >> n >> k;
vector<int> arr(n), a;
for(auto it: m1[k])
a.push_back(it);
for(i = 0; i < n; i++)
cin >> arr[i];
for(i = 0; i < n; i++){
if(isSubsetSum(a, a.size(), arr[i]))
cout << "YES" << endl;
else
cout << "NO" << endl;
}
}
return 0;
}
Kittu's beautiful code in C++👆👆👆.....
.......................................................
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int S, N, M;
cin >> S >> N >> M;
if (((N * 6) < (M * 7) && S > 6) || M > N)
cout << "-1"<<endl;
else {
int days = (M * S) / N;
if (((M * S) % N) != 0)
days++;
cout << days << endl;
}
}
return 0;
}
Book code in C++ lang👆👆....
......................................................
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
string str;
cin >> str;
int n = str.length(), ans = 0;
for(int i = 0; i < n; i++){
if(str[i] == '5' || str[i] == '6')
continue;
else
ans += 1;
}
cout << ans << endl;
}
return 0;
}
Beautiful number code in C++
Telegram:1:
@placementupdatess
1) Java Servlets Interview Questions - http://bit.ly/32h0jef
2) JDBC Interview Questions - http://bit.ly/2HB7CWa
3) Spring Interview Questions - http://bit.ly/329X88i
4) Hibernate Interview Questions - http://bit.ly/32fcubB
5) Python Interview Questions - http://bit.ly/30zxPvO
6) Core Java Interview Questions - http://bit.ly/2ZpVlhw
7) SQL Interview Questions - http://bit.ly/2TQ3XvX
8) Top 30 HR Interview Questions with Answers For Freshers - http://bit.ly/2WjSnpU
For All Interview Questions - http://bit.ly/2TTIges
Amcat Sample Paper with Answer
Amcat Sample Paper - 1 - http://bit.ly/2ZyCNMo
Amcat Sample Paper - 2 - http://bit.ly/2Zu8CWg
Join Us on Telegram for Instant Job Notification - https://t.me/jobscoupe
def main(edges):
res= []
v = [-1] * len(edges)
for i in range(0, len(edges)):
New = []
K= i
while cell > -1 and v[K] == -1:
v[K] = i
New.append(K)
K = edges[K]
if K > -1 and v[K] == i:
_idx = New.index(K)
New = New[_idx:]
if len(New) > len(res):
res = New
return res,len(res)
size =int(input())
edges =list(map(int,input().split())) [ : size]
print(main(edges))
Python3 ✅🙂🤞
Maximum weight node!
Coding Help, [03.07.21 16:17]
Siemens coding answer
#include<stdio.h>
#include<string.h>
#include<map>
#include<algorithm>
#define ll long long
using namespace std;
map<ll,ll>mp[30][30];
int n,m,t;
ll k,num[30][30],ans;
// double-ended dfs, one from (1,1) to (i,j) and i+j==max(n,m)
void dfs1(int x,int y,ll tt)
{
if(x+y==t)
{
Mp[x][y][tt]++; //Use the map to record the number of tt when searching for points (y, x)
return ;
}
if(x+1<=n)
dfs1(x+1,y,tt^num[x+1][y]);
if(y+1<=m)
dfs1(x,y+1,tt^num[x][y+1]);
}
// second from (n,m) to (i,j) and (i+j)==max(n,m)+1
void dfs2(int x,int y,ll tt)
{
if(x+y==t+1)
{
if(x-1>=1)
ans+=mp[x-1][y][tt^k];
if(y-1>=1)
ans+=mp[x][y-1][tt^k];
return ;
}
if(x-1>=1)
dfs2(x-1,y,tt^num[x-1][y]);
if(y-1>=1)
dfs2(x,y-1,tt^num[x][y-1]);
}
int main()
{
while(~scanf("%d%d%lld",&n,&m,&k))
{
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
mp[i][j].clear();
scanf("%lld",&num[i][j]);
}
}
ans=0;
t=max(n,m);
if(1+1<=t)
{
dfs1(1,1,num[1][1]);
dfs2(n,m,num[n][m]);
}
else
{
if(num[1][1]==k)
ans++;
}
printf("%lld\n",ans);
}
}
XoR paths
int solve (int n, int m, vector<int> v) {
long long l = 1, r = *max_element(v.begin(), v.end()), mid, cnt;
while(l<r){
mid = (l+r)/2;
cnt = 0;
for(auto& i: v) cnt += ceil(1.0*i/mid);
if(cnt>n) l = mid+1;
else r = mid;
}
return l;
}
Seimens vaccination distribution code
Telegram::: @placementupdatess
#include<bits/stdc++.h>
using namespace std;
int minOps(string& A, string& B)
{
int m = A.length(), n = B.length();
if (n != m)
return -1;
int count[256];
memset(count, 0, sizeof(count));
for (int i=0; i<n; i++)
count[B[i]]++;
for (int i=0; i<n; i++)
count[A[i]]--;
for (int i=0; i<256; i++)
if (count[i])
return -1;
int res = 0;
for (int i=n-1, j=n-1; i>=0; )
{
while (i>=0 && A[i] != B[j])
{
i--;
res++;
}
if (i >= 0)
{
i--;
j--;
}
}
return res;
}
int main()
{
string A = "0101";
string B = "1111";
cout << "Minnos" << minOps(A, B);
return 0;
}
C++
String Transformation
Telegram :: @placementupdatess
1.1
2.a
3.a
4.c
5.a
6.d
7.a
8.d
9.a
10.a
11.c
12.a
13.b
14.d
15.d
16.b
17.d
18.d
19.c
20. D
22 - B
23 - D
24 - A
25 - C
26 - C
27 -C
28 - B
29 - B
30 -A
JUSPAY Answer
Telegram::: @placementupdatess
