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DXC Hiring Fresher For Associate Software Engineeer
Location : Bangalore
Qualification : Bachelor's degree or equivalent combination of education
Work Experience : Fresher
CTC : 4 LPA
Apply Before Expired
Apply Link : https://bit.ly/3xzGPkt
Telegram Group Link : https://t.me/placementupdatess
Deloitte Off Campus Hiring Fresher For Analyst
Location : Hyderabad
Qualification : BE/B.Tech
Work Experience : Fresher
CTC : 6 LPA
Apply Before Expired
Apply Link :https://jobs2.deloitte.com/ui/en/job/E22ALLIABA-Offcampus
Telegram Group Link : https://t.me/placementupdatess
Helping hands group for all drive exams join this Chanel
https://t.me/joinchat/mtHiBeR0vFNkMTY1
💢💢💢💢💢💢💢
ADOBE IS HIRING
Eligibility: B. Tech(CS/IT/ECE/EEE)
Batch: 2021
Job Location: Noida/ Bengaluru
Percentage Criteria: 10th & 12th 60%, B-Tech - 70%
Apply: https://cntr.click/5y6b1fM
Join @placementupdatess on telegram for more useful material and off campus updates 🔥🔥🔥
In a shop every item code
CPP program to count number of visible boxes.
#include <bits/stdc++.h>
using namespace std;
// return the minimum number of visible boxes
int minimumBox(int arr[], int n)
{
queue<int> q;
// sorting the array
sort(arr, arr + n);
q.push(arr[0]);
// traversing the array
for (int i = 1; i < n; i++) {
int now = q.front();
// checking if current element
// is greater than or equal to
// twice of front element
if (arr[i] >= 2 * now)
q.pop();
// Pushing each element of array
q.push(arr[i]);
}
return q.size();
}
// driver Program
int main()
{
int arr[] = { 4, 1, 2, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << minimumBox(arr, n) << endl;
return 0;
}
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Box inside a box code
def max_clique(g):
n = 0
for x in g: n = max(n, len(x))
l = [set() for _ in range(n + 1)]
s = [0] * len(g)
for i, x in enumerate(g):
ll = len(x)
l[ll].add(i)
s[i] = ll
m = 0
for _ in range(len(g)):
for i in range(n + 1):
if len(l[i]) > 0:
x = l[i].pop()
m = max(m, i)
s[x] = 0
for k in g[x]:
if s[k] > 0:
l[s[k]].remove(k)
s[k] -= 1
l[s[k]].add(k)
break
return m + 1
def dfs(s, b, u):
s.add(u)
b[u] = True
for v in g[u]:
if b[v]: continue
dfs(s, b, v)
for _ in range(int(input())):
n, m = map(int, input().split())
g = [list() for _ in range(n)]
for _ in range(m):
u, v = map(int, input().split())
g[u-1].append(v-1)
g[v-1].append(u-1)
print(max_clique(g))
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👉the hugging distance code
Lynda has three children. She wantsgo distribute N candies among all three in such a way that exactly one childe gets the maximum number of candies among all three and each child gets at least one candy The task is to find the total number of ways distribute N candies among the three children.
Code:-
def countWays(N):
if(N<4):
return 0
ans = ((N-1) * (N-2)) // 2
s = 0
for i in range (2, N - 2,1):
for j in range(1, i, 1):
if(N==2*i+j):
s+=1
if(N%3==0):
s=3*s+1
else:
s=3*s
return ans-s
N = int(input())
print (countWays(N))
Example 1:-
Input:- 6
Output:- 9
@placementupdatess
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int Ngarden(int n, vector<int>& ranges) {
int ret = 1;
vector<int> v(n + 1, -1);
for (int i = 0; i <= n; i++) {
int u = max(i - ranges[i], 0);
int e = min(n, i + ranges[i]);
v[u] = max(v[u], e);
}
if (v[0] == -1)
return -1;
int curr = v[0];
int i = 0;
int next = 0;
while (curr < n) {
while (i <= curr) {
next = max(next, v[i]);
i++;
}
if (next == curr)
return -1;
curr = next;
ret++;
}
return ret;
}
};
main(){
int arr;
Solution ob;
vector<int> v ;
cout << (ob.Ngarden(arr, v));
}
Gardlen question
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Today Tcs code
For hiring car code
#include <bits/stdc++.h>
using namespace std;
int main()
{
long r1,n,r2,x,h,res=0;
cin>>r1;
cin>>n;
cin>>r2;
cin>>x;
h=n;
n=x;
x=n/60;
if(n%60)x++;
if(h>=x)
{
res=x*r1;
}
else
{
res=r1*h+r2*(x-h);
}
cout<<res;
return 0;
}
Telegram:- @placementupdatess
Opening for Freshers !!
Company name- Manuh Solutions India Pvt Ltd
Job Role: Trainee Engineer.
Degree Needed: B.E/B.Tech/M.E/M.Tech -2019, 2020, 2021 pass out.
10th/12th/Degree percentage: 70% and above.
Work Location(s): Bangalore.
Kindly share Cv to careers@manuhsolutions.com
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