IBM Kyndryl Off Campus Drive 2022

Post Name: Associate – Technical Engineer

Degree : B.E/B.Tech/M.E/M.Tech/B.Sc/M.Sc/BCA/MCA/Diploma

Experience : freshers

CTC : ₹4.5 LPA*

Location: Across India

Apply Link :


https://krb-sjobs.brassring.com/TGnewUI/Search/home/HomeWithPreLoad?partnerid=26059&siteid=5096&PageType=JobDetails&jobid=595844#jobDetails=595844_5096

bool issafe(int i, int j, vector<vector<int>> &a)
{
int rows = a.size();
int cols = a[0].size();
if (i < 0 or j < 0 or i >= rows or j >= cols or a[i][j] == 0)
{
return false;
}
return true;
}
int distanceTraversed(vector<vector<int>> a)
{
int rows = a.size();
int cols = a[0].size();
if (rows == 0)
{
return 0;
}
int finaldist = -1;
if (a[0][0] == 9)
{
return 0;
}
unordered_map<string, bool> visited;
queue<pair<int, pair<int, int>>> q;
int dist;
q.push({0, {0, 0}});
string s = to_string(0) + "@" + to_string(0);
visited[s] = true;
while (!q.empty())
{
pair<int, pair<int, int>> node = q.front();
q.pop();
if (a[node.second.first][node.second.second] == 9)
{
return node.first;
}
int i = node.second.first;
int j = node.second.second;
dist = node.first;

if (issafe(i - 1, j, a))
{
string str = to_string(i - 1) + "@" + to_string(j);
if (visited.find(str) == visited.end())
{
visited[str] = true;
q.push({dist + 1, {i - 1, j}});
}
}

if (issafe(i, j - 1, a))
{
string str = to_string(i) + "@" + to_string(j - 1);
if (visited.find(str) == visited.end())
{
visited[str] = true;
q.push({dist + 1, {i, j - 1}});
}
}

if (issafe(i, j + 1, a))
{
string str = to_string(i) + "@" + to_string(j + 1);
if (visited.find(str) == visited.end())
{
visited[str] = true;
q.push({dist + 1, {i, j + 1}});
}
}

if (issafe(i + 1, j, a))
{
string str = to_string(i + 1) + "@" + to_string(j);
if (visited.find(str) == visited.end())
{
visited[str] = true;
q.push({dist + 1, {i + 1, j}});
}
}
}
if (finaldist == -1)
{
return -1;
}
return dist;
}


Approach :


we are using BFS here and we know that even if we are traversing
a particular level, the distance will always be incremented by 1
only once, ultimately the distance would be minimum.
and if integer representing the minimum distance traversed to remove the
obstacle else return -1.
Time Complexity:- O(mn)
Space Complexity:- O(mn) // for queue

Amazon ✅

If interested then send your resume over to shreya.a@caratlane.com

Back to Basics ❤

❗️ColorTokens Off Campus Drive❗️

🎖 Batch - 2021 & 2022
💰 CTC - 7LPA

📝 Link - match.myanatomy.in/sc/627c05dc3a1eaf3b0b5daced/n

Extract user message from transcript
IBM✅
C++ language

GFG hiring part time mentors.

JD:

https://drive.google.com/file/d/1HVSGsm_dgLtT_6fMMMLbkm8dALiSKObN/view


Application form:

https://docs.google.com/forms/d/e/1FAIpQLSfCq-vlTGUS7FwL7tshFVkGs2OyOatlNcuAwIH6_JxnpNe_7w/viewform

https://www.linkedin.com/posts/activity-6944672619572932608-Fy6H?utm_source=linkedin_share&utm_medium=android_app

Give it a try.

5 to 7 LPA

Qualification : B.E/BTech
Specialization : Computer Science/Information Technology
Graduation year of passing: 2018, 2019, 2020, 2021 & 2022
Percentage : 70% above throughout (10th, 12th and Graduation)
https://match.myanatomy.in/corporate/customCampaign/view?publicLink=6ba517be793e1558cf2a962a3791e473%3B339a8b7a86213a807db7aa00fd3eda4a66aa869ff841f7db87257d34d2eb3911473280ccf7f8bcee3db3c4f21669b2a76efc3a82450ce14b798917fa6a7c76de2bccc121dc05458551d93b52d758cc19dd2ab10603e9942d9533aea72e1fb8933d9416f2b1cab26ee322e789aaa374e4e77163d05cad23f3816a0357471508a3f23ec79d5876685784a1fd8b5052d330d21039f28b45a374ecf146da916e1aa07b65115d2c6aa9ad47cbb2e31a3b620f44bdd075336b0d2c426f15b14d8f63ae&source

1642. Furthest Building You Can Reach
You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.
You start your journey from building 0 and move to the next building by possibly using bricks or ladders.
While moving from building i to building i+1 (0-indexed),
If the current building's height is greater than or equal to the next building's height, you do not need a ladder or bricks.
If the current building's height is less than the next building's height, you can either use one ladder or (h[i+1] - h[i]) bricks.
Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.
Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7
*/
class Solution
{
public:
int furthestBuilding(vector<int> &heights, int bricks, int ladders)
{
priority_queue<int> maxB;
int i = 0, diff = 0;
for (i = 0; i < heights.size() - 1; i++)
{
diff = heights[i + 1] - heights[i];
if (diff <= 0)
{
continue;
}
bricks -= diff;
maxB.push(diff);
if (bricks < 0)
{
bricks += maxB.top();
maxB.pop();
ladders--;
}
if (ladders < 0)
break;
}
return i;
}
};

Amazon (leetcode) ✅

Cognologix is hiring
Role: Full Stack enginner
Batch eligible: 2021 and 2020 passouts.

P.S. 2022 grad can also apply as in requirements they mentioned 0-2 years of experience, but I am not very much sure.

Link: https://careers.cognologix.com/jobs/tu2qwnvs5I3B/full-stack-engineer-entry-level

Hexaview interview experience✅
1) self intro
2) asked 3 coding questions
1 given array and values x
Check weather pair sum equal to x in o(n)
2) given string
Reverse each word maintaing order in o(n)
Ex i love books
Result i evol skoob
3 ) given arrays O's 1's 2's
Sort array without sorting function in o(n)
One logical question
Given two rod
One rod takes 60 mints to burn totally
Measure time taken for 45 mints

Cisco Ideathon 2022

Eligible criteria: 2023 passouts only

Apply Link:

https://lnkd.in/dzJ_HTwf

Smart Dictionary (Nagarro)
Python 3✅