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Career Opportunities for IT & CS Graduates at Sun Life across all India offices
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Clumio
Role: MTS Automation / MTS UI Engineer

Batch eligible: Only 2022 passouts ✅

Link:

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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 :


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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

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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)
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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;
}
};

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