32-culture/experience
33-twitter,LinkedIn
34-a,bd,e
35-one whose decision making
36-caterogization
37-b,e
38-attending events
39-escalations
40-a,c
41-false
42-we need to maintain
43-a,c
44-self conscientious
46- b,c
47-it help us to look
48-never end up making
50- text based networking
51-flipping a coin
52-true
53-so we can complete more
54-identifying a problem
55-b,d,e
56-b,e
57-bas that is implicit

Biz Skills Track 1 Slot 2 (3 Pm) Answers (TCS) ✅

1)false
2)a, c
Confident, accessibility
3)b
Talk about the weather
4)a
Organizational band
5)c, e, f
Reguraly c
You can get e
You can postion f
6)f
How
7)a b d
Mentor rele a
Mentor helps b
Mentor mene d
8)b
Know there sytle
9)c
Compassion
10)a b c
A every activitity a
Takes breaks b
Indify c
11)the amount of overlap
12)q)offering help raise profile
True
A
13)a d
Remote a
Quick nd eassy d
14)d
Uncertainty
15)b
Define goal
17)b
Equal opportunites

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Help us grow 🔥

I talked with Hexaview recruiter. He will hire some good ones from here who are graduating in 2022 Only, at 9PM will post one Google form link over LinkedIn (He need only resume)

✅Only first 100 response will entertain.

✅All the details will share on that post.

So, if interested please be online at 9PM Sharp😊

Hexaview is hiring for below roles (There is a bond of 2 years)

✅Only first 100 resumes will entertain.

✅Please see below the details :

On-Job Training
Duration - 6 months
Stipend - 20,000 INR for Application Engineer and Member of Technical Staff
Stipend - 35,000 INR for Software Development Engineer

Post-Training

✅CTC for 2022 passouts:
Software Development Engineer : 9 LPA
Member of technical staff : 7 LPA
Application Engineer : 5 LPA



The Job profile preference is in the hands of management based upon your performance in interview.

Job Description-
Employment Type: Full-time

Work week: 5 days

Location Noida, Pune

Service Level Agreement: Period- 24 months (includes 6 months of training)

Service Agreement Penalty- 3 LPA

✅Apply Link: https://forms.gle/qdHMXLX7PvFDewbY8

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If we get 5.5k subscribers today I'll share all answers with you .
No need to pay anyone ✌️

/*
167. Two Sum II - Input Array Is Sorted
Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.
Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.
The tests are generated such that there is exactly one solution. You may not use the same element twice.
Your solution must use only constant extra space.
Example 1:
Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore, index1 = 1, index2 = 2. We return [1, 2].
*/
class Solution
{
public:
vector<int> twoSum(vector<int> &nums, int target)
{
int i = 0, j = nums.size() - 1;
while (i < j)
{
int sum = nums[i] + nums[j];
if (sum == target)
return {i + 1, j + 1};
else if (sum > target)
j--;
else
i++;
}
return {}; // Target not found
}
};
// Time Complexity : O(N)
// Space Complexity : O(1)



Guys this is very very important questions for Amazon interview 🔥

Publicis Sapient
Role: Data Analytics Intern Batch Eligible: Only 2022 Graduation


https://docs.google.com/forms/d/e/1FAIpQLSeCSZluPV_q2C9lpJ42mBOVVsm57T4VVr_JqlAHs9XJKA3LDA/viewform

RoutePairs
Python 3language✅(Amazon)

Findresturnets. / Diveryplan
Python 3✅(Amazon)

Approach: we need to calculate the forward route list + return route list 2nd point and we will find the maximum of the list after that I will compare the forward route list and +return list if they are equal then I will push in 1 D vector then I will push 2nd vector in answer and finally we will return 2D vector answer


Complexity Analysis:
time complexity O(N*M)
N is the size of the forward route list and return route list
space complexity O(N*M)
and here N and M is the size of our 2D vector answer

'''
Find K Closest Points to the Origin

Approach: The idea is to calculate the Euclidean distance from the origin for every given point and sort the array according to the Euclidean distance found. Print the first k closest points from the list.

Algorithm :
Consider two points with coordinates as (x1, y1) and (x2, y2) respectively. The Euclidean distance between these two points will be:

√{(x2-x1)2 + (y2-y1)2}
Sort the points by distance using the Euclidean distance formula.
Select first K points form the list
Print the points obtained in any order.


Complexity Analysis:

Time Complexity: O(n log n).
Time complexity to find the distance from the origin for every point is O(n) and to sort the array is O(n log n)
Space Complexity: O(n).
As we are making an array to store distance from the origin for each point.

'''
def deliveryPlan(allLocation, NumDeliveries):

allLocation.sort(key = lambda NumDeliveries: NumDeliveries[0]2 + NumDeliveries[1]2)

return allLocation[:NumDeliveries]

Amazon ✅

Electrostatics
Virtusa

Palindrome count

S = input().split()
C = 0
for i in S:
if 'a' in S:
C += 1
print(c)