Media.net Interview Experience 2021

06 Aug, 2021
Online Assessment:
Given a tree rooted at 1 with n nodes, q queries are given. In each query, d, e are given as input. You need to find the maximum value of e^x where x is one of the ancestors of d or d itself in the tree.n<=10^5
q<=3*10^5
e<=3*10^5
1<=d<=nA bus stops at n bus stops, each bus stop having a[i] people. The bus needs to take in all the people on the bus. People from 1 bus stop get down before the next bus stop arrives. They use a resizing tech which allows the bus to be resized to whatever capacity they want. This action can be done only b times at max. The uselessness of the bus is the summation of a total number of unoccupied seats across n stops. Find the minimum uselessness the bus can achieve if the resizing tech is used optimally. 1<=a[i]<=10^6, 1<=b<=n<=400Ex 1:
a = [10 20] b = 0
Ans:10Explanation – the resizing tech cannot be applied. hence the capacity of the bus is 20 initially. in the first stop, 20-10 seats were unused. in the second stop 20 – 20 seats are unused. Total unused seats = 10Ex 2:
a = [10 20 30] b = 1
Ans: 10Explanation – the resizing tech can be applied only once. The capacity of the bus is 10 initially. in the first stop 10-10 seats unused = 0. in the second stop, the tech is used to resize to 30. 30 – 20 seats unused.In the third stop,  30-30 seats unusedTotal unused seats = 10.You will be given n points in a 2D plane which represents the corona orange zone. On the i-th day, Corona will spread to all the locations which are within i euclidean distance from each Corona orange zone. A zone will become red, if it coincides with atleast ‘x’ orange zones. Given the n  pairs and x, find the day in which the first red zone occurs.1<=n<=100, 1<=b<=n, 
for each point, 
1<=x<=10^9, 1<=y<=10^9
Example-
(9,4),(10,3) , x=2.
Ans : 1In point (9,3) both the zones would have been affected after day 1. Hence it will become a red zone after day 1.
Given a tree rooted at 1 with n nodes, q queries are given. In each query, d, e are given as input. You need to find the maximum value of e^x where x is one of the ancestors of d or d itself in the tree.n<=10^5
q<=3*10^5
e<=3*10^5
1<=d<=n
n<=10^5
q<=3*10^5
e<=3*10^5
1<=d<=n
A bus stops at n bus stops, each bus stop having a[i] people. The bus needs to take in all the people on the bus. People from 1 bus stop get down before the next bus stop arrives. They use a resizing tech which allows the bus to be resized to whatever capacity they want. This action can be done only b times at max. The uselessness of the bus is the summation of a total number of unoccupied seats across n stops. Find the minimum uselessness the bus can achieve if the resizing tech is used optimally. 1<=a[i]<=10^6, 1<=b<=n<=400Ex 1:
a = [10 20] b = 0
Ans:10Explanation – the resizing tech cannot be applied. hence the capacity of the bus is 20 initially. in the first stop, 20-10 seats were unused. in the second stop 20 – 20 seats are unused. Total unused seats = 10Ex 2:
a = [10 20 30] b = 1
Ans: 10Explanation – the resizing tech can be applied only once. The capacity of the bus is 10 initially. in the first stop 10-10 seats unused = 0. in the second stop, the tech is used to resize to 30. 30 – 20 seats unused.In the third stop,  30-30 seats unusedTotal unused seats = 10.
Ex 1:
a = [10 20] b = 0
Ans:10
Explanation – the resizing tech cannot be applied. hence the capacity of the bus is 20 initially. in the first stop, 20-10 seats were unused. in the second stop 20 – 20 seats are unused. Total unused seats = 10
Ex 2:
a = [10 20 30] b = 1
Ans: 10
Explanation – the resizing tech can be applied only once. The capacity of the bus is 10 initially. in the first stop 10-10 seats unused = 0. in the second stop, the tech is used to resize to 30. 30 – 20 seats unused.
In the third stop,  30-30 seats unused
Total unused seats = 10.
You will be given n points in a 2D plane which represents the corona orange zone. On the i-th day, Corona will spread to all the locations which are within i euclidean distance from each Corona orange zone. A zone will become red, if it coincides with atleast ‘x’ orange zones. Given the n  pairs and x, find the day in which the first red zone occurs.1<=n<=100, 1<=b<=n, 
for each point, 
1<=x<=10^9, 1<=y<=10^9
Example-
(9,4),(10,3) , x=2.
Ans : 1In point (9,3) both the zones would have been affected after day 1. Hence it will become a red zone after day 1.
1<=n<=100, 1<=b<=n, 
for each point, 
1<=x<=10^9, 1<=y<=10^9
Example-
(9,4),(10,3) , x=2.
Ans : 1
In point (9,3) both the zones would have been affected after day 1. Hence it will become a red zone after day 1.
Interview Round 1: I was asked to introduce myself. After a short introduction, I was directly given a problem. The problem is as follows:
You are given 4 strings w,x,y,z. You can permute each of the strings however you want. You have to fix 1 permutation for each of the 4 strings such that when you add all the 4 strings into a trie, the number of nodes created in the trie is minimized.Example-
w = abaa
x = aaaa
y = acca
z = abca
For permutaion:
w = abaa
x = aaaa
y = acca
z = abcaNumber of nodes in Trie – 1 (number of new nodes for first character of all strings) + 3(for second character ) + 4(for third character ) + 4(for fourth character ) = 12minimum number of trie nodes when:w = aaab
x = aaaa
y = aacc
z = aacd
Number of nodes : 
1 + 1 + 2 + 4 = 8I gave a O(2^ (number of words (4 in this case)) * number of characters(26 in this case)) using bitmasks. The interviewer was convinced.
You are given 4 strings w,x,y,z. You can permute each of the strings however you want. You have to fix 1 permutation for each of the 4 strings such that when you add all the 4 strings into a trie, the number of nodes created in the trie is minimized.Example-
w = abaa
x = aaaa
y = acca
z = abca
For permutaion:
w = abaa
x = aaaa
y = acca
z = abcaNumber of nodes in Trie – 1 (number of new nodes for first character of all strings) + 3(for second character ) + 4(for third character ) + 4(for fourth character ) = 12minimum number of trie nodes when:w = aaab
x = aaaa
y = aacc
z = aacd
Number of nodes : 
1 + 1 + 2 + 4 = 8I gave a O(2^ (number of words (4 in this case)) * number of characters(26 in this case)) using bitmasks. The interviewer was convinced.
Example-
w = abaa
x = aaaa
y = acca
z = abca
For permutaion:
w = abaa
x = aaaa
y = acca
z = abca
Number of nodes in Trie – 1 (number of new nodes for first character of all strings) + 3(for second character ) + 4(for third character ) + 4(for fourth character ) = 12
minimum number of trie nodes when:
w = aaab
x = aaaa
y = aacc
z = aacd
Number of nodes : 
1 + 1 + 2 + 4 = 8
I gave a O(2^ (number of words (4 in this case)) * number of characters(26 in this case)) using bitmasks. The interviewer was convinced.
Interview Round 2: There was a small introduction. Then the interviewer directly jumped into a problem.
You are given a 2D array of integers with dimension n X m and a value ‘k’. Find if there exists a square submatrix whose sum is equal to k.Example-
n = 3, m = 3, k = 10
1 2 3
2 3 4
3 2 6
Output: trueExplanation: The square starting from (1,0) to (2,1) (Zero based indexing) has a sum 2 + 3 + 2 + 3 = 10 which is equal to k.I first gave O(m*n*min(m,n)) solution using DP and optimised it O(m*n*log(min(n,m))) using binary search. But the expected solution was O(m*n) using 2 pointers.
You are given a 2D array of integers with dimension n X m and a value ‘k’. Find if there exists a square submatrix whose sum is equal to k.Example-
n = 3, m = 3, k = 10
1 2 3
2 3 4
3 2 6
Output: trueExplanation: The square starting from (1,0) to (2,1) (Zero based indexing) has a sum 2 + 3 + 2 + 3 = 10 which is equal to k.I first gave O(m*n*min(m,n)) solution using DP and optimised it O(m*n*log(min(n,m))) using binary search. But the expected solution was O(m*n) using 2 pointers.
Example-
n = 3, m = 3, k = 10
1 2 3
2 3 4
3 2 6
Output: true
Explanation: The square starting from (1,0) to (2,1) (Zero based indexing) has a sum 2 + 3 + 2 + 3 = 10 which is equal to k.
I first gave O(m*n*min(m,n)) solution using DP and optimised it O(m*n*log(min(n,m))) using binary search. But the expected solution was O(m*n) using 2 pointers.
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