A legend among a group of n people is a person who is top rated in all respects.The task is to identify a legend by asking a single question of the form "who deserves it?" Design an efficient algorithm to identify a legend or determine if the group has no such person.How many questions does your algorithm need in the worst case??

Given a string array， return all groups of strings that are anagrams

Convert Prefix expression to postfix expression

While travelling in a cycle race, there are several pit stops. A cyclist can travel upto 50 km without stopping at any pit stop, where he can have a health drink and refill his energy. Given a start point ’t’ and let ’k’ be the distance needed to be travelled by the cyclist, find an efficient algorithm to reach the destination ’d’ so as to have minimum pit stops and prove your algorithm is the most optimal one.

You have given n numbers from 1 to n. You have to sort numbers with increasing number of set bits.

for ex: n=5.
output: 1,2,4,3,5
Note: If you have two number with equal number of set bits, then number with lowest value come first in the output.

In a binary tree replace each node's value (except leaf node) with the sum of left and right subtree's value.

You have N computers and [Ca, Cb] means a is connected to b and this connectivity is symmetric and transitive. then write a program which checks that all computers are interconnected and talk two each other

I Believe there are multiple way to solve this problem 1. Here is my approach since graph is connected there will be path from each source to destination vertex , isn't it ? so we can simply run a dfs or bfs for all computers to check if all such path exist , finally we can return & of all operation that means all computer are inter connected and can talk to each other . PS.Since graph is undirected if there exist path from u to v then vice-verse also true so we wont calculate it again.

Write a function to convert Hexadecimal number (String) to Decimal Number (Int)

Maximum Sum Subsequence in an unsorted array of size n

We are given an array A having n integers. What we want to find is a contiguous subseqnence (A[i], A[i+1], …, A[j]) such that the sum of the elements in that subsequence is maximized. (Note that, in general, there may be more than one such optimal subsequence.)

Note that if all elements in A are non-negative, we could directly return the entire array as the required optimal subsequence.

APPROACH 1

A simple brute-force method could be to compute the sums of all possible subseqnences in A. How many subsequences are there? There are  possible subsequences. Computing the sum of a subsequence takes O(n) in the worst-case. Hence, the running time of the algorithm would be .

In C++, we could write the following function to do what was explained above:

// start and end are the starting and ending indices of an optimal subsequence.

void f ( int* A, int n, int &start, int& end)

{

int sum, max = A[0];

for (int i = 0; i < n ; i++)

for (int j = i; j < n; j++)

{

sum = 0;

for (int k = i; k <=j; k++)

sum+= A[k];

if (sum >= max)

{

start = i;

end = j;

max = sum;

}

}

}

————

APPROACH 2:

We can improve the running time to  by being a bit more clever in computing the sums of different subsequences. We observe that the sum of the subsequence A[i ... j+1] = A[j+1] + sum of the subsequence A[i ... j].

In C++, we could write as follows:

void f (int *A, int n, int &start, int &end)

{

int sum, max = A[0];

for (int i = 0; i < n; i++)

{

sum = 0;

for (int j = i; j < n; j++)

{

sum + = A[j];

if (sum >= max)

{

start = i;

end = j;

max = sum;

}

}

}

}

———–

APPROACH 3:

Using dynamic programming, we can solve the problem in linear time.

We consider a linear number of subproblems, each of which can be solved using previously solved subproblems in constant time, this giving a running time of .

Let  denote the sum of a maximum sum contiguous subsequence ending exactly at index .

Thus, we have that:

 (for all )

Also, S[0] = A[0].

——–

Using the above recurrence relation, we can compute the sum of the optimal subsequence for array A, which would just be the maximum over  for .

Since we are required to output the starting and ending indices of an optimal subsequence, we would use another array  where  would store the starting index for a maximum sum contiguous subsequence ending at index i.

In prseducode form, the algorithm would look thus:

Create arrays S and T each of size n.

S[0] = A[0];

T[0] = 0;

max = S[0];

max_start = 0, max_end = 0;

For i going from 1 to n-1:

// We know that S[i] = max { S[i-1] + A[i], A[i] .

If ( S[i-1] > 0)

S[i] = S[i-1] + A[i];

T[i] = T[i-1];

Else

S[i] = A[i];

T[i] = i;

If ( S[i] > max)

max_start = T[i];

max_end = i;

max = S[i];

EndFor.

Output max_start and max_end.

———–

The above algorithm takes  time and  space.

——–

We can however improve upon the space requirements, reducing it to . The key idea is this: for computing S[i] and T[i], all we need are the values of S[i-1] and T[i-1] apart from the given array A. Hence, there is no need to store all  values of the S and T arrays.

We could proceed as follows:

max = A[0];

max_start = 0;

max_end = 0;

S = A[0];

T = 0;

For i going from 1 to n-1:

// S = max { S + A[i], A[i] )

if ( S > 0)

S = S + A[i];

Else

S = A[i];

T = i;

If ( S > max)

max_start = T;

max_end = i;

max = S;

End For.

Output max_end and max_start.

———–

The above algorithm takes  time and  space.

How much tottal water can be stored in buildings

Given array of height 5 1 3 6 1 2 

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