Contest Experience : GeeksforGeeks Mega Job-A-Thon [05-07-23]

About Contest:

• This contest consists of 4 parts:
  • 3 DSA coding problems
  • 5 MCQ on programming logic
  • 5 MCQ on logical reasoning
  • 5 MCQ on quantitative aptitude
• This contest is held to shortlist programmers by specific companies based on their own criteria.
• Languages supported for the DSA problems were C++, Java, and Python.
• There was also 5% penalty for each wrong submission.
• Link of the contest !!

Coding problems:

There was total 3 coding problems, all of them based on DSA.

Overview Table:

Problem Name	Difficulty	Pre-requisite	Approx. time to solve	Number of Submissions
Magic and Toy shop	Easy	Greedy	5-8 mins	1
Count beautiful Strings	Medium-Hard	Hashing	20- 25 mins	3 (2 TLE)
Maximum good length	Medium	Binary Search ,Pre-sum	15- 20 mins	1

Question 1 : Magic and toy shop

Observations:

It was an easy problem in which we just have to think about in which index we have to perform the operation first to get most benefit , after some observation we can easily find out the greedy approach will work just fine.

Approach :

• Calculate the sum of magical_price for all elements.
• if it is greater than M than it is not possible to buy all the toys so returning -1.
• Else create a priority queue of in which we store = ‘ price[i] – magical_price[i] ‘
• It will store the maximum profitable element we can get , at the front after the operation.
• Calculate the current total price.
• while our sum of current total price is greater than M we do the following:
  • We increase the operation count.
  • We decrease the current total price by front element and pop that element from queue.
• Return the operation count.

Result : Accepted!!

Question 2 : Count beautiful Strings

Observations:

This was a very good problem , Initially my thought was we can always select two strings having same odd characters and merge them it will create a string having all characters with even frequencies , but how should I find out that one odd occurring character ? after some thought I have the idea about:

Brute force Approach [TLE] :

Well I saw the problem and thought of a fairly easy brute force approach which was:

• find every pair in the string and check weather we can make a beautiful string with the help of them or not.

Obviously it will work but it will surely give us TLE because of constraints.

Optimized Approach :

• Create a 26 size vector for each string and store the even or odd frequency of the characters , and store it in a map.
• Initialize the answer variable = 0
• Traverse through the map and for each time increase the value of answer by:
  • ans = ans + freq*(freq-1)/2 . // taking two vectors of same kind of character appears odd times and create a new string.
    • where freq = number of times that perticular vector occur .
• Till now we have calculated the strings having all characters even times.
• Now for “atleast one character is odd” :
• iterate through all the vectors in map and check for each lowercase alphabet , weather it can be that odd character or not ?
• if we have already present a vector in map where all other characters parity are same [ odd and even ] only the current characters parity is different then we can make a string by selecting these two strings also so :
  • update the answer by multiplication of those two vectors frequency .
• return the answer.

Result: TLE … again !! But why ??

Reason: we are searching the vector again and again in the map which causes this problem … now how can we improve it ??

Improved Optimized Approach:

• In this approach we are doing same thing as above just replacing the vector for each string by a number .
• Idea :
  • We will traverse in the string and if any character present odd times we will set the i-th bit of that number where i will be index of the character in the English alphabet [ like: a->0 , b->1 ,c->2 … z->25].
  • ex if string is : S = “abc” then our number will be : (111)₂ -> 7.
  • now we will store these values in map and do the operation just like above and we will save a lot of time by converting the vector into a number .
• in the end return the answer.

Result: Accepted!!

Question 3 : Maximum Good Length

Observations:

Because we have to tell the maximum size of L for which has atleast one L*L submatrix having minimum value >= L . And most of the maximum of minimum problems can be solved by Binary Search , So I applied it.

Approach :

• Apply binary search on the Length ‘L’ and for each L we calculate the following:
  • make a temporary_matrix and store its value matrix[i][j] >= L ? 1 : 0 .
  • and do the 2D pre-sum , it will help us to find out a valid submatrix in O(n*m) time.
• Store the value of L in the answer and update value of low and high accordingly.
• In the end return the answer .

Result: Accepted !!