Flipping the Matrix : Solution to Hackerrank Challenge

[@monkeydog8681]

I got this as a mock exam for day 2. 💀💀 Was given 24 mins to solve. Like wth.

[@AkeemJonesajonesrock]

awesome video and great use of excel to show a graphical representation.

[@untitled6483]

How did you reach this solution? I cannot even think about how to brute force it?

[@seanclarke-1]

My only problem with this is the BIG assumption being made here. You're assuming that in all cases, you can get the maximum number to it's desired cell position. The game involves flipping entire columns and rows. This means that you could be displacing an already maximised cell to achieve the maximisation of another cell. I know that it must be possible, since the problem is too large otherwise, but is there a proof you could add to show that it is always possible to achieve the maximum colour value in the upper left quadrant by only flipping rows and columns?

[@shabrishpraveen1]

Wow. I am happy someone asked me this This is exactly the question I had in my mind when I started. I wrote a brute force program and verified. Imagine a rubix. We will always be be able to bring the desired max value to specific cell.

[@paladattadri7465]

@@shabrishpraveen1 really same doubt which i got like @Sean Clarke, i didn't understand why this much complicated task for basic mock test, i thought to leave the test. Thank for your explanation.

[@kumaranil181]

kzbin.info/www/bejne/Z5uqcqiqo62oj68 please check this for the extension where the example is also with 6*6 matrix and helps in imagining

[@ktcoder]

@@shabrishpraveen1 Agree, in your example the flipping sequence will be C1 + R0 + C1

[@karthiknaik7056]

Good one and thanks Praveen! Liked the detailed explanations with examples. Hoping to see more such videos.

[@shabrishpraveen1]

Thank you, Sure karthik :)

[@missbluesky2512]

Awesome video, thank you so much!

[@helmetman216]

Great video Praveen, it is helping me a lot to understand. Keep up the good work

[@trainguyen]

sir, your explaination so fantastc. thanksss

[@himanshubakshi4176]

But 1 issue in this approach will be. When we are flipping the matrix for 2nd value, in that case the 1st value depending upon the row flipped will also change. Thus making the sum incorrect.

[@yowpop3677]

thank you sir! i was able to understand the problem and how to solved it!

[@arturtelezhkin632]

Nice explanation. Thank you

[@shabrishpraveen1]

Glad it was helpful!

[@LearnerH]

thank you sir maza aa gya.

[@prabinbasnet4741]

Thank you!

[@shabrishpraveen1]

You're welcome!

[@TheSilentLooters]

def flippingMatrix(matrix): n = len(matrix) // 2 total_sum = 0 for i in range(n): for j in range(n): total_sum += max(matrix[i][j], matrix[i][2 * n - 1 - j], matrix[2 * n - 1 - i][j], matrix[2 * n - 1 - i][2 * n - 1 - j]) return total_sum written in python3

[@tadweber8140]

I need help with this, Im pretty sure my thought process was exactly yours.. ```int currentSum = matrix[i][j] + matrix[i][n-j-1] + matrix[m-i-1][j] + matrix[m-i-1][n-j-1];``` looping through every opposing value in the matrix but i keep coming up with 401 instead of 414...

[@md.shafaatjamilrokon8587]

wow. genius