BS-27. Median in a Row Wise Sorted Matrix 

tereko uske job chodne ya na chodne se kya farak pdta hai

@@22gauravkumar70 Hes appreciating his hardwork, nothing else. Managing both and providing this quality education is just extra ordinary dedication and spirit

15 element hi hai dhyan se gin

agle saal aeega wo wala playlist

@@cdnggk6455 really or kidding?

why?@@cdnggk6455

Strings me kya pdna h? vo toh question practice krte-krte hi aayega.

@@zanies6288Improving problem solving of Strings in a brute, better, optimal is much necessary for beginners of DSA.. I would like to remind that it's A TO Z BEGINNER COURSE...🙂

yes

Even I have the same doubt. Lemme know if you get that.

++ Is there some kind of proof for the same? Please let me know once you get it.

@@aletivardhan6679 @palakkalsi8595 @Ajay-cv1zs , striver has replied on this issue on one of the comment, let me paste here, "this is why we keep moving low and high, and we don't stop. Its bound to stop at a number which is answer. Because if the numbers are [1,5, 6, 7, 11] It can never stop at 9 because 8,9,10 all of them will have same number of elements on left."

See if some element in not present and it has

if (smallEqual

🙂

aranthamaintharaja

have you got your issue resolved yet or not

because I'm thinking the same thing for half hour

I also thought the same.

See the idea lies here 1 1 1 3 3 3 9 9 9 Assume that these are elements when we make the matrix into 1D array. Number - Frequencey 1 frequency - 3 2 frequency - 3 3 frequency - 6 4 frequency - 6 5 frequency - 6 6 frequency - 6 7 frequency - 6 8 frequency - 6 9 frequency - 9 low = 1 and high=9 => mid = 5 => freq(5) =6 => move high = mid -1 = 4 low = 1 and high =4 => mid = 4 => freq(4) =6 => move high = mid -1 = 3 low = 1 and high =3 => mid = 2 => freq(2) =3 => move low= mid +1 = 3 low = 3 and high = 3 => mid = 3 => freq(3) =6 => move high = mid -1 = 2 low = 3 (holds on and loop break) You got the pattern. Low will point to the element which is always present in the list. It will always point to the first element which has same frequency of the all the elements present in that range. In our case 3, 4,5,6,7 all have same frequency 6 but ends up the first (3). 3 is present in the list. Hence low cannot point to a number which is not present in the list.

yes correct 👍

this is why we keep moving low and high, and we don't stop. Its bound to stop at a number which is answer. Because if the numbers are [1,5, 6, 7, 11] It can never stop at 9 because 8,9,10 all of them will have same number of elements on left.

@@takeUforward What if array is [[1,2,3],[1,2,3],[1,2,3]] . In that case how will the req = (n*m)/2 will hold ?

The TC originally is n*m*log(2nm) because , you will use O(n*m) to traverse all elements in the 2d array, and then you will sort the n*m elements using inbuilt function, so TC will be O(n*m + n*m*log(2nm))

Also the final answer/code for this problem isn't available in the description. Can you please look into it?

same 😂

No those videos are (Trees, DP and Graph) pretty new. And bhaiyaa will not be replacing those videos.

@@bishalkundu7592 is hsi playlist completely updated and complete? Are all the videos arranged in serial?

@@ultimatescience3842 what do u mean by hsi playlist?

@@bishalkundu7592 sorry bro, I meant "His". Matlab poora complete hai dsa? aur agar playlist ko follow karein A to Z toh kya saare videos serial mein hain? Plus mai ekdom new hoon programming mein, abhi third year start hua. Mai c++ soch raha seekhne ko toh wo kaha se seekhun? c++ with dsa karne ki soch raha hoon. Please bro thoda detail mein bata dena:)

bhai striver ne poori complete karwa diya hai kya dsa playlist? aur kya videos sequence mein hain? tumko kitna fayda hua? kitna padhe?

​@@ultimatescience3842pura complete nhi hua hai........sequence mat Haan jaisa a-z DSA playlist me hai waise padhna start Karo bahut faida hoga......Maine personally padha hai islia bata sakta hu

Bhai humari bhi yahi doubt h ! Ki ye khdse aana toh impossible lagtah

If you are final year student, don't waste on this course. This is for beginners and people who have a lot of time. Just read the articles on takeuforward solve some problems on leetcoee u will learn more in less time.

it still holds because, 1 1 1, 2 2 2, 3 3 3 the median is 2. Irrespective of the input left and right of the median has (n*m)/2 these many elements.

Instead of k, take a variable named as count & initialise with ( n * m + 1 ) / 2, which is the middle element of the sequence when you arrange matrix in linear ascending order. Let’s say m = 3 and n = 5, then count = 8. It means 8th element is the middle element, i.e the median of the matrix. Now we are sure that there must be 7 elements on left side of the 8th Element & 7 elements on right side of 8th element. Now when black box returns value which is less than count we sure that current mid can’t be the answer, but when black box returns value which is greater than or equal to count, it may or may not be the answer, so we store it in ans variable hoping it as answer. changes I suggest for better understanding cnt = (n + m)/2; while ( L < R) { mid = (L + R)/2; val = blackbox(mat, mid); if ( val < cnt){ L = Mid + 1; } else{ R = mid - 1; ans = mid; // might be answer }

Yes, in that case last index of second row should be the largest value Eg [[1,2,3],[4,4,4]]

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