Відео

LOL, I don't need to see any solution for these kind of questions. 😂

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Not possible bro. I have submitted the solution in Video. Also, you can check the solution link in the description

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Yes,Trie is an optimal solution. I will upload the Trie solution very soon. I have not included the Trie solution in this video due to busy schedule. Hope you Understand. Thank you

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isn't the time complexity O(N * maxLengthofWord * 26) ?

Right, Thanks for pointing out. You can take maxLengthofWord or precisely it is AverageLengthOfWord

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No worries bro. It happens. Hope you liked it. Thank you

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Thanks bhai . Please share the channel among your friends

Thank you

Sorry for that. Let me explain. We need to convert the string to 1. Binary representation of 1 is 001,right ? We need to worry about the bits till msb only. So, msb is already a setbit (i.e 1) After doing all possible operations just before msb(from right to left) there can be two possible cases:- 1) Either the carry is 1 -> it means that when this 1 will get add to msb, it will generate an 1 in left and an extra 0 to it right and which we need to remove, so for remove that 0 leads to 1 operation Ex-> 0010 -> suppose carry is 1 on the msb part ,so after addition it will becomes to 0100 -> so, we get an extra 0 which we need to remove by dividing by 2,so thats leads to an extra operation 2) Or the carry is 0 -> It means that if we add 1 with 0 ,it will lead to 1,so ultimately we need only 1 ,so we don't need to any operation so, for implementation i have just added carry to the final ans because finally it depends on the carry value that we will need an extra operation or not. Hope you understand it. Thank you

Some top companies might ask you these types of questions in Interview.

Sure Bro.

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Thank you

Glad you liked it

Sorry about that 😂. My roommate was doing it.

Glad it was helpful!

I think you have not watched the full video. I suggest you to watch again. Don't go directly to the implementation part. Thank you

@@CodeWithARB No, this is not true, you have not explained the proof of your implementation, there are two ways to reach the target, and both yield the same result. However, in interviews, it's important to consider both approaches and provide proof for our chosen method.

Here is a proposed proof, though I think its fairly clear from the explanation that the solution is optimal. Its a proof by induction, with induction on columns. The base case of 1st column is not discussed here, its optimality is discussed in the video. Let r_{i,j} denote the score by considering i^th row and only considering its first 'j' columns. Let mat[i][j] denote the value in binary matrix at location (i, j). If score obtained till 'k'^th column is optimal, then for the (k+1)st column we have score of a row as: r_{i,k} * 2 + mat[i][k+1]. And total score as: r_{0,k} * 2 + mat[0][k+1] + r_{1,k} *2 + mat[1][k+1] + ...+ r_{m -1, k} * 2 + mat[m-1][k+1] = 2 * ( r_{0, k} + r_{1, k} + ... + r_{m-1,k} ) + ( mat[0][k+1] + mat[1][k+1] + ..+ mat[m-1][k+1] ) The sum in the first bracket is optimal by induction. The sum from the second bracket would be optimal if number of '1''s in (k +1)st columns is maximized. Since the algorithm precisely does it the score obtained by considering only first k + 1 columns using this algorithm is optimal. Hence the score by considering all n+1 columns using this algorithm is optimal.

Are you referring to Time or Space Complexity ?

@@CodeWithARB time

@@mendor0 And how will you do that ? Are you referring to the Radix sort approach ?

@@CodeWithARB start with nth_element to find top k, then counting sort for remaining elements valued less then k (elements over k should be executed in _ANY_ order first)

Yeah right. So, Xor function is equivalent to addition/subtraction under modulo 2. For subtraction, if you try to subtract any two number and see its binary representation that will be equals to Xor operation between them Suppose we subtract 4 from 6. 6+(-4) = 10 6 -> 110 4 -> 100 6^4 = 010 = 2 here modulo 2 is to show that when there are even number of occurences then the result will be always 0 (false) and for odd, it will be 1(true) which is the equivalent to XOR properties.

Sorry for that. The problem was that the original video which i made was not working properly due to some reasons so i made a just go through explaination.

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