Score After Flipping Matrix - Leetcode 861

Score After Flipping Matrix - Leetcode 861 - Python

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Күн бұрын

Пікірлер: 27

@ay5960 6 ай бұрын

There is a nuance, for the leftmost column we could either flip rows to make it all 1s, or we could flip rows to make it all 0s and then flip the first column to make it all 1s. Now the question is whether the two lead to different answers? The answer is no but this kind of elucidates that why the fact that greedy approach is working is not trivial. How do you know you get the global optimum solution rather than a local optimum one? The answer is that there is only one type of nuance and that is exactly what I mentioned above. Another question, what if we remove the fact that rows represent binary numbers and just simply want to maximize the number of 1s? I have no idea how to do this one.

@Yaqk 6 ай бұрын

I hate it when I get so close to the answer but miss some small step🤦‍♂ Thanks for making these videos again, it really helps to hear someone speak and go over the problem

@ronbuchanan5432 6 ай бұрын

I gotta say your ability to explain your thought process here was really good !

@skshaheen6309 6 ай бұрын

I clicked the logic at 5:56 then I paused the video and wrote code to implement the logic. And it was shocking that I perfectly wrote the correct code with 100% beats and readability. And it is recommended to all watchers that don't watch any solution video complete. You should stop watching when you click the logic and try to implement code yourself.

@MP-ny3ep 6 ай бұрын

Thank you so much again. Great explanation as always.

@existanz 6 ай бұрын

We can use xor for line 9-10 like: cnt += 1^grid[r][0]^grid[r][c]

@vinayaksuthar2472 6 ай бұрын

"Thank you, it was helpful."

@EranM 6 ай бұрын

yo neet, use a 2^n lookup table and reverse it. That will make the index consistent instead of COL -1 - m or whatever

@connies_slice_of_life 6 ай бұрын

thanks for the awesome explanation!

@mralix9566 6 ай бұрын

The NEET Streak Saver 🔥

@krateskim4169 6 ай бұрын

Thank you for the solution

@rajsh3285 6 ай бұрын

why are we not considering the case where we flip the most significant bit column wise then the we don't have to check the most significant value while calculating the number of Zeros/ones for other columns

@jhumpadas147 6 ай бұрын

Great explanation🙏

@ahmmedabdullahsaquif559 6 ай бұрын

Thank you sir

@priyanshagarwal8490 6 ай бұрын

2055. Plates Between Candles... Next Please..

@SanjayB-vy4gx 6 ай бұрын

Thanks man❤

@MegaZz-j9c 6 ай бұрын

why flip rows first not column first?

@chien-yuyeh9386 6 ай бұрын

Nice🎉🎉

@swanv951 6 ай бұрын

🙏🙏 for method 2

@aashishbathe 6 ай бұрын

I didn't get the intuition at first, but as soon as you said first column check for 0, to flip row, I understood the problem. Anyways, here's my code - class Solution: def matrixScore(self, grid: List[List[int]]) -> int: ROWS, COLS = len(grid), len(grid[0]) res = (2 ** (COLS - 1)) * ROWS for r in range(ROWS): if grid[r][0] == 0: for c in range(COLS): if grid[r][c] == 0: grid[r][c] = 1 else: grid[r][c] = 0 for c in range(1, COLS): count1 = 0 for r in range(ROWS): if grid[r][c] == 1: count1 += 1 if count1 > (ROWS // 2): res += count1 * (2 ** (COLS - c - 1)) else: res += (ROWS - count1) * (2 ** (COLS - c - 1)) return res

@aashishbathe 6 ай бұрын

This is basically the same way I thought of it, but then damn your use of ternary conditions makes some lines of code so much shorter!

@sathishn6708 6 ай бұрын

you can use grid[r][c] ^ 1 one line to flip

@aashishbathe 6 ай бұрын

@@sathishn6708 yeah that also works!