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Join date: 15 de mai. de 2022

###### Sobre helix academy 1 the new kid helix academy new kid picture helix academy picture helix academy animated gif helix academy A: I am assuming that you are trying to solve it using dynamic programming, where you are trying to find best results for all of the dimensions that we can get. In this case, you can find these results for every cell by iterating over all the edges that can be taken from this cell. However, if you have a linear array where every cell would be connected to only a single cell, then you can just look at the neighboring cells and the best option for all of them is just to look at the cell with the maximum score. In this case, you would just iterate over all the rows and find the maximum score in each row. If this maximum score is more than the score in the same row in the previous step, then update it with the score in the current row. I assume that if this can be solved in n^2, you can find it in O(n^3). The following code snippet is in Python 3, but the logic is the same in any programming language: def get_row(a, n): if n == 0: return a else: return min(get_row(a[1:], n - 1), get_row(a[:-1], n - 1), a[1:n-1], a[-1:n-1], key=lambda x: x[n-1]) def get_max_cell(a): a.sort() return get_row(a, a[-1]) For example, get_max_cell([[1,2,3],[4,5,6],[7,8,9]]) == 3 Q: How to write the following fragment of code more effectively My objective is to write a fragment of code that will replace a single letter from a string using a for loop. The replacement will be performed on the same string. Also, the same replacement will be made for the single letter in the string. I can do it, but I would like to know if there

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