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ZigZag traversal

This is one of the question from algoexpert and this is my solution involving math

def zigzagTraversal(arr):
out = []
test = 1
out.append(arr[0][0])
while test <= len(arr)+len(arr[0]):
    for j in range(len(arr[0])):
        for i in range(len(arr)):
            if test % 2 == 1:
                if i + j == test:
                    # print(arr[i][j])
                    out.append(arr[i][j])
            else:
                if i + j == test:
                    out.append(arr[j][i])
    test += 1
return out
print(zigzagTraversal([[1, 3, 4, 10],
                   [2, 5, 9, 11],
                   [6, 8, 12, 15],
                   [7, 13, 14, 16]]))
Tanishka Kohli
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  • I haven't busted out the calculus in a long time, but... You are traversing a 2D array, and you are doing row + col times. If your array is n x n in size your complexity is 2n x n^2, which if I remember my reducing rules comes out to roughly o(n^2). I am curious to see if someone does full analysis for you. – codingCat Apr 13 '21 at 14:58
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    @codingCat if the OP isn't interested in feedback about any or all facets of the code then it wouldn't really be [on-topic](https://codereview.stackexchange.com/help/on-topic) for CR. Please refer to [this meta answer](https://codereview.meta.stackexchange.com/a/8374/120114) (and its siblings) for more information. – Sᴀᴍ Onᴇᴌᴀ Apr 13 '21 at 15:52

1 Answers1

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Let N=len(arr), M=len(arr[0]). This code has O(NM^2+N^2M) complexity (or O(N^3) for square input matrix). Here's why: outer loop (while) is fully equivalent to for loop (because you increment test at the end and never change it elsewhere) and will be executed N+M times. Then you enter the second loop which is executed exactly N times. Then inner loop - M times. Inner loop will be repeated for each step of middle loop, the same with outer. So we have to multiply all this counts together. All inner operations are not atomic, of course, but for small data will be O(1) on average. If your input is large, I suggest pre-allocating out (creating it as out = [None for _ in range(N*M)]) to avoid costs of its growth and keeping current first free index as additional variable.

STerliakov
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