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Deleting a character and appending another one corresponds to adding a number and subtracting a number in our hashing algorithm. The complexity of the algorithm is O(n).
Let's go through an example. If we S and P to be strings of digits and chose a to be 10, our problem maps to finding numbers in a string of digits.
If S and P are strings of digits and a is 10, our problem maps to finding numbers in a string of digits.
Let's go through an example.
== code(c) |
Let P = 53424, S = 3249753424234837 and a = 10
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