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Hi everyone, we have successfully coded the second method.

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Now let's quickly take a sample input deaf, deaf and walk through what's happening over here.

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Initially we create an empty array over here.

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New string to compare.

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So this is an empty array in this case.

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And we traverse the given string this one from uh right to left in this direction right.

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First we take the rightmost character and we insert it into the empty array.

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And we keep doing that again and again.

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Right.

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We are moving the pointer to the left.

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We take one character at a time and insert it into the empty array.

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So that's what we are doing over here with Dot.

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Push right with this the.

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And remember this is a constant time operation.

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And at the end we are joining this and making it into a string.

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Right.

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That's happening over here.

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Right?

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So once we have done that, we are just checking whether this is the same as the given string.

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Now in this case it's not the same.

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Therefore we will return false and we have our solution.

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Now we we've seen that this is this has a time complexity of O of n.

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Now there are two things.

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First of all we have this for loop over here which is an O of N time operation.

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Right.

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Why is that so?

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Because we are traversing the given string once.

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Now remember we are pushing each character to the empty array and then to first it's empty and later

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on it's not empty.

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We are keeping on pushing characters to it.

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But the interesting thing over here is pushing an element into an array is a constant time operation,

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so it's not costly.

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So the net time complexity of this solution is O of N.

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And we have seen that the space complexity of this solution is also O of n because we are creating this

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empty array, and then we are changing it into a string.