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Hi everyone, we have successfully coded the third method.

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Now let's take a sample input and quickly walk through what's happening and see what's happening in

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the code.

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So let's say the input string is d e, f, e, d and the indices of the string.

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Let's quickly write that it's 01234.

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Now what we are doing is initially I is zero and j is equal to string dot length minus one.

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So the length of this string is five.

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So five minus one is equal to four right.

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So I is pointing to index zero and j is pointing to index four.

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Now while I is less than equal to j we keep repeating what's there inside the while loop right.

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So we check whether these two characters are the same.

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And over here we have D.

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And over here also we have D because they are the same we move forward.

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If not, we would have returned false.

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Now because they are the same.

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What we are doing is we are incrementing I and Decrementing J.

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So I becomes one and j becomes four.

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And again we check whether the characters over here are the same again because they are the same.

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We increment I and decrement j.

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Now both are pointing to index two.

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Therefore definitely they will be the same.

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So we won't come into this part.

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We'll come again to else.

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Now at this point J will become one and I will become three.

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Right now at this point we break because I is no longer less than or equal to J.

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Or you can see that I is greater than J.

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So we break and we come over here and we return true.

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Right.

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So that's how we solve this question.

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And we have seen that the time complexity over here is O of n and the space complexity is O of one.

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The time complexity is O of n because we are just traversing the given string once, right?

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In fact, we could say that it's o of n by two, but then this is just a constant.

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So we remove it.

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So the time complexity is O of n and we are not using any additional space.

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So the space complexity of this solution is O of one.