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Let's now get into the code for the iterative approach that we have discussed in the previous video.

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Okay.

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So initially we will say let's survivor is equal to zero.

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And finally we will do something over here to find the exact index of the survivor.

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And then we will go ahead and return survivor plus one okay.

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So that's the skeleton of the solution.

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Now over here what we are going to do is we are going to iterate from two up to n, where n is the size

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of the problem which is given to us.

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Okay.

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And again we have discussed why we are going to start at two because if n is equal to one, then we

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just need to return zero plus one which is equal to one.

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So we don't even need to get into this for loop.

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So that's why over here we are going to start to iterate from two.

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So I can say for let I is equal to two and we are going to go up to n.

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So I less than equal to n and I plus plus okay.

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Now for each value of I we are going to find the survivor using the previous value of the survivor.

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So I can say survivor is equal to survivor plus k.

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And then we will have to do modulo the size of the problem at that particular instance, which is given

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by the value of I.

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So over here I will do modulo I.

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Okay.

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So again that's it.

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So by the time we are out of this for loop we will have found the survivor.

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And then we will just have to add one to it.

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Because the survivor that we are getting is going to be zero indexed okay.

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So that's it.

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Now this is the solution.

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Now let's go ahead and run this code and see whether it's passing all the test cases.

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And you can see that it's passing all the test cases.