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Let's now get into the code for the second approach that we have discussed for solving the sorted squared

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array question.

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Okay.

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So we are going to start by creating a new array.

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So let me say const new array is equal to new array.

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And this new array will have the same size as the input array which is what is given us over here.

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Okay.

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So over here and say array dot length.

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And what we're going to do is we're going to fill every cell in the new array with the value zero.

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So over here I'll say dot fill zero.

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Okay.

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Now we have seen that the highest value will either be the squared of the leftmost element or the rightmost

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element in the array, which is given to us.

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So let's have two pointers.

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So I'll say let pointer left.

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Is equal to zero, which is the left most index.

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And also I'll say let pointer right.

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Is equal to array dot length minus one.

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Okay, now I'm going to iterate over every index in the new array which we have created over here.

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But we will start at the last index and go up to zero.

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So over here I'll say for let I is equal to array dot length minus one and I greater than equal to zero.

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Because we will be going till I equal to zero and I minus minus.

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Okay.

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Now we will find the squared value of the leftmost element and the rightmost element.

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So I'll say const left squared.

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Is equal to math dot pow array at pointer left okay.

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And we are squaring it.

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So comma two over here.

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And similarly const right squared.

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Is equal to math dot pow array.

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At point or write.

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And again comma two okay.

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Because we are squaring it.

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Now we are going to compare these two values.

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So I will say if left squared greater than right squared.

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So if this is the case then we will say new array at index I is equal to left squared.

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And because we have used up the left squared or the element at which the left pointer is pointing at

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the current moment, we are going to change that pointer by incrementing it.

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So I will say pointer left plus plus okay.

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Now if this is not the case.

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What we are going to do is we'll say new array at index I is equal to write squared okay.

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And then because we have used the element at which the right pointer was pointing, we are going to

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shift the right pointer by decrementing it.

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So I will say pointer right minus minus okay.

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So that's it.

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So this should solve the question.

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Now all that remains is we'll have to return the new array.

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So here we'll say return new array.

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And that's it.

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

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

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And again do make use of the user logs.

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So for every test case you can see the input, the expected output and the actual output.

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When you take a look at the user logs for each individual test case.

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Also do make use of this.

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If you want to debug your code, you can just have console log statements and you can print various

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things which will help you debug your code in case you're stuck.