You are given an array a of size n and a number c we define...

You are given an array A of size N and a number C. We define the beauty number of a number X as: -> We have a universal array U where all the integers from -10^18 to 10^18 are present in sorted order (U=[-10^18,1-10^18,....0,1,2...10^18-1,10^18]). -> Now the beauty number of X is the number of subarrays of U with the sum X. Find the number of elements in array A whose beauty number is greater than C. In other words, find the number of i (0 <= i< N) such that the (beauty number of A[i]) > C. Constraints - 1 <= N <= 10^5 1 <= C <= 10^12 1 <= A[i] <= 10^5 Sample test case: Input: 3 Provide me solution with explanation and code in java.

Transcript

00:01 In this question, we have to find the array e that stores the closest elements from array b to each element in array a.

00:12 So for that, let us look at the logic or the steps that will be required.

00:17 So firstly, what we'll do is initialize an empty array c.

00:36 So after doing this for each element in array ai, we have to do certain steps.

00:42 So in each element in a, that is array a, we have to do these things.

01:01 So i'll write them with abc.

01:03 We have to initialize, first of all, a variable that is minpiff to positive infinity.

01:28 Then next thing we'll do is we have to initialize a variable closest element to none.

01:49 We have to initialize variable that is closest element to none.

02:09 Then what we have to do is we have to iterate through each element bj in array b.

02:18 We have to iterate through each element in array by.

02:41 So we have to calculate the absolute difference between ai and bj.

02:47 So that we have to do.

02:49 And then we have to give the absolute difference is less than minimum difference.

02:54 So we have to update the minimum difference to absolute difference and the closest element to bj.

03:01 And then we append and then we return.

03:08 Return array c when all the operations are carried on it.

03:15 So this algorithm iterates through each element in array a and finds the element in array b that has the minimum absolute difference with it.

03:25 So the closest element is then stored in the corresponding position in array c.

03:30 But now as we can see that time complexity that is there, time complexity of this algorithm is given by 0 n cap 2.

03:53 So where n is the size of the input arrays.

04:10 So this is because for each element in array a, we iterate through all the elements in array b to find the closest one.

04:20 So in the worst case, this results in n iterations for n elements leading to a quadratic time complexity.

04:30 So however, there is room for optimization by using sorting and binary search...

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