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Most optimized Solution Required
By go_rob, history, 7 years ago, In English

Given an array of n elements. ‘k’ operations need to be performed on the array.

For each operation start_index, end_index,trim_value is given. One has to trim the value starting from the start_index to end_index inclusive.

If the value at that index is more than or equal to trim value then make it equal to trim_value else leave it as it is. for each A[i] for i between start and end , A[i] = min(A[i],h) .

After performing, ‘k’ such operations find the maximum value of the array.

I have a solution of O(n + k*log(n)). I think it can be optimized to O( n*log(n) ) at least, if not better.

Note: Please try not to use Segment tree or BIT.

Constraints -> n<=10^6 (sure) , A[i] <= 10^6 (Not quite sure of this though)

For Eg. Given array = 7, 2, 8, 5, 1, 6, 4 and k = 3

Following are the k operation (start, end, value) -

1 3 4 => Now the array = 4, 2, 4, 5, 1, 6, 4

3 5 5 => Now the array = 4, 2, 4, 5, 1, 6, 4

4 7 4 => Now the array = 4, 2, 4, 4, 1, 4, 4

So the Maximum value is 4 in the array.

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I can think of solution using only sets. The constraint on size of each element is not necessary.

You process elements of A one by one, and maintain the operations applied to the current element in a set S. For every element of A, you do the following operations to construct the updated array B, one by one:

  1. For each operation (i, e, m), add (m, e) to S.
  2. Assign B[i] = min(A[i], max(S).first) (if S is empty, just put B[i] = A[i])
  3. For each operation (b, i, m), remove (m, i) from the set.

Now you have the array B that you can use to answer any queries.

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