How to find number of elements greater than that elements for all elements of the array on the right side.
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How to find number of elements greater than that elements for all elements of the array on the right side.
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You can use policy based data structure ( Ordered set ) . Then you can traverse the array from the right and keep checking no. of elements using order_of_key(a[i]) . o(nlogn) tc
I guess ordered set not work in case of duplicate element.
It works
4 types of modification you can do Greater Greater Equal Less Less Equal
Also ig it can be done using Divide and conquer ( Merge sort )
Here similar problem on leetcode :
https://leetcode.com/problems/count-of-smaller-numbers-after-self/description/
You can also use Segment Tree, if array value is large in that case you have to normalize it down to <=n range.
Can you explain more as I am not much familiar with segment tree.
you can sort the array in decreasing order keeping their index then you build segment / fenwick tree ans for index i is i — sum(0..j) j denotes index of element in original array then update the value of segment / fenwick tree. sum(0..i) denotes cnt of elements having value less <= element at index i and index < j
Caculate i < j and a_i < a_j. Enumerate from n to 1. For position i, you only need to caculate how many greater than a_i. Use BIT, query(n) — query(a[i]) is answer for position i.
Also, you should normalize a[i] to the range [1,n]
Can you please explain this in a bit more detail and with clarity? Also what is the time complexity of this approach?
i guess find the next greater element for every element to its right and add 1 to the answer* of that element to get the answer for this element.
I guess you are trying to find count of inversions in array which is
if i < j and a[i] > a[j]
This can be done using merge sort along with a simple modification.
Link to understand sol
Statement's really unclear, can you clarify what the problem is?