Count Smaller elements

Given an array Arr of size N containing positive integers. Count number of smaller elements on right side of each array.

• Time : O(nlogn)
• Space : O(n)

c++

struct Seg {
    int mi, ma, count;
    Seg *l, *r;
    Seg(vector<int>& A, int le, int ri) : mi(A[le]), ma(A[ri]), count(0), l(nullptr), r(nullptr){
        if(le == ri) return;
        int m = le + (ri - le) / 2;
        l = new Seg(A, le, m);
        r = new Seg(A, m + 1, ri);
    }
    int query(int s, int e) {
        if(mi > e or ma < s) return 0;
        if(s <= mi and ma <= e) return count;
        return (l ? l->query(s, e) : 0) + (r ? r->query(s, e) : 0);
    }
    int insert(int n) {
        if(mi <= n and n <= ma) {
            count++;
            if(l) l->insert(n);
            if(r) r->insert(n);
        }
    }
};

class Solution{
    Seg* seg;
    int mi;
    void build(vector<int> A) {
        sort(begin(A), end(A));
        A.erase(unique(begin(A), end(A)), end(A));
        mi = A[0];
        seg = new Seg(A, 0, A.size() - 1);
    }
public:
    vector<int> constructLowerArray(int *arr, int n) {
        vector<int> res(n), A;
        for(int i = 0; i < n; i++) A.push_back(arr[i]);

        build(A);

        for(int i = n - 1; i >= 0; i--) {
            res[i] = seg->query(mi, A[i] - 1);
            seg->insert(A[i]);
        }
        return res;
    }
};