Merge pull request #519 from MANISH-BEESA/day9-q2

added day 9 q2 solution

Author: Dheeraj-06

Problems/Data-structures/Day-09/sol/BEESA-MANISH/Solution2.cpp

@@ -0,0 +1,72 @@
+/*
+
+Submission link: https://codeforces.com/contest/61/submission/356395307
+
+TC- o(n log n)
+SC - o(n)
+
+Approach:
+Take each element as one by one and treat that element as middle element,count number of elements greater than that element on left side of that element and count
+number of elements smaller than that element on right side of that element. 
+
+Use a Fenwick Tree (Binary Indexed Tree) to efficiently count elements in ranges, Since array values can be as large as 10^9, first compress the values to ranks 1..n
+
+Do the left to right pass in Fenwick tree to count number of elements greater than current element on left side.
+
+Reset Fenwick tree and do right to left pass to count number of elements smaller than current element on right side.
+
+Finally, for each element multiply the two counts and sum them up to get the final answer.
+
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long ll;
+const int N = 1000005;
+int n;
+int a[N], t[N];
+int l_big[N], r_small[N];
+
+void add(int i, int v) {
+    for (; i <= n; i += i & -i) t[i] += v;
+}
+
+int sum(int i) {
+    int s = 0;
+    for (; i > 0; i -= i & -i) s += t[i];
+    return s;
+}
+
+int main() {
+    ios::sync_with_stdio(0); cin.tie(0);
+    cin >> n;
+    vector<int> v;
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+        v.push_back(a[i]);
+    }
+    sort(v.begin(), v.end());
+
+    for (int i = 0; i < n; i++) {
+        int rk = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
+        l_big[i] = i - sum(rk);
+        add(rk, 1);
+    }
+
+    memset(t, 0, sizeof(t));
+
+    for (int i = n - 1; i >= 0; i--) {
+        int rk = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
+        r_small[i] = sum(rk - 1);
+        add(rk, 1);
+    }
+
+    ll total = 0;
+    for (int i = 0; i < n; i++) {
+        total += (ll)l_big[i] * r_small[i];
+    }
+    cout << total << endl;
+
+    return 0;
+}