Solved Issue #508: Enemy is weak (Codeforces 61E)

Author: Krishna200608

Problems/Data-structures/Day-09/sol/Krishna200608/Solution2.cpp

@@ -0,0 +1,127 @@
+/*
+Submission Link:
+https://codeforces.com/contest/61/submission/356381771
+*/
+
+/*
+Problem: Enemy is weak
+Link: https://codeforces.com/contest/61/problem/E
+Author: Krishna200608
+
+Short Problem Statement:
+    Given distinct integers, count how many triplets (i, j, k) exist
+    such that i < j < k and the values strictly decrease:
+    a[i] > a[j] > a[k].
+
+Approach:
+        We first compress the values to handle large numbers easily.
+        Treat each element as the middle of a triplet.
+        For every position, count bigger elements on the left   .
+        and count smaller elements on the right.
+        A Fenwick Tree helps do these counts efficiently .
+        The final answer is the sum of left multiplied by right for all positions.
+
+Time Complexity: O(N log N)
+Space Complexity: O(N)
+*/
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define MOD 1000000007
+const int M = 1e9 + 7;
+const double PI = acos(-1.0);
+#define INF 1e18
+#define pb push_back
+#define ppb pop_back
+#define ll long long
+#define no cout << "NO" << endl
+#define yes cout << "YES" << endl
+#define ff first
+#define ss second
+#define inn(x) int x; cin >> x
+#define ill(x) ll x; cin >> x
+#define all(x) x.begin(), x.end()
+#define in(a) for (int i = 0; i < (int)a.size(); i++) cin >> a[i]
+#define out(a) for (int i = 0; i < (int)a.size(); i++) cout << a[i] << " "
+typedef vector<int> vi;
+typedef vector<ll> vll;
+#define ceil_div(n, x) (((n) % (x) == 0) ? ((n) / (x)) : ((n) / (x) + 1))
+#define debug(x) cout << "x -> " << x << endl
+#define outt(x) cout << x << endl
+#define endl "\n"
+
+const int sz = 1000005;
+int bit[sz];
+int nn;
+
+void upd(int i, int x) {
+    for (; i <= nn; i += i & -i)
+        bit[i] += x;
+}
+
+int qry(int i) {
+    int res = 0;
+    for (; i > 0; i -= i & -i)
+        res += bit[i];
+    return res;
+}
+
+void solve() {
+    inn(n);
+    nn = n;
+
+    vi a(n);
+    in(a);
+
+    vi b = a;
+    sort(all(b));
+
+    for (int i = 0; i < n; i++) {
+        a[i] = lower_bound(all(b), a[i]) - b.begin() + 1;
+    }
+
+    vll le(n), ri(n);
+
+    for (int i = 0; i < n; i++) {
+        int smaller = qry(a[i]);
+        int total = i;
+        le[i] = total - smaller;
+        upd(a[i], 1);
+    }
+
+    memset(bit, 0, sizeof(bit));
+
+    for (int i = n - 1; i >= 0; i--) {
+        int smaller = qry(a[i] - 1);
+        ri[i] = smaller;
+        upd(a[i], 1);
+    }
+
+    ll ans = 0;
+    for (int i = 0; i < n; i++) {
+        ans += le[i] * ri[i];
+    }
+
+    outt(ans);
+}
+
+signed main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+    cout.tie(nullptr);
+
+    auto begin = chrono::high_resolution_clock::now();
+
+    int t = 1;
+    while (t--) {
+        solve();
+    }
+
+    auto end = chrono::high_resolution_clock::now();
+    auto elapsed = chrono::duration_cast<chrono::nanoseconds>(end - begin);
+    cerr << "Time measured: " << elapsed.count() * 1e-6 << "ms";
+
+    return 0;
+}