Update day09sol02.cpp

Author: suzzzal

Problems/Data-structures/Day-09/sol/day09sol02.cpp

@@ -1,75 +1 @@
-//submission link- https://codeforces.com/contest/61/submission/356380170
 
-
-/*
-1. First, we change the big numbers into simple ranks (1, 2, 3...) so they are easy to count.
-2. We treat each number as the middle person and look at its neighbors.
-3. We count how many numbers to the left are bigger than our middle number.
-4. We count how many numbers to the right are smaller than our middle number.
-5. We multiply the bigger-left count by the smaller-right count to get the total triplets.
-6. We add up the triplets for every middle number to get the final answer.
-
-Time Complexity: O(N \log N)
-Space Complexity: O(N)
-*/
-
-
-#include <bits/stdc++.h>
-using namespace std;
-
-int bit[1000005];
-int a[1000005];
-long long left_greater[1000005];
-int n;
-
-void update(int idx, int val) {
-    while (idx <= n) {
-        bit[idx] += val;
-        idx += idx & -idx;
-    }
-}
-
-int query(int idx) {
-    int sum = 0;
-    while (idx > 0) {
-        sum += bit[idx];
-        idx -= idx & -idx;
-    }
-    return sum;
-}
-
-int main() {
-    cin >> n;
-    
-    vector<int> temp;
-    for (int i = 0; i < n; i++) {
-        cin >> a[i];
-        temp.push_back(a[i]);
-    }
-
-    sort(temp.begin(), temp.end());
-
-    for (int i = 0; i < n; i++) {
-        a[i] = lower_bound(temp.begin(), temp.end(), a[i]) - temp.begin() + 1;
-    }
-
-    for (int i = 0; i < n; i++) {
-        left_greater[i] = i - query(a[i]);
-        update(a[i], 1);
-    }
-
-    for (int i = 0; i <= n; i++) {
-        bit[i] = 0;
-    }
-
-    long long ans = 0;
-    for (int i = n - 1; i >= 0; i--) {
-        long long right_smaller = query(a[i] - 1);
-        ans += left_greater[i] * right_smaller;
-        update(a[i], 1);
-    }
-
-    cout << ans << endl;
-
-    return 0;
-}