day 9 q2

Author: Aiyaan-Mahajan

Problems/Data-structures/Day-09/sol/Aiyaan_Mahajan/q2sol.cpp

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+/*
+DAY 8
+Q2 Enemy Is Weak
+
+*/
+/*
+
+
+For each position j (as middle element), count how many elements on its left are greater than a[j] and how many on its right are smaller than a[j]. Multiply these two counts - this gives triplets with j as middle. Sum this for all positions.
+Use Fenwick Tree (BIT) to efficiently count elements in ranges. Since values go up to 10^9, compress them to 1..n first. Two passes: left-to-right for greater left counts, right-to-left for smaller right counts.
+
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+int ft[1000005];
+int n;
+
+void upd(int idx, int val){
+    for(; idx<=n; idx+=idx&-idx)
+        ft[idx]+=val;
+}
+
+int qry(int idx){
+    int ans=0;
+    for(; idx>0; idx-=idx&-idx)
+        ans+=ft[idx];
+    return ans;
+}
+
+int main(){
+    cin>>n;
+    vector<int> a(n);
+    for(int i=0;i<n;i++) cin>>a[i];
+    
+    // compress coordinates 
+    vector<int> sorted_vals=a;
+    sort(sorted_vals.begin(),sorted_vals.end());
+    map<int,int> mp;
+    for(int i=0;i<n;i++){
+        mp[sorted_vals[i]]=i+1;
+    }
+    
+    vector<int> compressed(n);
+    for(int i=0;i<n;i++){
+        compressed[i]=mp[a[i]];
+    }
+    
+    vector<long long> left_cnt(n), right_cnt(n);
+    
+    // left pass
+    memset(ft,0,sizeof(ft));
+    for(int i=0;i<n;i++){
+        int pos=compressed[i];
+        left_cnt[i]=qry(n)-qry(pos);
+        upd(pos,1);
+    }
+    
+    // right pass
+    memset(ft,0,sizeof(ft));
+    for(int i=n-1;i>=0;i--){
+        int pos=compressed[i];
+        right_cnt[i]=qry(pos-1);
+        upd(pos,1);
+    }
+    
+    long long ans=0;
+    for(int i=0;i<n;i++){
+        ans+=left_cnt[i]*right_cnt[i];
+    }
+    
+    cout<<ans<<"\n";
+    
+    return 0;
+}
+
+//Tc =O(nlogn) per test  SC = O(n)
+/*
+My submission : https://codeforces.com/contest/61/submission/356391525
+*/