Add solution for Codeforces problem 1538 C

Author: ViMo018

Problems/Binary-Search/Day-07/sol/vishva/soln1.cpp

@@ -0,0 +1,127 @@
+#include<bits/stdc++.h>
+#include<queue>
+using namespace std;
+typedef unordered_map<int, int> umii;
+typedef unordered_map<long long, long long> umll;
+typedef unordered_map<char, long long> umci;
+typedef vector<pair<int, int>> vpi;
+typedef vector<int> vi;
+typedef long long ll;
+typedef vector<long long> vll;
+typedef unordered_map<int , bool> umib;
+#define sum(v) accumulate(v.begin(), v.end(), 0)
+#define endl '\n'
+#define f0(i, n) for(long long i = 0; i < n; i++)
+#define f1(i, n) for(long long i = 1; i < n; i++)
+#define as(v) sort(v.begin(), v.end())
+#define all(x) (x).begin(), (x).end()
+#define pb push_back
+template<class T> umll frequency(vector<T> &v) {umll freq;for(auto &x:v) freq[x]++; return freq;}
+template<class T> umci S_frequency(vector<T> &v) {umci freq;for(auto &x:v) freq[x]++; return freq;}
+template <class T> void input(vector<T> &v){for(auto &x:v)cin>>x;}
+ll power(ll x, ll y){ ll res = 1; while (y > 0){ if (y & 1) res = (ll)(res*x); y = y>>1; x = (ll)(x*x); } return res; }
+void pvll(const vector<long long> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+void pvi(const vector<int> &arr){for(auto it : arr){cout << it << " ";}cout << endl;}
+
+
+// submission link :- 
+//https://codeforces.com/contest/1538/submission/352943894
+// C_Number_of_Pairs.cpp
+
+
+
+
+
+
+void solve(){
+
+//-------------INPUT-------------
+    ll n; 
+    cin >> n;
+    ll l,r;
+    cin >> l >> r;
+    vll v(n); input(v);
+    map<ll,ll>mp1;
+    map<ll,ll>mp2;
+    map<ll,ll>mp;
+    ll count=0;
+    as(v);
+    vll temp;
+    
+    for(ll i=0;i<n;i++)
+    {
+        ll req1 = l-v[i];
+        ll a,b;
+        ll req2 = r-v[i];
+        // apply lowerbound to find req1
+        auto it = lower_bound(temp.begin(),temp.end(),req1);
+
+        // using binary search to find the req1 index
+    
+
+        if( it!=temp.end() && *it>=req1)
+        {
+            a = mp[*it];
+        }
+        else
+        {
+            a =-1;
+        }
+        
+        //cout << a << endl;
+
+        // if got a particular a then try to find b
+
+        if(a!=-1 && !temp.empty())
+        {
+            
+            auto it = lower_bound(temp.begin(),temp.end(),req2);
+            //cout << req2 << " " << mp2[*it] << " " << *it << endl;
+            //cout << mp1[*it];
+            if(it == temp.end() && !temp.empty())
+            {
+                b = mp2[*(it-1)]; 
+                //cout << "j";
+            }
+            else if(*it>req2)
+            {
+                b = mp[*it]-1;
+                //cout << "l";
+            }
+            else{
+                b= mp2[*it];
+                //cout << "lf";
+            }
+            if(b>=a)
+            {
+                // this will be total pairs
+                count = count+b-a+1;
+            }
+        }
+        //cout << "b" << " " << b << endl;
+
+        temp.push_back(v[i]);
+        if(mp[v[i]]==0 && v[i]!=v[0])
+        {
+
+            mp[v[i]] = i;
+        }
+        mp2[v[i]] = i;
+    }
+  
+    cout << count << endl;
+
+
+
+//time complexity o(nlogn)
+//-------------CODE--------------
+    
+
+
+}
+
+
+int main(){
+    int tt; cin >> tt; while(tt--)
+{solve();};
+}