Add solution for B.S. Day 07 q1

Implement a solution to find pairs of integers within a specified sum range.

Author: Abh-igyan

Problems/Binary-Search/Day-07/sol/Abhigyan/Solution1.cpp

@@ -0,0 +1,44 @@
+// https://codeforces.com/contest/1538/submission/356243914
+// Given an array of n integers,find all pairs (ai,aj): i<j where  l≤ai+aj≤r
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+#define all(x) (x).begin(), (x).end()
+
+const int MOD = 1e9 + 7; 
+const ll INF = 1e18;
+
+void solve() {
+    ll n,l,r;
+    cin>>n>>l>>r;
+    vector<ll> a(n);
+    for(auto &x: a) cin>>x;
+    //sort the array for Binary search
+    sort(all(a));
+    ll prs=0;
+    //first find all pairs A with sum <= r then those pairs B with sum less than l
+    for(int i=0;i<n;i++){
+        int p =upper_bound(all(a),r-a[i])-a.begin();
+        int q =upper_bound(all(a),l-1-a[i])-a.begin();
+        prs+=p-1; prs-=q-1;
+        if(a[i]+a[i]>=l && a[i]+a[i]<=r) prs--; //to remove self pair e.g. (ai,ai) absurd
+    }
+    //ans is (A-B)/2, divided by 2 to remove repeated pairs as order doesn't matter
+    cout<<prs/2<<endl;
+}
+// T.C.: O(n*logn); S.C.: O(1)
+int main() {
+    // Fast I/O
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}