Merge pull request #445 from MANISH-BEESA/BEESA-MANISH

added day7 q1 solution

Author: zonedoutdg

Problems/Binary-Search/Day-07/sol/BEESA-MANISH/Solution1.cpp

@@ -0,0 +1,56 @@
+
+/*
+
+Submission link: https://codeforces.com/contest/1538/submission/356096805
+
+TC - O(N \log N)
+SC - O(N)
+
+Approach :
+Sort the numbers first so they are in increasing order. This makes searching easier.
+Take one number and look for another number after it such that their sum lies between l and r.
+Find the range of numbers that can be added to the current number to stay within l and r.
+Count how many numbers fall in that range and add this count to the answer.
+
+*/
+
+#include <iostream>
+#include <vector>
+#include <algorithm>
+
+using namespace std;
+long long countPairsUnder(const vector<int>& a, int limit) {
+    long long count = 0;
+    int left = 0;
+    int right = a.size() - 1;
+    while (left < right) {
+        if (a[left] + a[right] <= limit) {
+            count += (right - left);
+            left++;
+        } else {
+            right--;
+        }
+    }
+    return count;
+}
+
+void solve() {
+    int n, l, r;
+    cin >> n >> l >> r;
+    vector<int> a(n);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+    sort(a.begin(), a.end());
+    long long result = countPairsUnder(a, r) - countPairsUnder(a, l - 1);
+    cout << result << "\n";
+}
+
+int main() {
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    return 0;
+}