Merge pull request #439 from Aaryan-Degama/main

d7q1

Author: zonedoutdg

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

@@ -0,0 +1,71 @@
+//link : https://codeforces.com/problemset/submission/1538/356089286
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio ios::sync_with_stdio(false); cin.tie(NULL);
+
+using ll = long long;
+using vi = vector<ll>;
+
+void solve() {
+    int n;
+    ll l, r;
+    cin >> n >> l >> r;
+
+    vi a(n);
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+    }
+
+    sort(a.begin(), a.end());
+
+    ll ans = 0;
+
+    for (int i = 0; i < n; i++) {
+        ll low = l - a[i];
+        ll high = r - a[i];
+
+        int left = lower_bound(a.begin() + i + 1, a.end(), low) - a.begin();
+        int right = upper_bound(a.begin() + i + 1, a.end(), high) - a.begin();
+
+        ans += max(0, right - left);
+    }
+
+    cout << ans << '\n';
+}
+
+int main() {
+    fastio
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+
+    return 0;
+}
+
+
+
+/*
+
+Explaination:
+1. Sort the array.
+2. For each index i, we want to count indices j > i such that:
+      l ≤ a[i] + a[j] ≤ r
+   This can be rewritten as:
+      l - a[i] ≤ a[j] ≤ r - a[i]
+3. Since the array is sorted, use binary search:
+   - lower_bound to find the first j where a[j] ≥ (l - a[i])
+   - upper_bound to find the first j where a[j] > (r - a[i])
+4. The number of valid pairs for index i is (right - left).
+5. Sum this for all i.
+
+Complexity:
+- Time complexity: O(n log n)
+- Space complexity: O(n)
+
+*/