Merge pull request #423 from Aaryan-Degama/main

d6q1

Author: Dheeraj-06

Problems/Binary-Search/Day-06/sol/Aaryan/Solution1.cpp

@@ -0,0 +1,86 @@
+// link : https://codeforces.com/contest/2014/submission/356025598
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+
+void solve() {
+    int n;
+    cin >> n;
+
+    vector<ll> a(n);
+    ll sum = 0;
+    for (int i = 0; i < n; i++) {
+        cin >> a[i];
+        sum += a[i];
+    }
+
+    if (n <= 2) {
+        cout << -1 << "\n";
+        return;
+    }
+
+    sort(a.begin(), a.end());
+
+    ll l = 0, r = 1e18, ans = -1;
+
+    while (l <= r) {
+        ll mid = (l + r) / 2;
+
+        long double avg = (long double)(sum + mid) / n;
+        long double half = avg / 2.0;
+
+        int cnt = 0;
+        for (ll v : a) {
+            if ((long double)v < half)
+                cnt++;
+        }
+
+        if (cnt > n / 2) {
+            ans = mid;
+            r = mid - 1;
+        } else {
+            l = mid + 1;
+        }
+    }
+
+    cout << ans << "\n";
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+
+    return 0;
+}
+
+
+/*
+Approach:
+We are allowed to add a non-negative value x to the total sum of the array.
+After adding x, let the new average be:
+    avg = (sum + x) / n
+We want more than half of the elements to be strictly less than:
+    avg / 2
+If n <= 2, this is impossible, so we directly output -1.
+
+Since adding more x only increases avg, the condition becomes monotonic.
+  Hence, we can binary search the minimum x.
+For a fixed x:
+    * Compute avg and avg/2
+    * Count how many elements are < avg/2
+    * If count > n/2, the condition is satisfied.
+
+Binary search over x in the range [0, 1e18] and keep the minimum valid x.
+
+Time Complexity: O(n log n)
+
+Space Complexity: O(n) 
+*/