Merge pull request #428 from amansharma264/day-6-sol1

Implement solution for "Robin Hood in Town" problem

Author: zonedoutdg

Problems/Data-Structures/Day-06/sol/amansharma264/solution1.cpp

@@ -0,0 +1,102 @@
+/*
+Problem Name: Robin Hood in Town
+
+Short Problem Statement:
+There are n people in a town, each having some amount of gold.
+The richest person finds an extra pot of gold containing x gold and adds it
+to their wealth.
+
+A person is considered unhappy if their wealth is strictly less than
+half of the average wealth of all people.
+
+If strictly more than half of the population becomes unhappy,
+Robin Hood appears.
+
+The task is to determine the minimum value of x such that Robin Hood appears.
+If it is impossible, output -1.
+
+Approach:
+1. If n <= 2, it is impossible for more than half of the people to be unhappy,
+   so return -1.
+2. Sort the array of wealth values.
+3. Compute the initial total sum of wealth.
+4. Use Binary Search on x (extra gold added to the richest person):
+   - For a candidate x, compute the new total wealth.
+   - Count how many people have wealth strictly less than half of the average.
+   - If unhappy people > n/2, the condition is satisfied.
+5. Minimize x using binary search.
+
+Binary search works because increasing x increases the average wealth,
+which monotonically increases the number of unhappy people.
+
+Time Complexity:
+O(n log M) per test case  
+(where M is the search range for x, up to 1e18)
+
+Space Complexity:
+O(n)
+
+Question Link:
+https://codeforces.com/contest/2014/problem/C
+
+Submission Link:
+https://codeforces.com/submissions/amansharma264
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+#define fastio() ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr)
+#define ll long long
+#define endl '\n'
+
+int main() {
+    fastio();
+
+    ll t;
+    cin >> t;
+    while (t--) {
+        ll n;
+        cin >> n;
+
+        vector<ll> a(n);
+        for (ll i = 0; i < n; i++) {
+            cin >> a[i];
+        }
+
+        if (n <= 2) {
+            cout << -1 << endl;
+            continue;
+        }
+
+        sort(a.begin(), a.end());
+
+        ll sum = 0;
+        for (ll x : a) sum += x;
+
+        ll low = 0, high = 1e18, ans = -1;
+
+        while (low <= high) {
+            ll mid = low + (high - low) / 2;
+            ll totalSum = sum + mid;
+
+            ll unhappy = 0;
+            for (ll i = 0; i < n; i++) {
+                if (2LL * n * a[i] < totalSum) {
+                    unhappy++;
+                }
+            }
+
+            if (unhappy > n / 2) {
+                ans = mid;
+                high = mid - 1;
+            } else {
+                low = mid + 1;
+            }
+        }
+
+        cout << ans << endl;
+    }
+
+    return 0;
+}