Fixed the Code.

Author: Krishna200608

Problems/Binary-Search/Day-06/sol/Krishna200608/Solution1.cpp

@@ -1,107 +1,75 @@
 /*
 Problem: Robin Hood in Town
 Link: https://codeforces.com/contest/2014/problem/C
-Author: Krishna Sikheriya (Krishna200608)
+Author: Krishna200608
 
-Short Problem Statement:
-We have a population of n people with given wealths. We want to give x gold to one person 
-(the richest) such that strictly more than half of the population becomes "unhappy".
-A person is unhappy if their wealth is strictly less than half of the average wealth.
-Find the minimum non-negative integer x to achieve this. If impossible, output -1.
+Short Problem Statement:=>
+Find minimum x to add to the richest person so that strictly more than half the population
+has wealth less than half the average.
 
 Approach:
-We need to find the minimum integer x such that the count of people with wealth < (new_average / 2) is > n / 2.
-1. If n <= 2, it is impossible to have strictly more than half the population satisfy the condition (since the max wealth will always be >= average). Output -1.
-2. Sort the wealth array to easily count or check conditions.
-3. The condition is monotonic: adding more gold increases the average, making it easier for existing wealths to fall below half the average.
-4. We can Binary Search for x in the range [0, 1e18].
-   - Calculation: New Sum = Current Sum + mid.
-   - Check: We need strictly more than n/2 people to be poor. 
-     In a sorted array, this means the person at index n/2 (0-indexed) must be poor.
-     Condition: a[n/2] < (Sum + mid) / (2 * n)
-     To avoid floating point errors, we use cross-multiplication:
-     2 * n * a[n/2] < Sum + mid
-5. The smallest valid 'mid' is our answer.
+    If n <= 2, it's impossible (-1).
+    Sort the array. The target person to become "poor" is at index n/2.
+    Binary Search for x in range [0, 1e18].
+    Condition: 2 * n * a[n/2] < current-sum + x.
 
-Time Complexity: O(n log n) due to sorting. Binary search takes O(log(1e18)) steps, each check is O(1).
-Space Complexity: O(n) to store the wealths.
-
-Example I/O:
-Input:
-2
-1
-1
-3
-2 3 3
-
-Output:
--1
-3
+Time : O(N log N)
+Space Complexity: O(1)
 */
 
-#include <iostream>
-#include <vector>
-#include <algorithm>
-#include <numeric>
+/*
+Submission Link:
+https://codeforces.com/contest/2014/submission/356175570
+*/
 
+#include <bits/stdc++.h>
 using namespace std;
-
-// Use long long to prevent overflow
 using ll = long long;
+#define in(a) for(int i = 0; i<a.size(); i++) cin>>a[i];
+typedef vector<int> vi;
+typedef vector<ll> vll;
 
 void solve() {
     int n;
-    if (!(cin >> n)) return;
+    cin >> n;
+    vll a(n);
 
-    vector<ll> a(n);
     ll sum = 0;
-    for (int i = 0; i < n; ++i) {
+    for(int i = 0; i < n; i++) {
         cin >> a[i];
         sum += a[i];
     }
 
-    // If n is 1 or 2, it's impossible to have >50% population strictly less than half average
-    if (n <= 2) {
-        cout << -1 << "\n";
+    if(n <= 2) {
+        cout << "-1\n";
         return;
     }
 
     sort(a.begin(), a.end());
 
-    // We binary search for x. Range [0, 2e18] is sufficient.
-    ll low = 0, high = 2e18;
+    ll l = 0, r = 1e18; 
     ll ans = -1;
 
-    while (low <= high) {
-        ll mid = low + (high - low) / 2;
-
-        // Check condition: a[n/2] < (sum + mid) / (2 * n)
-        // Equivalent to: 2 * n * a[n/2] < sum + mid
-        if (2LL * n * a[n/2] < sum + mid) {
+    while(l <= r) {
+        ll mid = l + (r - l) / 2;
+        if(2LL * n * a[n/2] < sum + mid) {
             ans = mid;
-            high = mid - 1; // Try smaller x
+            r = mid - 1;
         } else {
-            low = mid + 1;  // Need larger x
+            l = mid + 1;
         }
     }
-
     cout << ans << "\n";
 }
 
 int main() {
-    ios_base::sync_with_stdio(false);
+    ios::sync_with_stdio(false);
     cin.tie(NULL);
-
     int t;
-    if (cin >> t) {
-        while (t--) {
-            solve();
-        }
+    cin >> t;
+    while(t--) {
+        solve();
     }
     return 0;
 }
 
-/*
-SUBMISSION LINK:
-https://codeforces.com/contest/2014/submission/356082645
-*/