Merge pull request #462 from Krishna200608/feature/issue-405-robin-hood

Solved Issue #405: Robin Hood in Town (Binary Search)

Author: Dheeraj-06

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

@@ -0,0 +1,75 @@
+/*
+Problem: Robin Hood in Town
+Link: https://codeforces.com/contest/2014/problem/C
+Author: Krishna200608
+
+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:
+    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 : O(N log N)
+Space Complexity: O(1)
+*/
+
+/*
+Submission Link:
+https://codeforces.com/contest/2014/submission/356175570
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+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;
+    cin >> n;
+    vll 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; 
+    ll ans = -1;
+
+    while(l <= r) {
+        ll mid = l + (r - l) / 2;
+        if(2LL * n * a[n/2] < sum + mid) {
+            ans = mid;
+            r = mid - 1;
+        } else {
+            l = mid + 1;
+        }
+    }
+    cout << ans << "\n";
+}
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(NULL);
+    int t;
+    cin >> t;
+    while(t--) {
+        solve();
+    }
+    return 0;
+}
+