Merge pull request #449 from ayush2005k/ayush2005k

Ayush2005k

Author: zonedoutdg

Problems/Binary-Search/Day-06/sol/ayush2005k/solution.py

@@ -0,0 +1,55 @@
+"""
+Problem: C. Robin Hood in Town
+Link: https://codeforces.com/contest/2014/problem/C
+
+Short Problem Statement:
+There are n people with given wealth.
+The richest person finds extra gold.
+A person is unhappy if their wealth is less than half of the average.
+Find the minimum gold required so that more than half the population becomes unhappy.
+
+Approach:
+- If n <= 2, it is impossible.
+- Sort the wealth array.
+- To make strictly more than n/2 people unhappy, it is sufficient to
+  make the person at index n//2 unhappy (0-based).
+- Derive the inequality directly and compute the minimum required gold.
+
+Time Complexity:
+O(n log n)
+
+Space Complexity:
+O(n)
+
+Example:
+Input:
+4
+1 2 3 4
+
+Output:
+15
+
+Submission Link:
+https://codeforces.com/contest/2014/submission/356103335
+"""
+
+import sys
+input = sys.stdin.readline
+
+t = int(input())
+
+for _ in range(t):
+    n = int(input())
+    a = list(map(int, input().split()))
+
+    if n <= 2:
+        print(-1)
+        continue
+
+    a.sort()
+    total = sum(a)
+
+    k = n // 2
+    x = 2 * n * a[k] - total + 1
+
+    print(max(0, x))