code(java) : Solved que 1 of day-6

Author: dwivediprashant

Problems/Binary-Search/Day-06/sol/prashantdwivedi/Solution1.java

@@ -0,0 +1,90 @@
+// -------------------- Problem --------------------
+/*
+There are n people in a town and each person has some gold.
+
+The richest person suddenly finds extra gold (x coins) and adds it
+to their own wealth.
+
+A person is called unhappy if their wealth becomes strictly less than
+half of the average wealth of all people.
+
+If more than half of the people are unhappy, Robin Hood shows up.
+
+For every test case, we need to find the minimum value of x that
+makes this happen. If it can never happen, print -1.
+*/
+
+// -------------------- How this works --------------------
+/*
+Adding gold only increases the average, so the richest person will
+never be unhappy.
+
+The only way Robin Hood appears is when more than half of the people
+fall below half of the average wealth.
+
+So after sorting, we only care about the point where more than half
+of the population crosses this limit.
+
+That point is the person at index n/2 (0-based). If this person
+becomes unhappy, then strictly more than half are unhappy.
+
+From the inequality, we directly compute how much gold is needed.
+If the value comes out negative, zero is enough.
+
+For n = 1 or 2, it is impossible to make more than half unhappy.
+*/
+
+// -------------------- Complexity --------------------
+/*
+Sorting dominates the solution.
+Time: O(n log n)
+Space: O(n)
+*/
+
+// -------------------- Submission --------------------
+/*
+https://codeforces.com/contest/2014/submission/355974166
+*/
+
+// -------------------- Code -------------------------
+
+import java.io.*;
+import java.util.*;
+
+public class Solution1 {
+    public static void main(String[] args) throws Exception {
+        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+        StringBuilder sb = new StringBuilder();
+
+        int t = Integer.parseInt(br.readLine());
+
+        while (t-- > 0) {
+            int n = Integer.parseInt(br.readLine());
+            StringTokenizer st = new StringTokenizer(br.readLine());
+
+            long[] a = new long[n];
+            long sum = 0;
+
+            for (int i = 0; i < n; i++) {
+                a[i] = Long.parseLong(st.nextToken());
+                sum += a[i];
+            }
+
+            if (n <= 2) {
+                sb.append("-1\n");
+                continue;
+            }
+
+            Arrays.sort(a);
+
+            int idx = n / 2;
+            long x = 2L * n * a[idx] - sum + 1;
+
+            if (x < 0) x = 0;
+
+            sb.append(x).append('\n');
+        }
+
+        System.out.print(sb.toString());
+    }
+}