Add Day-08 Mathematics solution for Sasha and the Apartment Purchase

Author: Ayush Singh

Problems/Mathematics/Day-08/sol/ayush2005k/Solution1.py

@@ -0,0 +1,58 @@
+"""
+Problem: B. Sasha and the Apartment Purchase
+Link: https://codeforces.com/contest/2098/problem/B
+
+Short Problem Statement:
+There are n bars located on a street at given house numbers.
+At most k bars can close.
+For a house x, define f(x) as the sum of distances from x to all open bars.
+Sasha can buy a house x if it is possible to close at most k bars such that
+f(x) is minimal among all houses.
+The task is to count how many different houses Sasha can potentially buy.
+
+Approach:
+The sum of absolute distances is minimized at the median.
+After closing at most k bars, the number of remaining bars is m = n - k.
+For all subsets of size m, the possible medians form a continuous range.
+After sorting the bar positions:
+- The minimum possible median is at index (m - 1) // 2
+- The maximum possible median is at index (n - m) + (m // 2)
+Any integer house number between these two values (inclusive) is valid.
+The answer is the length of this interval.
+
+Time Complexity:
+O(n log n) per test case due to sorting.
+
+Space Complexity:
+O(n)
+
+Example:
+Input:
+4 0
+1 2 3 4
+
+Output:
+2
+
+Submission Link:
+https://codeforces.com/contest/2098/submission/356283081
+"""
+
+import sys
+input = sys.stdin.readline
+
+t = int(input())
+
+for _ in range(t):
+    n, k = map(int, input().split())
+    a = list(map(int, input().split()))
+
+    a.sort()
+
+    m = n - k
+
+    left_index = (m - 1) // 2
+    right_index = (n - m) + (m // 2)
+
+    result = a[right_index] - a[left_index] + 1
+    print(result)