Add Solution1.cpp for Maths day-8 p1

Implement solution for Codeforces problem 2098, calculating the range of medians after removing up to k bars.

Author: Abh-igyan

Problems/Mathematics/Day-08/sol/Abhigyan Tiwari/Solution1.cpp

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+// submission: https://codeforces.com/contest/2098/submission/356344573
+
+// we have houses numbered from 1 to 1e9 and n bars in houses a1,a2,.....,an (there can be multiple bars in single house)
+//and also an int k: at max k bars can be closed
+// For any house with number x, define f(x) as the sum of |x−y| over all open bars y (that is, after closing some bars).
+// sasha can buy apartment in a house with no. x if f(x) is minimal among all houses considering k constraint
+// Find the no. of different houses sasha can take apartment in
+
+
+#include <bits/stdc++.h>
+using namespace std;
+
+using ll = long long;
+#define all(x) (x).begin(), (x).end()
+
+const int MOD = 1e9 + 7; // 998244353
+const ll INF = 1e18;
+// It was simply finding the range of medians possible after removing the k elements. The range difference is the answer
+void solve() {
+    int n,k;
+    cin>>n>>k;
+    vector<ll> a(n);
+    for(auto &x: a) cin>>x;
+    //sort the array for finding median
+    sort(all(a));
+    // find the worst case median after removing k bars from right
+    int idx=(n-k-1)/2;
+    cout<<a[n-1-idx]-a[idx]+1<<endl; //range difference is the answer
+}
+//T.C: O(t*n*logn) ; S.C.: O(1)
+
+int main() {
+    // Fast I/O
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+    
+    return 0;
+}