Merge pull request #496 from Naman2251/main2

day8sol1

Author: Dheeraj-06

Problems/Mathematics/Day-08/sol/Naman2251/Solution1.cpp

@@ -0,0 +1,25 @@
+//#Submission Link: https://codeforces.com/contest/2098/submission/356258492
+
+/*
+If n-k is odd, the median is a single specific element in the middle of the subset. So, we calculate median by taking first(n-k) 
+houses and then last (n-k) houses. All houses betwen them an them are possible. so ans is 
+(median of last n-k)-(median of first n-k)+1.
+Similarly for even case, but in this case we have two medians so we take bigger one for last n-k and smaller one for first n-k.
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+    long long t;
+    cin>>t;
+    while(t--) {
+        long long n, k;
+        cin>>n>>k;
+        vector<long long> a(n+1);
+        for(int i=1; i<=n; i++) cin>>a[i];
+        sort(a.begin(), a.end());
+        if((n-k)%2!=0) cout<<a[(n+k+1)/2]-a[(n-k+1)/2]+1<<endl;
+        else cout<<a[(n+k)/2+1]-a[(n-k)/2]+1<<endl;
+    }
+}